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Primes decide everything
A STORY
The heap and the trays
Kabir and Priya pack mangoes at the school stall. The heap holds $60$ mangoes. Every tray must
take the same number, with none left over.
They use trays of $12$. One tray stands full already. A second tray is only part filled. A third
tray is still empty.
Priya checks the full tray against her notes. “Twelve fits into sixty five times, exactly,” she
says.
“Could a tray of $15$ work too?” Kabir asks, lifting another mango from the heap. “Or would some
mangoes be left over?”
Priya works it out. “$60$ divided by $15$ is $4$,” she says. “That divides evenly too.”
“So two different tray sizes both divide the heap exactly,” Kabir says. “How many sizes can do
that?”
Priya has no answer yet. Neither do you. How many tray sizes split sixty exactly, with nothing
left over?
Kabir and Priya’s sixty mangoes are packed twelve to a tray in five trays, and fifteen to a tray in four trays, none left over.
Let us break $60$ into smaller pieces. Split it once, then break each piece again, until only primes are left.
We started two different ways. We reached the same four primes: $2$, $2$, $3$ and $5$. Try a third way yourself, starting with $60 = 6 \cdot 10$. You will reach the same four primes again.
This always happens. Every whole number bigger than $1$ breaks into primes in one way only. You can write the primes in a different order. You cannot get different primes. This fact has a name: the Fundamental Theorem of Arithmetic.
In this chapter we use this one fact for two jobs.
Job one: finding the HCF and LCM of two numbers from their primes.
Job two: proving that $\sqrt{2}$, $\sqrt{3}$ and $\sqrt{5}$ can never be written as a fraction.
Both jobs rest on the same fact. A number has one prime factorisation, and no other.
Both trees start at 60 and split it differently, and both finish on the same four primes, 2, 2, 3 and 5.
A SCHOLAR INDIA REMEMBERS
Kanada lived in north India more than two thousand years ago. He taught that all matter is built from tiny parts that cannot be split. Two parts join, or three parts join, and bigger things are made. Let us do the same with a number. We split $60$ until no piece splits again: $60 = 2 \times 2 \times 3 \times 5$. Those last pieces are primes.
Kanada · the sparrow
Check yourself
The Fundamental Theorem of Arithmetic states that every integer greater than $1$
factors into primes in exactly one way, including a fixed order
can be factored into primes in more than one way if the number is large enough
factors into primes in exactly one way, apart from the order of the factors
is itself always a prime number
Check your answer
factors into primes in exactly one way, including a fixed order — The theorem is silent on order — only the primes and their powers are fixed.
can be factored into primes in more than one way if the number is large enough — Uniqueness holds for every integer greater than 1, however large.
✓ factors into primes in exactly one way, apart from the order of the factors — (C) The theorem’s uniqueness holds regardless of the order the primes are written in.
is itself always a prime number — The theorem is about factorising into primes, not about the number itself being prime.
This chapter uses the Fundamental Theorem of Arithmetic to prove that $\sqrt{2}$ cannot be written as a fraction. Which property of the theorem does that proof actually need?
that every number has at least one prime factor
that prime numbers are infinite in number
that a number’s prime factorisation is fixed
that composite numbers can always be factored eventually
Check your answer
that every number has at least one prime factor — Existence of a prime factor is not enough; the proof needs the factorisation to be the ONLY one.
that prime numbers are infinite in number — How many primes exist has no role in the contradiction the proof builds.
✓ that a number’s prime factorisation is fixed — (C) The proof needs uniqueness — a fixed factorisation that a contradiction can be measured against.
that composite numbers can always be factored eventually — Being able to factor a number at all is not the same as that factorisation being the only one.
The same “fixed factorisation” idea used to prove $\sqrt{2}$ irrational could, in principle, also prove
$\sqrt{7}$ is irrational, using the same contradiction method with the prime $7$
$\sqrt{4}$ is irrational, since $4$ is also a whole number
$22/7$ is irrational, since it approximates $\pi$
$2 + 3$ is irrational, since both numbers appear inside square roots elsewhere in the chapter
Check your answer
✓ $\sqrt{7}$ is irrational, using the same contradiction method with the prime $7$ — (A) $7$ is prime, so the whole contradiction argument carries over unchanged with $7$ in place of $2$.
$\sqrt{4}$ is irrational, since $4$ is also a whole number — $\sqrt{4} = 2$, a whole number and therefore rational — the method proves nothing false here.
$22/7$ is irrational, since it approximates $\pi$ — $22/7$ is a ratio of integers, so it is rational by definition, however close it sits to $\pi$.
$2 + 3$ is irrational, since both numbers appear inside square roots elsewhere in the chapter — $2+3=5$ is an ordinary integer; where $2$ and $3$ appear elsewhere is irrelevant.
Four panels write the primes of 36, of 60 and 40, of 4 and 6, and of 90 as tokens to answer each check.
You already know what a prime number is. Now numbers break apart in a new way. Try each check below. It takes a minute.
A prime number’s factors (Class 8, Ganita Prakash Part 1, page 3). Here: every factorisation, like $60 = 2^2 \cdot 3 \cdot 5$, uses only primes. Check: $7$ has only two factors, $1$ and $7$.
A number’s factorisation (Class 8, Ganita Prakash Part 1, page 9). Here: every number breaks into primes in only one way, in any order. Check: $36 = 2^2 \cdot 3^2$ is a perfect square, since each power is even.
The HCF of two numbers (Class 8, Ganita Prakash Part 1, page 161). Here: the HCF keeps the smallest power of each shared prime. Check: $60$ and $40$ have HCF $20$.
The LCM of two numbers (Class 8, Ganita Prakash Part 1, page 121). Here: the same LCM comes from two numbers’ largest prime powers. Check: $4$ and $6$ have LCM $12$.
Lowest terms of a fraction (Class 9, Ganita Manjari, page 58). Here: the root 2 proof assumes the fraction is written this same way. Check: $3/20$ has no common factor other than $1$.
A number divisible by 10 (Class 8, Ganita Prakash Part 1, page 123). Here: a number ending in zero has both $2$ and $5$ as factors. Check: $90$ ends in $0$, and $90 = 2 \cdot 3^2 \cdot 5$.
If any of these felt new, read the page named before going on.
Seven tiles can be pushed into one rectangle and no other, so seven has only two factors. Twelve tiles make three, so twelve has six.
Take the number $7$. Which numbers divide $7$ exactly, with no remainder left over? Try $2$: $7 \div 2 = 3$ remainder $1$, so $2$ does not work. You try $3$, $4$, $5$ and $6$ the same way. None of them work either. Only $1$ and $7$ divide $7$ exactly. So $7$ has exactly two factors: $1$ and $7$. A number with exactly two factors is called PRIME.
Now take $12$. Which numbers divide $12$ exactly? $1$, $2$, $3$, $4$, $6$ and $12$ all do. That is six factors, more than two. A number with more than two factors is called COMPOSITE. Every composite number can be built by multiplying primes together. $12 = 2 \cdot 2 \cdot 3$. We also write this as $12 = 2^2 \cdot 3$. The small $2$ in $2^2$ means $2$ is used twice.
What about $1$? Only $1$ divides $1$ exactly. That is one factor, not two. So $1$ is not prime. It is not composite either, since it does not have more than two factors. $1$ is the one number that is neither.
When we write a number using only primes multiplied together, we call it a PRIME FACTORISATION. $60 = 2^2 \cdot 3 \cdot 5$ is a prime factorisation of $60$. But $60 = 4 \cdot 15$ is not a prime factorisation, because $4$ and $15$ are not primes. $4 = 2 \cdot 2$, and $15 = 3 \cdot 5$, so both break apart further.
Your turn: is $15$ prime or composite? Find all its factors first. (Answer: $1$, $3$, $5$ and $15$. Four factors, so $15$ is composite.)
A SCHOLAR INDIA REMEMBERS
What kind of number is $1$, really? Let us check. A prime has exactly two factors. $7$ has two factors, $1$ and $7$, so $7$ is prime. A composite number has more than two. $1$ has only one factor, itself. So $1$ is not prime, and it is not composite. It is a kind of its own.
Kanada · the sparrow
Check yourself
A composite number is a positive integer that
cannot be divided evenly by any smaller number
has more than two positive factors
is always an even number
has exactly two positive factors
Check your answer
cannot be divided evenly by any smaller number — That description fits a prime number — a composite number CAN be divided evenly by smaller numbers.
✓ has more than two positive factors — (B) A composite number’s defining property is having more than the two factors a prime has.
is always an even number — Composite numbers can be odd — $9 = 3 \cdot 3$ and $15 = 3 \cdot 5$ are both composite and odd.
has exactly two positive factors — Exactly two positive factors is what makes a number prime, not composite.
A student claims: “the number $1$ is the smallest prime number, since it only divides itself.” What is wrong with this claim?
$1$ is actually composite, not prime
$1$ has only one factor
the claim is correct — $1$ is indeed the smallest prime
primes must be even, so $1$ cannot be prime
Check your answer
$1$ is actually composite, not prime — $1$ fails the composite definition too — it does not have more than two factors.
✓ $1$ has only one factor — (B) A prime needs exactly two factors; $1$ has only one, so it fails the definition on both sides.
the claim is correct — $1$ is indeed the smallest prime — $1$ has only one positive factor, so the definition of a prime number does not fit it.
primes must be even, so $1$ cannot be prime — Primes are not required to be even — $3$, $5$, $7$ are all odd primes; the real issue is $1$’s factor count.
Every composite number can always be built by
adding two smaller composite numbers together
dividing a larger prime by a smaller one
multiplying two or more primes together
squaring any prime number
Check your answer
adding two smaller composite numbers together — Prime factorisation is built by multiplying primes, not by adding composite numbers.
dividing a larger prime by a smaller one — Dividing one prime by another does not produce a composite integer in general.
✓ multiplying two or more primes together — (C) A composite number breaks down into smaller primes multiplied together, by definition of prime factorisation.
squaring any prime number — Squaring a prime gives only one shape of composite number, such as $9$ or $25$ — most composites need several different primes.
In the factorisation $60 = 2^2 \cdot 3 \cdot 5$, the pieces $2$, $3$ and $5$ are all prime because
the numbers involved happen to be small
any factorisation of $60$ must use exactly three factors
a prime factorisation may mix in composite factors as long as the product is correct
a prime factorisation is only allowed to use prime numbers as its factors
Check your answer
the numbers involved happen to be small — Size is irrelevant — a prime factorisation uses only primes whether the number is small or enormous.
any factorisation of $60$ must use exactly three factors — There is no rule fixing how many prime factors a number must have — $60$ happens to need three distinct primes.
a prime factorisation may mix in composite factors as long as the product is correct — Mixing in a composite factor would no longer be a PRIME factorisation, even if the product still came out right.
✓ a prime factorisation is only allowed to use prime numbers as its factors — (D) By definition, a prime factorisation’s pieces are always prime — that is what makes it a PRIME factorisation.
The same four shaded tiles read as four of six, then divide by 2 to read as two of three.
Take $3/4$. It is a fraction of two integers, so it is RATIONAL. Now take $5$. Write it as $5/1$ yourself, and check that it fits the same pattern. Any number we can write as $p/q$, for integers $p$ and $q$ with $q \neq 0$, is rational.
Now take $4/6$. Let us simplify it. Both $4$ and $6$ share the factor $2$. We divide it out: $4/6 = 2/3$. $2/3$ has no factor left to divide out, so it is in LOWEST TERMS. Its numerator and denominator are then COPRIME. *A fraction is in lowest terms when its numerator and denominator share no factor except $1$. $2/3$ is in lowest terms; $4/6$ is not.*
A number that cannot be written as $p/q$ at all, in any form, is IRRATIONAL.
Two more facts about rationals. Add, subtract or multiply two rationals and the answer is rational again. Divide them and it is rational again too, with one catch: you may never divide by $0$. So take a rational $r$ that is not $0$. Then $1/r$ is rational — turn $p/q$ upside down and you get $q/p$, a fraction of two integers again. And $s/r$ is rational for any rational $s$, since it is $s$ times $1/r$.
When we meet a rational number in a proof, we assume it is already written in lowest terms. This is not a small detail to skip past. It is what makes a contradiction proof work. Without the assumption, there is nothing left for the contradiction to catch.
Your turn: is $6/9$ in lowest terms? (Answer: No. $3$ divides both $6$ and $9$. In lowest terms it is $2/3$.)
Both names sit at the same point, so 4/6 and 2/3 are the same number.
Check yourself
A rational number, by definition, can be written as
$p/q$ for integers $p$ and $q$ with $q \neq 0$
$p/q$ for integers $p$ and $q$ with $p \neq 0$
$\sqrt{p}$ for any integer $p$
a terminating decimal only
Check your answer
✓ $p/q$ for integers $p$ and $q$ with $q \neq 0$ — (A) The definition requires an integer numerator and a nonzero integer denominator.
$p/q$ for integers $p$ and $q$ with $p \neq 0$ — It is the denominator, $q$, that cannot be zero — the numerator $p$ may be zero.
$\sqrt{p}$ for any integer $p$ — A square root of an integer is not the form a rational number takes — most such square roots are irrational.
a terminating decimal only — Repeating decimals, such as $0.333\ldots$, are rational too — the definition is not limited to terminating decimals.
A fraction $p/q$ is in lowest terms when
$p$ and $q$ are both prime numbers
$q$ is as small as possible, regardless of $p$
$p$ and $q$ share no factor but $1$
$p$ is greater than $q$
Check your answer
$p$ and $q$ are both prime numbers — $p$ and $q$ need not be prime themselves — $4/9$ is in lowest terms though neither $4$ nor $9$ is prime.
$q$ is as small as possible, regardless of $p$ — Lowest terms is about shared factors, not about $q$ being the smallest possible denominator.
✓ $p$ and $q$ share no factor but $1$ — (C) Lowest terms means the numerator and denominator are coprime, sharing only the trivial factor $1$.
$p$ is greater than $q$ — Whether $p$ is greater than $q$ has nothing to do with whether they share a common factor.
The irrationality proofs in this chapter always assume $a/b$ is written in lowest terms. Why does that assumption matter?
it makes the numbers smaller and therefore easier to square
without it, $a$ and $b$ could not be integers at all
it guarantees $a$ and $b$ are both prime numbers
it lets the proof reach a contradiction
Check your answer
it makes the numbers smaller and therefore easier to square — Lowest terms is not chosen for ease of arithmetic — it is what makes the final contradiction actually a contradiction.
without it, $a$ and $b$ could not be integers at all — Any rational number can be written as SOME fraction of integers, lowest terms or not — the assumption is a choice, not a necessity for being an integer.
it guarantees $a$ and $b$ are both prime numbers — Coprime means $a$ and $b$ share no common factor — it says nothing about either one being prime.
✓ it lets the proof reach a contradiction — (D) The proof’s whole payoff is showing coprime $a$ and $b$ stop being coprime, which is only a contradiction because lowest terms was assumed first.
Every composite number factors into primes, uniquely
▲CONCEPT
Sixty’s four primes are 2, 2, 3 and 5, and the 24 arrangements are four choices, then three, then two, then one.
Let us state the Fundamental Theorem of Arithmetic plainly. Every composite number breaks into primes in exactly one way. Write the primes in a different order, and it is still the same set of primes. No second, different set exists.
Try it on $1400$. Factor it one way: $1400 = 2^3 \cdot 5^2 \cdot 7$. Now factor it a different way. Pull out a $5$ first instead of a $2$. You still land at $2^3 \cdot 5^2 \cdot 7$. No other set of primes multiplies out to $1400$.
This says more than just that primes exist for every number. It says the primes are fixed. Break $1400$ apart any way you like. The primes that appear, and how many times each one appears, come out the same every time.
Your turn: could $84$ have two different prime factorisations? (Answer: No. $84 = 2^2 \cdot 3 \cdot 7$ is its only prime factorisation.)
A SCHOLAR INDIA REMEMBERS
Will the smallest parts come out the same every time? Test a new number. Split $84$ as $4 \times 21$. That gives $2 \times 2 \times 3 \times 7$. Now split $84$ as $2 \times 42$. That gives $2 \times 2 \times 3 \times 7$ again. Two different starts, and the same four primes.
Kanada · the sparrow
Check yourself
The Fundamental Theorem of Arithmetic says every composite number factors into primes
in exactly one way, apart from order
in exactly one way, including a fixed order
in more than one way for some large numbers
only if the number is even
Check your answer
✓ in exactly one way, apart from order — (A) The uniqueness is exact, once order is set aside.
in exactly one way, including a fixed order — Order is explicitly set aside by the theorem — only the primes and their powers are fixed.
in more than one way for some large numbers — Uniqueness holds for every composite number, large or small, with no exceptions.
only if the number is even — The theorem applies to every composite number, even and odd alike.
$1400 = 2^3 \cdot 5^2 \cdot 7$. According to the Fundamental Theorem of Arithmetic, what can be said about any OTHER prime factorisation of $1400$?
no other set of primes multiplies to $1400$
a second factorisation exists using different primes, but it is rarely found
a second factorisation exists only if $1400$ is written in a different number base
the theorem does not apply to numbers with three distinct prime factors
Check your answer
✓ no other set of primes multiplies to $1400$ — (A) Uniqueness rules out any second, different set of primes multiplying to the same number.
a second factorisation exists using different primes, but it is rarely found — No second factorisation exists to be found — uniqueness is exact, not merely rare to violate.
a second factorisation exists only if $1400$ is written in a different number base — A number’s prime factorisation does not change with the base it is written in — $1400$ has the same primes regardless.
the theorem does not apply to numbers with three distinct prime factors — The theorem places no limit on how many distinct primes a number’s factorisation may contain.
A student factors $1400$ twice — once starting by dividing by $2$, once by dividing by $7$ — and gets two different lists of primes. What must be true?
an arithmetic error was made somewhere
both factorisations are correct, since the starting prime was different
$1400$ is an exception where two factorisations exist
one of the factorisations used a composite number by mistake, which is a valid shortcut
Check your answer
✓ an arithmetic error was made somewhere — (A) Uniqueness guarantees a single correct answer, so two different results mean one of the two computations went wrong.
both factorisations are correct, since the starting prime was different — The order factors are pulled out in never changes the final answer — it must come out the same either way.
$1400$ is an exception where two factorisations exist — The theorem has no exceptions — every composite number, $1400$ included, has exactly one prime factorisation.
one of the factorisations used a composite number by mistake, which is a valid shortcut — Using a composite number inside a “prime” factorisation is exactly the kind of error that produces a wrong, mismatched list.
If a number’s prime factorisation could change depending on the order factors were pulled out, which chapter method would immediately fail?
the definition of a composite number, which does not depend on factorisation order
the HCF/LCM method
the test for whether $9^n$ ends in $0$, since that argument only tracks one prime
nothing would fail — every other chapter result is independent of unique factorisation
Check your answer
the definition of a composite number, which does not depend on factorisation order — A composite number is defined by its factor COUNT, not by any particular factorisation, so this definition would be unaffected.
✓ the HCF/LCM method — (B) HCF/LCM by prime factorisation only works because each number has exactly one factorisation to read powers off.
the test for whether $9^n$ ends in $0$, since that argument only tracks one prime — That test also depends on $9^n$ having a fixed, unique set of prime factors — it would not survive either, though HCF/LCM fails first and most directly.
nothing would fail — every other chapter result is independent of unique factorisation — HCF/LCM by prime factorisation is built entirely on reading powers off ONE fixed factorisation — it would fail immediately without uniqueness.
$378 / 2 = 189$ Start with the smallest prime that goes in. $378$ is even, so that prime is $2$.
$189 / 3 = 63$, $63 / 3 = 21$, $21 / 3 = 7$ $189$ is odd, so no second $2$ comes out. Move up to the next prime, $3$, and keep dividing while it still goes in. Here it goes in three times.
$7 / 7 = 1$ $7$ is prime, so it divides itself. The quotient is $1$ and the chain stops.
$378 = 2 \cdot 3 \cdot 3 \cdot 3 \cdot 7 = 2 \cdot 3^3 \cdot 7$ Collect the divisors you used, in the order you used them. The three $3$s are written $3^3$.
$3^3 = 27$, $2 \cdot 27 \cdot 7 = 378$ Multiply the primes back yourself. You started at $378$, so you must land on $378$.
Find the prime factorisation of any number, by repeated division.
Divide by the smallest prime For $378$, that prime is $2$, because $378$ is even. $378 / 2 = 189$.
Keep going with that prime $189$ is odd, so $2$ is finished. The next prime to try is $3$. $189 / 3 = 63$, $63 / 3 = 21$, $21 / 3 = 7$. Now $3$ is finished too.
Stop at a quotient of one Now $7$ is prime, so it divides itself. $7 / 7 = 1$, and there is nothing left to divide.
Collect the divisors, then check The divisors used were $2, 3, 3, 3, 7$, so $378 = 2 \cdot 3^3 \cdot 7$. Multiply them back to check. $3^3 = 27$ and $2 \cdot 27 \cdot 7 = 378$. It matches, so the factorisation is right.
The same reading works for any two repeating times a problem gives you, not only 24 and 36 minutes.
The HCF sits at or below every number, and the LCM sits at or above every number, for both groups shown.
Take $8$ and $12$. First write each one as primes.
$8 = 2 \cdot 2 \cdot 2$
$12 = 2 \cdot 2 \cdot 3$
Now let us find the HCF. The HCF is the biggest number that divides both $8$ and $12$. Look at the prime $2$. It appears three times in $8$ and two times in $12$. Take the smaller count, two. The prime $3$ is only in $12$, not in $8$, so it is not shared. Leave it out. So $\text{HCF} = 2 \cdot 2 = 4$. Check: $4$ divides $8$. $4$ divides $12$. Yes.
Now the LCM. The LCM is the smallest number that both $8$ and $12$ divide into. This time take every prime that appears in either list, and take the bigger count. The prime $2$: three times in $8$, two times in $12$. Take three. The prime $3$: once, in $12$. Take one. So $\text{LCM} = 2 \cdot 2 \cdot 2 \cdot 3 = 24$. Check: $24 \div 8 = 3$, and $24 \div 12 = 2$. Both exact.
What if the two numbers share no prime at all? Take $8 = 2 \cdot 2 \cdot 2$ and $15 = 3 \cdot 5$. Go looking for a prime on both lists and there is none. So there is nothing to multiply together, and the HCF is $1$: $1$ divides every number, and here nothing bigger divides both. Two numbers whose HCF is $1$ are COPRIME. The LCM is read off exactly as before, from every prime in either list: $\text{LCM}(8, 15) = 2 \cdot 2 \cdot 2 \cdot 3 \cdot 5 = 120$.
You can use the same method for three numbers or more. Nothing changes. It also answers a question like this: one bell rings every $8$ minutes, and another every $12$ minutes. When do they ring together again? The answer is the LCM, $24$ minutes.
*Rule: the HCF uses the smaller count of each prime the numbers share. The LCM uses the bigger count of every prime that appears. For $8$ and $12$, $\text{HCF} = 4$ and $\text{LCM} = 24$.*
Your turn: a prime appears in only one of two numbers. Does it appear in their HCF? (Answer: No. The HCF only uses primes that both numbers share.)
A SCHOLAR INDIA REMEMBERS
Before you look for the HCF, make a clean list of parts. $12 = 2 \times 2 \times 3$. $18 = 2 \times 3 \times 3$. Which primes are on both lists? One $2$ and one $3$. So the HCF is $2 \times 3 = 6$. Now tell me: which primes will the LCM need?
Kanada · the sparrow
Check yourself
To find the HCF of two numbers from their prime factorisations, take
the largest power of every prime the two numbers share
the smallest power of every prime the two numbers share
the smallest power of every prime appearing in either number
the product of all shared primes, ignoring their powers
Check your answer
the largest power of every prime the two numbers share — Taking the LARGEST shared power is the LCM rule, not the HCF rule.
✓ the smallest power of every prime the two numbers share — (B) HCF takes the smallest shared power of each common prime.
the smallest power of every prime appearing in either number — HCF only looks at primes BOTH numbers share — a prime appearing in only one number plays no part.
the product of all shared primes, ignoring their powers — Powers matter — HCF needs the smallest matching power of each shared prime, not just the primes themselves.
To find the LCM of two numbers from their prime factorisations, take
the largest power of every prime appearing in either number
the largest power of every prime the two numbers share
the smallest power of every prime appearing in either number
the sum of all powers of every shared prime
Check your answer
✓ the largest power of every prime appearing in either number — (A) LCM takes the largest power of every prime that shows up anywhere in either number.
the largest power of every prime the two numbers share — LCM includes every prime appearing in EITHER number, not only the shared ones.
the smallest power of every prime appearing in either number — Taking the SMALLEST power is the HCF rule, not the LCM rule.
the sum of all powers of every shared prime — LCM takes the LARGEST power present, not a sum of powers.
The prime-factorisation method for HCF and LCM extends to three or more numbers by
finding the HCF and LCM of the first two numbers, then ignoring the rest
the smallest/largest power of each prime, across every number
averaging the powers of each prime across all the numbers
only working when all the numbers share exactly the same primes
Check your answer
finding the HCF and LCM of the first two numbers, then ignoring the rest — Every number in the group must be included — none can be left out of the smallest/largest comparison.
✓ the smallest/largest power of each prime, across every number — (B) The same smallest/largest rule simply runs across every number in the group, not a pair at a time.
averaging the powers of each prime across all the numbers — The method takes the SMALLEST or LARGEST power present, never an average of the powers.
only working when all the numbers share exactly the same primes — The method works even when the numbers do not all share the same primes — a prime missing from one number simply contributes power $0$ there.
Two bells ring every $12$ minutes and every $18$ minutes. Finding when they next ring together needs
the HCF of $12$ and $18$, since that is the smaller shared unit
the LCM of $12$ and $18$, read off their prime factorisations
the product $12 \cdot 18$, since both events must occur
the average of $12$ and $18$, since the bells ring at different rates
Check your answer
the HCF of $12$ and $18$, since that is the smaller shared unit — HCF finds the largest number dividing both — it does not answer when two repeating events coincide; that is the LCM’s job.
✓ the LCM of $12$ and $18$, read off their prime factorisations — (B) A “next time both repeat” problem is answered by the LCM of the repeat times.
the product $12 \cdot 18$, since both events must occur — The plain product overcounts — the LCM already accounts for the factors $12$ and $18$ share.
the average of $12$ and $18$, since the bells ring at different rates — Averaging two rates has no bearing on when their cycles line up — only the LCM answers that.
The prime tokens of 8 and 12 split into the shared tokens for the HCF, 4, and the rest for the LCM, 24.
Take $8$ and $12$ again. Their HCF is $4$. Their LCM is $24$. Let us multiply the HCF by the LCM, then multiply the two original numbers.
$4 \cdot 24 = 96$
$8 \cdot 12 = 96$
The two answers match. This is not a coincidence for this one pair.
For any two positive integers $a$ and $b$: $\text{HCF}(a, b) \cdot \text{LCM}(a, b) = a \cdot b$. Once you know any three of HCF, LCM, $a$ and $b$, this relation finds the fourth. You only need to divide. No fresh factorising needed.
Here it is, run backwards. The HCF of $24$ and $36$ is $12$. What is their LCM? Put the three numbers you know into the relation, then divide.
Check it the long way if you like: $24 = 2^3 \cdot 3$ and $36 = 2^2 \cdot 3^2$, so the LCM is $2^3 \cdot 3^2 = 72$. The same answer, and the short way never factorised anything.
The relation works for a PAIR of numbers only. For three integers, the product of all three is usually NOT equal to the product of their HCF and LCM. Check a third number by prime factorisation directly. Do not carry this two-number shortcut over.
Your turn: for $4$ and $6$, the HCF is $2$ and the LCM is $12$. Does $\text{HCF} \cdot \text{LCM}$ equal $4 \cdot 6$? (Answer: Yes. $2 \cdot 12 = 24$, and $4 \cdot 6 = 24$.)
Check yourself
For any two positive integers $a$ and $b$, $\text{HCF}(a, b) \cdot \text{LCM}(a, b)$ equals
$a^2 \cdot b^2$
$a \cdot b$
$a + b$
$a - b$
Check your answer
$a^2 \cdot b^2$ — The relation is a plain product, $a \cdot b$ — squaring each number first is not part of it.
✓ $a \cdot b$ — (B) The product of the HCF and LCM of two numbers always equals the product of the numbers themselves.
$a + b$ — The relation is a product relation — it equals $a \cdot b$, not $a+b$.
$a - b$ — The relation equals $a \cdot b$, not a difference of the two numbers.
If $\text{HCF}(a, b) = 4$ and $a \cdot b = 96$, the relation $\text{HCF} \cdot \text{LCM} = a \cdot b$ gives
$\text{LCM}(a, b) = 384$, found by multiplying $96$ by $4$
$\text{LCM}(a, b) = 92$, found by subtracting $4$ from $96$
$\text{LCM}(a, b) = 24$, found by dividing $96$ by $4$
$\text{LCM}(a, b) = 4$, the same as the HCF
Check your answer
$\text{LCM}(a, b) = 384$, found by multiplying $96$ by $4$ — The relation is solved by DIVIDING $a \cdot b$ by the HCF, not multiplying — the LCM here is $24$, not $384$.
$\text{LCM}(a, b) = 92$, found by subtracting $4$ from $96$ — The relation is a product relation, solved by division — subtracting the HCF from the product is not a valid step.
✓ $\text{LCM}(a, b) = 24$, found by dividing $96$ by $4$ — (C) The relation is solved for LCM by dividing the known product by the known HCF.
$\text{LCM}(a, b) = 4$, the same as the HCF — HCF and LCM are equal only when $a=b$; here dividing $96$ by $4$ gives $24$, not $4$.
A student computes the HCF and LCM of three numbers $a, b, c$ and claims $\text{HCF}(a,b,c) \cdot \text{LCM}(a,b,c) = a \cdot b \cdot c$, extending the two-number relation. What is wrong?
nothing is wrong — the relation extends to any number of integers
the relation only fails when one of $a, b, c$ is prime
the relation is specific to two numbers
the student should have used addition instead of multiplication
Check your answer
nothing is wrong — the relation extends to any number of integers — The chapter states directly that this relation does NOT extend beyond two numbers.
the relation only fails when one of $a, b, c$ is prime — The relation fails for three numbers in general, not only in the special case where one happens to be prime.
✓ the relation is specific to two numbers — (C) The product relation was proved only for a pair, and it does not carry over to three or more numbers.
the student should have used addition instead of multiplication — Changing the operation does not fix the underlying issue — the two-number relation simply has no three-number counterpart.
Knowing $\text{HCF}(a, b) \cdot \text{LCM}(a, b) = a \cdot b$ is most useful when
neither HCF nor LCM has been computed yet, and both are needed
one of HCF, LCM is known and the other is needed fast
the numbers $a$ and $b$ are themselves prime
three or more numbers are involved instead of two
Check your answer
neither HCF nor LCM has been computed yet, and both are needed — The relation needs at least one of HCF or LCM already known — it cannot produce both from nothing.
✓ one of HCF, LCM is known and the other is needed fast — (B) The relation’s value is as a shortcut from one known quantity to the other.
the numbers $a$ and $b$ are themselves prime — Whether $a$ and $b$ are prime does not change when this shortcut is worth using.
three or more numbers are involved instead of two — This relation is specific to exactly two numbers — it offers no shortcut once a third number joins.
$12 = 2^2 \cdot 3$, $18 = 2 \cdot 3^2$ Factor each number into primes first. Then you can compare them, prime by prime.
$\text{HCF} = 2^1 \cdot 3^1 = 6$ Take the smaller power of every prime the two numbers share. Here that means $2^1$ against $2^2$, and $3^1$ against $3^2$.
$\text{LCM} = 2^2 \cdot 3^2 = 36$ Take the larger power of every prime that appears in either number.
$6 \cdot 36 = 216$, $12 \cdot 18 = 216$ Check the product relation yourself. Multiply your HCF by your LCM. Then multiply the two original numbers. Both give $216$.
Find the HCF and LCM of any two numbers, from their primes.
Factor both numbers Write $12$ and $18$ as primes. $12 = 2^2 \cdot 3$. $18 = 2 \cdot 3^2$.
List every prime The primes that appear are $2$ and $3$.
Take the smaller power For $2$, the powers are $2^2$ and $2^1$. Take the smaller, $2^1$. For $3$, the powers are $3^1$ and $3^2$. Take the smaller, $3^1$. $\text{HCF} = 2^1 \cdot 3^1 = 6$.
Take the bigger power For $2$, take the bigger power, $2^2$. For $3$, take the bigger power, $3^2$. $\text{LCM} = 2^2 \cdot 3^2 = 36$.
Check by multiplying $\text{HCF} \cdot \text{LCM} = 6 \cdot 36 = 216$. $12 \cdot 18 = 216$ too. They match, so the answer checks out.
the shorter plank
the longer plank
Dhruv checks two wooden planks laid side by side, one clearly longer than the other.
Check yourself
Factoring $12 = 2^2 \cdot 3$ and $18 = 2 \cdot 3^2$, the HCF is found by
taking $2^2$ and $3^2$, the larger powers, giving $36$
taking $2^1$ and $3^2$, mixing the two numbers’ larger powers, giving $18$
taking $2^1$ and $3^1$, the smaller powers, giving $6$
adding the exponents of $2$ and $3$ across both numbers, giving $2^3 \cdot 3^3$
Check your answer
taking $2^2$ and $3^2$, the larger powers, giving $36$ — Taking the larger powers gives the LCM, $36$ — the HCF needs the smaller matching power of each prime.
taking $2^1$ and $3^2$, mixing the two numbers’ larger powers, giving $18$ — Each prime’s power must be compared between the two numbers, not mixed and matched — the correct pairing gives $6$, not $18$.
✓ taking $2^1$ and $3^1$, the smaller powers, giving $6$ — (C) HCF pulls the smaller matching power of each shared prime: $2^1 \cdot 3^1 = 6$.
adding the exponents of $2$ and $3$ across both numbers, giving $2^3 \cdot 3^3$ — Exponents are compared to find the smaller (or larger) one, never added together.
After computing $\text{HCF}(12,18)=6$ and $\text{LCM}(12,18)=36$, a check confirms $6 \cdot 36 = 12 \cdot 18$. If the check had instead given $6 \cdot 36 \neq 12 \cdot 18$, what would that mean?
the product relation is only approximately true, so small mismatches are normal
an arithmetic error was made in the factorisation or the HCF/LCM
$12$ and $18$ would be an exception to the product relation
the HCF and LCM would need to be added instead of multiplied to fix it
Check your answer
the product relation is only approximately true, so small mismatches are normal — The relation is exact for any two positive integers — a mismatch always points to an error, never to normal rounding.
✓ an arithmetic error was made in the factorisation or the HCF/LCM — (B) The product relation always holds for two numbers, so a mismatch signals a computational mistake, not an exception.
$12$ and $18$ would be an exception to the product relation — There are no exceptions to the two-number relation — $12$ and $18$ are ordinary integers, not a special case.
the HCF and LCM would need to be added instead of multiplied to fix it — Changing the check to addition does not address the real issue — a mismatch means the HCF or LCM was computed incorrectly.
To find the HCF and LCM of $20$ and $30$ using this worked example’s method, the first step is
guess likely HCF and LCM values, then check by dividing
list every factor of $20$ and $30$ and compare the two lists
multiply $20$ and $30$ together first
write each number’s prime factorisation first
Check your answer
guess likely HCF and LCM values, then check by dividing — The method starts from the factorisation, not from a guess checked afterwards.
list every factor of $20$ and $30$ and compare the two lists — Listing every factor is a different approach — this worked example’s method starts from prime factorisation instead.
multiply $20$ and $30$ together first — The product of $20$ and $30$ plays no role until the HCF or LCM is already known, if at all.
✓ write each number’s prime factorisation first — (D) The method always starts from each number’s own prime factorisation before comparing powers.
HCF of $9$, $15$ and $21$ is $3$, so I wrote $\text{LCM} = (9 \cdot 15 \cdot 21) / 3 = 2835 / 3 = 945$.
Second look
That rule only holds for a pair. Back to primes, $9 = 3^2$, $15 = 3 \cdot 5$, $21 = 3 \cdot 7$, so $\text{LCM} = 3^2 \cdot 5 \cdot 7 = 315$.
HCF times LCM equals the product only for a pair. For three numbers or more, build the LCM from the primes.
Does HCF times LCM equal the product, for three numbers too?
Weaker. For $9$, $15$ and $21$: the HCF is $3$. Carrying the two-number rule across, $\text{LCM} = (9 \cdot 15 \cdot 21) / 3 = 2835 / 3 = 945$.
That feels right — it reused a trusted shortcut. But check it: $945 / 3 = 315$ is still a common multiple of $9$, $15$ and $21$. If $945$ divides by $3$ and still works, $945$ was never the LEAST common multiple.
Stronger. The relation $\text{HCF} \cdot \text{LCM} = a \cdot b$ is proved for a PAIR only. For three numbers, go back to the primes: $9 = 3^2$, $15 = 3 \cdot 5$, $21 = 3 \cdot 7$. Take the largest power of each prime: $\text{LCM} = 3^2 \cdot 5 \cdot 7 = 315$.
Three numbers can share a prime in pairs without all three sharing it — $9$ and $21$ share an extra factor of $3$ that $15$ does not — and the product rule has no way to count that. $315$, not $945$, is the true LCM.
assume $9^n$ ends in the digit $0$ Start from the ending we want to rule out. See what it would force to be true.
a number ending in $0$ is divisible by $10 = 2 \cdot 5$ Any number ending in $0$ is a multiple of ten. Ten factors as $2$ times $5$.
so $9^n$’s factorisation would need both $2$ and $5$ in it If $10$ divides $9^n$, then $9^n$’s own factorisation must contain both of $10$’s prime factors.
$9 = 3^2$, so $9^n = 3^{2 n}$ $9$ is built from the single prime $3$. So every power of $9$ is too. No other prime can enter.
contradiction: the only prime in $9^n$ is $3$ The Fundamental Theorem of Arithmetic fixes which primes make up a number. $9^n$ can never also contain $2$ and $5$.
$9^n$ never ends in $0$, for any $n$ The assumed ending forced something impossible. So the assumption fails.
$9^1 = 9$, $9^2 = 81$, $9^3 = 729$, $9^4 = 6561$ Check the pattern yourself. None of these powers end in $0$. That matches what you just proved for every $n$.
The units digit alternates between 9 and 1 for every power of 9 shown, and 0 never appears.
Decide whether a number can end in a chosen digit, without checking every power.
Name the digit We want to know if $9^n$ can end in $0$.
Find the needed primes Ending in $0$ needs a factor of $10 = 2 \cdot 5$. So the number needs both $2$ and $5$ in its primes.
Factor the number $9 = 3^2$, so $9^n = 3^{2 n}$. Its only prime factor is $3$.
Compare and conclude $3$ is not $2$ and not $5$. So $9^n$ can never end in $0$, for any $n$.
Check yourself
$9^n = 3^{2n}$ for every natural number $n$. This shows $9^n$ can never end in $0$ because
ending in $0$ needs both $2$ and $5$; $9^n$ only has $3$
$9^n$ is always odd, and no odd number can end in $0$
$9$ itself does not end in $0$, so no power of it can either
$n$ would need to be negative for $9^n$ to end in $0$
Check your answer
✓ ending in $0$ needs both $2$ and $5$; $9^n$ only has $3$ — (A) Ending in $0$ means divisible by $10=2 \cdot 5$, and $9^n$ never carries either of those primes.
$9^n$ is always odd, and no odd number can end in $0$ — Oddness alone is not the reason given here — the chapter’s argument is that $9^n$ carries no factor of $2$ or $5$ at all.
$9$ itself does not end in $0$, so no power of it can either — A number’s own last digit does not determine its powers’ last digits in general — the real reason is the missing prime factors $2$ and $5$.
$n$ would need to be negative for $9^n$ to end in $0$ — The argument holds for every natural number $n$ — no value of $n$, negative or otherwise, changes $9^n$’s prime factor.
A student argues: “$9^n$ ends in $9, 1, 9, 1, \ldots$ alternately, so it can never end in $0$ — that is just a pattern, nothing to do with primes.” What does this argument miss?
the real reason is structural
the pattern claim is false; $9^n$ does sometimes end in other digits
the student is completely correct, and prime factorisation is not needed here
the argument is right for even $n$ but wrong for odd $n$
Check your answer
✓ the real reason is structural — (A) A pattern in the last digit is an observation, not the underlying reason the chapter’s method actually proves.
the pattern claim is false; $9^n$ does sometimes end in other digits — The alternating $9, 1$ pattern is correct — the gap is that it is observed, not explained by prime factors.
the student is completely correct, and prime factorisation is not needed here — An observed pattern is not a proof — the chapter’s method explains WHY through the missing primes $2$ and $5$, which the pattern alone does not.
the argument is right for even $n$ but wrong for odd $n$ — Neither the pattern nor the underlying prime-factor reason depends on whether $n$ is even or odd.
By the same reasoning as $9^n$, can $8^n$ ever end in $0$?
yes — $8$ is even, and even numbers can end in $0$
no — $8^n$ is always odd
no — $8 = 2^3$
yes, once $n$ is large enough for $5$ to appear as a factor
Check your answer
yes — $8$ is even, and even numbers can end in $0$ — Being even is not enough — ending in $0$ needs BOTH $2$ and $5$, and $8^n$ never has a factor of $5$.
no — $8^n$ is always odd — $8^n$ is always even, since $8$ itself is even — the real reason it never ends in $0$ is the missing factor of $5$.
✓ no — $8 = 2^3$ — (C) The same missing-prime argument applies, this time with $2$ present but $5$ permanently absent.
yes, once $n$ is large enough for $5$ to appear as a factor — No power of $8$ ever gains a factor of $5$ — $8^n$’s prime factorisation is always a power of $2$ alone, for every $n$.
$3 \cdot 5 \cdot 7 + 7$ Look for a factor every term already shares. Do not add the sum first.
$= 7 \cdot (3 \cdot 5 + 1)$ $7$ divides every term on the left. So we can pull it out the front.
$= 7 \cdot 16 = 112$ Finish the arithmetic inside the bracket. Then multiply.
$112$ is composite It is written as a product of two integers greater than $1$. That is exactly what composite means. No further factor check is needed.
$112 = 4 \cdot 28$ Check it a different way. $112$ splits into another pair of factors greater than $1$ too. That confirms it is composite by more than one route.
A sum whose terms share a factor is that many copies of it, so 3 · 5 · 7 + 7 is sixteen sevens, not prime.
Show a sum is composite, without checking any factors by hand.
Find a shared factor Every term in $3 \cdot 5 \cdot 7 + 7$ shares the factor $7$.
Pull it out front $3 \cdot 5 \cdot 7 + 7 = 7 \cdot (3 \cdot 5 + 1)$.
Finish the arithmetic $3 \cdot 5 + 1 = 16$. So the sum is $7 \cdot 16 = 112$.
Name it composite $112$ is a product of two integers greater than $1$. So $112$ is composite.
Check yourself
$3 \cdot 5 \cdot 7 + 7$ is shown to be composite by
computing the sum directly and checking it against a list of primes
noticing that $7$ appears in the expression, which alone makes it composite
factoring out the shared $7$: $7 \cdot (3 \cdot 5 + 1) = 7 \cdot 16$
dividing the whole expression by $3$ instead of $7$
Check your answer
computing the sum directly and checking it against a list of primes — The worked example never checks against a list of primes — it factors out the shared $7$ directly.
noticing that $7$ appears in the expression, which alone makes it composite — Spotting the shared $7$ is only the first step — it must actually be factored out to show the number is a product of two integers greater than $1$.
✓ factoring out the shared $7$: $7 \cdot (3 \cdot 5 + 1) = 7 \cdot 16$ — (C) Pulling out the shared factor turns the sum into an explicit product, which is what proves it composite.
dividing the whole expression by $3$ instead of $7$ — $7$ divides every term in the expression; $3$ does not divide the second term, $7$, so it cannot be factored out cleanly.
A student says: “this number is composite just because it is a sum, and sums are never prime.” What is wrong?
the student is right — every sum of integers is composite
the number is actually prime, and the worked example is wrong
sums CAN be prime (for example $2+3=5$)
the factor that matters is $3$, not $7$
Check your answer
the student is right — every sum of integers is composite — $2+3=5$ is prime, so sums are not always composite — the shared factor $7$ is the actual reason here.
the number is actually prime, and the worked example is wrong — $3 \cdot 5 \cdot 7 + 7 = 112 = 7 \cdot 16$ is genuinely composite — the worked example’s factoring is correct.
✓ sums CAN be prime (for example $2+3=5$) — (C) The real mechanism is the shared factor, not the mere fact that a sum was computed.
the factor that matters is $3$, not $7$ — $7$ is the factor every term shares — $3$ does not divide the second term, $7$, on its own.
By the same method, is $2 \cdot 3 \cdot 11 + 11$ composite?
no — $11$ is prime, so anything built from it stays prime
no — the expression is too large to check without a calculator
yes, but only because $2 \cdot 3 \cdot 11$ is even
Check your answer
✓ yes — factor out $11$: $11 \cdot (2 \cdot 3 + 1) = 11 \cdot 7$ — (A) The same shared-factor trick applies, this time pulling out $11$.
no — $11$ is prime, so anything built from it stays prime — A prime factor does not make the whole expression prime — $2 \cdot 3 \cdot 11 + 11 = 11 \cdot 7$ is composite despite $11$ being prime.
no — the expression is too large to check without a calculator — Size does not block the method — factoring out the shared $11$ works the same way regardless of how large the numbers are.
yes, but only because $2 \cdot 3 \cdot 11$ is even — Evenness plays no role here — the reason is that $11$ divides both terms, not that $2 \cdot 3 \cdot 11$ happens to be even.
The proof that root 2 is irrational reaches the line a squared is 2 b squared, and the condition at the top comes from there.
Here is a fact we will use many times. If $p$ is a prime number and $p$ divides $a^2$, then $p$ also divides $a$, for any integer $a$. We will not prove this fact. We will use it as a tool.
Check it on a small case. Take $a = 6$, so $a^2 = 36$. The prime $2$ divides $36$. $2$ also divides $6$. Now try $a = 10$ yourself: $a^2 = 100$. The prime $5$ divides $100$. Does $5$ divide $10$ too? Yes. You just checked two examples, and both confirm the fact.
Why is this true? It comes from the Fundamental Theorem of Arithmetic. A number’s prime factorisation never changes, no matter how it is written. The primes in $a^2$’s factorisation are exactly the primes already in $a$’s own factorisation, each one appearing twice. So if $p$ shows up in $a^2$, $p$ was already there in $a$.
*We use this one fact in every proof that a square root is irrational: $p = 2$ for $\sqrt{2}$, $p = 3$ for $\sqrt{3}$, $p = 5$ for $\sqrt{5}$.*
Check yourself
If $p$ is prime and $p$ divides $a^2$, the key fact used throughout this chapter’s proofs says
$p$ must also divide $a$ itself
$p$ must divide $a$ twice, since squares double every factor
$p$ divides $a^2$ only when $a$ is itself prime
$p$ divides $a$ only when $p$ is even
Check your answer
✓ $p$ must also divide $a$ itself — (A) The lemma’s whole content is that dividing the square forces dividing the number.
$p$ must divide $a$ twice, since squares double every factor — The lemma states that $p$ divides $a$ — it says nothing about dividing it “twice”.
$p$ divides $a^2$ only when $a$ is itself prime — The lemma holds for ANY integer $a$, not only when $a$ is prime.
$p$ divides $a$ only when $p$ is even — The lemma holds for any prime $p$, even or odd — $p=2$ is only one example used in this chapter.
This chapter states the “prime divides a square” fact WITHOUT proving it, but says it follows from
a separate rule about even numbers, unrelated to prime factorisation
the definition of a rational number in lowest terms
a property that holds only for the specific primes $2$, $3$ and $5$
the Fundamental Theorem of Arithmetic’s uniqueness
Check your answer
a separate rule about even numbers, unrelated to prime factorisation — The lemma has nothing to do with evenness specifically — it follows from unique prime factorisation for any prime $p$.
the definition of a rational number in lowest terms — Lowest terms is used LATER, inside the irrationality proofs — it is not where this lemma itself comes from.
a property that holds only for the specific primes $2$, $3$ and $5$ — The lemma holds for every prime, not only the three used as worked examples in this chapter.
✓ the Fundamental Theorem of Arithmetic’s uniqueness — (D) Uniqueness pins down exactly which primes can appear in $a^2$, which is what forces $p$ to already be one of $a$’s primes.
A student applies the fact to claim: “since $4$ divides $a^2$, $4$ must divide $a$.” What goes wrong?
nothing goes wrong — the fact holds for any divisor, prime or not
the fact only fails when $a$ itself is prime
$4$ would need to divide $a^2$ twice, not once, for the rule to apply
the fact only holds when the divisor is PRIME
Check your answer
nothing goes wrong — the fact holds for any divisor, prime or not — The lemma is stated only for a PRIME divisor — dropping that condition lets a false conclusion through, since $a=2$ gives $a^2=4$ divisible by $4$ while $4$ does not divide $2$.
the fact only fails when $a$ itself is prime — The condition belongs on the divisor, which must be prime — it says nothing about whether $a$ is prime.
$4$ would need to divide $a^2$ twice, not once, for the rule to apply — The lemma is about a single divisibility, not a repeated one — the real issue is that $4$ is not prime.
✓ the fact only holds when the divisor is PRIME — (D) The lemma’s condition is that the divisor be prime, and $4=2^2$ is not prime, so it cannot be applied here.
The fraction a over b, with both a and b even, halves again to c over d.
We have built three tools already.
Prime factorisation. Every whole number bigger than $1$ breaks into primes.
One factorisation only. A number’s prime factorisation never changes.
The lemma. A prime dividing a square must divide the number itself.
Let us put all three together. They prove something concrete.
*THEOREM. $\sqrt{2}$ is irrational.*
Here is the plan, before you read the numbered proof. Assume $\sqrt{2}$ is a fraction. Follow where that leads. Watch it collide with something impossible. That collision is the whole proof.
We call this proof by contradiction. You assume the opposite of what you want to prove. Then you show that assumption forces something impossible. The numbered proof below gives every step its own reason. Do not skip a step. Each one is a gap the proof cannot survive without.
The square on diagonal a splits into four triangles, and the square of side b splits into two, giving a squared equals 2 b squared.
Proof
Given. A rational number written in lowest terms is $p/q$. Here $p$ and $q$ are integers. They share no common factor except $1$. We also have the lemma from before. If a prime divides $a^2$, it divides $a$ too.
To prove. $\sqrt{2}$ is irrational.
assume, to the contrary, that $\sqrt{2}$ is rational We assume this. The rest of the proof will show it cannot be true.
$\sqrt{2} = a/b$ for integers $a$, $b \neq 0$, with $a$ and $b$ coprime Every rational number can be written this way, in lowest terms.
$2 = a^2/b^2$, so $a^2 = 2 b^2$ Square both sides of $\sqrt{2} = a/b$. Then multiply both sides by $b^2$.
$2$ divides $a^2$ $a^2 = 2 b^2$. This says $a^2$ is $2$ times a whole number. So $2$ divides $a^2$.
$2$ divides $a$ $2$ is prime, and $2$ divides $a^2$. By the lemma, $2$ must divide $a$ itself.
$a = 2 c$ for some integer $c$ If $2$ divides $a$, then $a$ is $2$ times some whole number. Call that number $c$.
$(2 c)^2 = 2 b^2$, so $b^2 = 2 c^2$ Put $a = 2 c$ into $a^2 = 2 b^2$. $(2 c)^2 = 4 c^2$, so $4 c^2 = 2 b^2$. Divide both sides by $2$. $b^2 = 2 c^2$.
$2$ divides $b^2$ $b^2 = 2 c^2$. This says $b^2$ is $2$ times a whole number. So $2$ divides $b^2$.
$2$ divides $b$ This is the same lemma again. $2$ is prime, and $2$ divides $b^2$. So $2$ divides $b$.
$2$ divides both $a$ and $b$ But $a$ and $b$ were assumed coprime. They share no factor but $1$. This is the contradiction the whole proof was built to reach.
$\sqrt{2}$ is irrational The assumption that $\sqrt{2}$ is rational led to something impossible. So the assumption is false. $\sqrt{2}$ must be irrational.
■
No fraction, no matter how large its denominator, ever closes this gap, which is the endless search one proof by contradiction replaces.
Check yourself
Step (2) of the $\sqrt{2}$ proof writes $\sqrt{2} = a/b$ with $a$ and $b$ coprime. Why must this fraction be assumed in LOWEST TERMS?
to make the arithmetic in step (3) simpler
because $\sqrt{2}$ is only rational when written in lowest terms
so that the later contradiction
lowest terms is not required; any $a/b$ would work equally well
Check your answer
to make the arithmetic in step (3) simpler — Squaring both sides in step (3) works the same regardless of lowest terms — the assumption’s real job is setting up the contradiction, not easing arithmetic.
because $\sqrt{2}$ is only rational when written in lowest terms — The proof ASSUMES $\sqrt{2}$ is rational in order to contradict that assumption — lowest terms is not a condition on $\sqrt{2}$ being rational.
✓ so that the later contradiction — (C) The contradiction only lands because coprimality was assumed at the start and then shown to fail.
lowest terms is not required; any $a/b$ would work equally well — Without lowest terms, step (10)’s conclusion — $2$ divides both $a$ and $b$ — would not contradict anything, since sharing a factor was never ruled out.
Suppose the proof skipped the coprime assumption in step (2) and just wrote $\sqrt{2} = a/b$ for any integers $a, b$. Which step stops working?
step (10), the contradiction that needs coprime $a, b$
step (3), where squaring both sides would no longer be valid
step (5), where the prime-divides-square lemma would no longer apply
no step fails — the proof reaches the same contradiction either way
Check your answer
✓ step (10), the contradiction that needs coprime $a, b$ — (A) Every earlier algebraic step still holds; only the final contradiction depends on the coprime assumption made in step (2).
step (3), where squaring both sides would no longer be valid — Squaring both sides is valid for any integers $a$ and $b$ — coprimality plays no role in that algebraic step.
step (5), where the prime-divides-square lemma would no longer apply — The lemma needs only that $2$ is prime and divides $a^2$ — it holds for any $a$, coprime with $b$ or not.
no step fails — the proof reaches the same contradiction either way — Without coprimality, $2$ dividing both $a$ and $b$ is unremarkable — the contradiction disappears entirely.
Step (5) concludes $2$ divides $a$ from “$2$ divides $a^2$”. This step is only valid because
$2$ is prime
$a^2$ is always even when $a$ is even, regardless of any lemma
step (4) already proved $2$ divides $a$ directly
$2$ is the only number this lemma works for
Check your answer
✓ $2$ is prime — (A) The move from dividing the square to dividing the number is exactly what the lemma licenses, and only for a prime divisor.
$a^2$ is always even when $a$ is even, regardless of any lemma — This assumes $a$ is already even, which is exactly what step (5) is trying to establish — the lemma is what licenses the move without assuming it first.
step (4) already proved $2$ divides $a$ directly — Step (4) shows $2$ divides $a^2$ — it is step (5), using the lemma, that concludes $2$ divides $a$.
$2$ is the only number this lemma works for — The lemma holds for any prime, not only $2$ — the $\sqrt{3}$ and $\sqrt{5}$ proofs reuse it with different primes.
Steps (7)-(9) repeat the lemma to show $2$ divides $b$. A student skips straight from “$b^2 = 2 c^2$” to “$2$ divides $b$” without mentioning the lemma. Is this a gap?
yes — the same prime-divides-square lemma is needed for $b$ too
no — once the lemma is used once in a proof, it can be assumed for the rest
no — “$b^2 = 2c^2$” already IS the statement that $2$ divides $b$, with no lemma needed
yes, but a different lemma is needed for $b$ than the one used for $a$
Check your answer
✓ yes — the same prime-divides-square lemma is needed for $b$ too — (A) The lemma must be invoked a second time, with $b$ in the role $a$ played the first time.
no — once the lemma is used once in a proof, it can be assumed for the rest — A lemma applies every time its condition is met — it must be invoked again here, for $b$, not assumed automatically.
no — “$b^2 = 2c^2$” already IS the statement that $2$ divides $b$, with no lemma needed — “$b^2=2c^2$” shows $2$ divides $b^2$ — getting from there to “$2$ divides $b$” is exactly what the lemma is needed for.
yes, but a different lemma is needed for $b$ than the one used for $a$ — The same lemma applies to $b$ as to $a$ — nothing about $b$ requires a different fact.
The proof’s final contradiction, in step (10), is that
$a^2 = 2b^2$ has no integer solutions at all
$\sqrt{2}$ is shown to equal a fraction after all
$2$ divides both $a$ and $b$ — not coprime
$b$ turns out to equal $0$, which is not allowed
Check your answer
$a^2 = 2b^2$ has no integer solutions at all — $a^2=2b^2$ does have integer solutions — the contradiction is that any such $a,b$ cannot also be coprime, not that none exist.
$\sqrt{2}$ is shown to equal a fraction after all — The proof concludes $\sqrt{2}$ is NOT a fraction — the contradiction is what forces that opposite conclusion.
✓ $2$ divides both $a$ and $b$ — not coprime — (C) The contradiction is exactly the coprime assumption from step (2) being broken by steps (5) and (9).
$b$ turns out to equal $0$, which is not allowed — Nothing in the proof forces $b=0$ — the actual contradiction is $a$ and $b$ sharing the factor $2$ despite being assumed coprime.
Let us try a shortcut. Guess a fraction for $\sqrt{2}$. Check a few candidates, like $7/5$ or $99/70$. Stop once none of them work.
That is not a proof. Infinitely many fractions are still untried. Checking a finite number of them can never rule out the rest. A fraction that happens to equal $\sqrt{2}$ could be sitting further down the list than you looked.
Only a proof by contradiction actually settles the question. Assume a lowest-terms fraction equals $\sqrt{2}$. Then show that assumption is impossible, the way the proof above does. Checking more fractions only makes you more sure. It never rules out the fractions still untried.
Your turn: you check $10$ fractions and none equals $\sqrt{2}$. Have you proved $\sqrt{2}$ is irrational? (Answer: No. Infinitely many fractions are still untried.)
Does checking a few fractions ever prove the square root of 2 is irrational?
Weaker. A student wants to know if $\sqrt{2}$ is irrational. They try $7/5$. They square it: $(7/5)^2 = 49/25 = 1.96$. Close to $2$, but not $2$. They try $99/70$. $(99/70)^2 = 9801/4900 \approx 2.0002$. Even closer, but still not $2$. They try eight more fractions. None of them works. “I have checked ten fractions,” the student says. “None of them equals $\sqrt{2}$. So $\sqrt{2}$ must be irrational.”
This feels like proof. It is not. Infinitely many fractions are still untried. Checking ten fractions, or a thousand, never rules out the rest. The next one you try could be the one that works.
Stronger. We assume, to the contrary, that some fraction $a/b$, in lowest terms, equals $\sqrt{2}$. Let us follow where that leads. Squaring gives $a^2 = 2 b^2$. So $2$ divides $a^2$. So $2$ divides $a$. The same steps then force $2$ to divide $b$ too. But $a$ and $b$ were coprime. They cannot both be divisible by $2$. This is impossible.
The assumption is what breaks, not the arithmetic. Since assuming a fraction equals $\sqrt{2}$ leads to something impossible, no fraction can equal $\sqrt{2}$. This checks every fraction at once, not one at a time.
Check yourself
A student checks $1.4^2$, $1.41^2$ and $1.414^2$, finds none exactly equal $2$, and concludes $\sqrt{2}$ is irrational. Is this a valid proof?
yes — checking three cases is enough to establish a mathematical fact
yes, as long as the numbers checked are close enough to $\sqrt{2}$
no — finitely many checks cannot rule out infinitely many fractions
no, but only because the student did not check enough decimal places
Check your answer
yes — checking three cases is enough to establish a mathematical fact — Three checked cases, or three thousand, still leave infinitely many fractions untried — no finite check settles the question.
yes, as long as the numbers checked are close enough to $\sqrt{2}$ — No degree of closeness closes off the infinitely many other fractions that were never checked.
✓ no — finitely many checks cannot rule out infinitely many fractions — (C) Checking finitely many fractions can never close off the infinitely many that remain unchecked.
no, but only because the student did not check enough decimal places — Adding more decimal places is still a finite search — the real issue is that no finite search can ever be exhaustive.
Which of these WOULD count as a valid proof that a number $x$ is irrational?
computing $x$ to $50$ decimal places and finding no repeating pattern
checking every fraction with denominator under $1000$ and finding none equal $x$
assuming $x=a/b$ in lowest terms and deriving a contradiction
asking whether $x$ “looks like” a nice number
Check your answer
computing $x$ to $50$ decimal places and finding no repeating pattern — Fifty decimal places is still a finite check — no amount of decimal expansion, by itself, proves irrationality.
checking every fraction with denominator under $1000$ and finding none equal $x$ — Denominators under $1000$ still leave infinitely many larger denominators unchecked.
✓ assuming $x=a/b$ in lowest terms and deriving a contradiction — (C) The contradiction method is exhaustive by construction — it rules out every possible fraction at once, not case by case.
asking whether $x$ “looks like” a nice number — An impression of “niceness” is not an argument — only a contradiction proof settles the question for every fraction at once.
The reason NO finite check can ever prove a number irrational is that
computers are not precise enough to check enough fractions
irrational numbers are, by definition, impossible to write down at all
some fractions are excluded from any check by convention
infinitely many fractions exist; a finite list cannot exhaust them
Check your answer
computers are not precise enough to check enough fractions — The issue is not precision or computing power — even a perfect, infinitely fast check would still only ever cover finitely many fractions at a time.
irrational numbers are, by definition, impossible to write down at all — Irrational numbers can be written down, as decimals or symbols such as $\sqrt{2}$ — what they cannot be written as is a fraction of integers.
some fractions are excluded from any check by convention — No fraction is excluded by convention — every fraction is a candidate, which is exactly why a finite check can never be exhaustive.
✓ infinitely many fractions exist; a finite list cannot exhaust them — (D) Irrationality is a claim about EVERY fraction, and only an argument covering all of them at once can establish it.
assume, to the contrary, $\sqrt{3} = a/b$ with $a$, $b$ coprime This is proof by contradiction. It reuses the $\sqrt{2}$ proof’s method with the prime $3$.
$a^2 = 3 b^2$ Square both sides. Then clear the denominator.
$3$ divides $a^2$, so $3$ divides $a$ The same lemma as before. Apply it with $p = 3$.
$a = 3 c$, so $9 c^2 = 3 b^2$, so $b^2 = 3 c^2$ Substitute $a = 3 c$. $(3 c)^2 = 9 c^2$, so $9 c^2 = 3 b^2$. Divide both sides by $3$.
$3$ divides $b^2$, so $3$ divides $b$ The same lemma again.
$3$ divides both $a$ and $b$ But $a$ and $b$ were assumed coprime. This is the same contradiction as the $\sqrt{2}$ proof, one prime later.
$\sqrt{3}$ is irrational The assumption fails. So its opposite holds.
$\sqrt{3} \approx 1.7320508$ Check it for yourself. This decimal never settles into a repeating block. That is exactly what an irrational number’s expansion looks like.
The same picture holds for the square root of any prime, not only 3 and 5.
Check yourself
The $\sqrt{3}$ proof reuses the $\sqrt{2}$ proof’s method by
substituting the prime $3$ everywhere $2$ appeared
starting completely from scratch with a different technique
using $3$ as a coprime pair with $2$
checking a few fractions close to $\sqrt{3}$
Check your answer
✓ substituting the prime $3$ everywhere $2$ appeared — (A) The whole argument carries over unchanged because the lemma applies to any prime, and $3$ is prime.
starting completely from scratch with a different technique — The proof reuses the exact same contradiction structure, only swapping in the prime $3$.
using $3$ as a coprime pair with $2$ — $3$ replaces $2$ as the prime the proof is built around — it is not paired with $2$ as a coprime partner.
checking a few fractions close to $\sqrt{3}$ — A finite check of nearby fractions was already ruled out as invalid — the proof instead reuses the contradiction method with $3$.
Could the same method prove $\sqrt{4}$ irrational, by substituting $4$ for the prime in the argument?
yes — the method works for substituting any integer, prime or not
yes, but only because $4$ is a perfect square
no, because $\sqrt{4} = 2$ is itself prime
no — $4$ is not prime
Check your answer
yes — the method works for substituting any integer, prime or not — The lemma needs a PRIME divisor — $4$ is composite, so the substitution fails at the very step that made the $\sqrt{2}$ and $\sqrt{3}$ proofs work.
yes, but only because $4$ is a perfect square — $4$ being a perfect square is the reason $\sqrt{4}=2$ is rational, not a reason the method should work.
no, because $\sqrt{4} = 2$ is itself prime — That $\sqrt{4}=2$ happens to be prime is beside the point — the method fails because $4$, not $2$, is not prime.
✓ no — $4$ is not prime — (D) The substitution only works for a prime, and $4=2^2$ fails that condition.
Using this method, proving $\sqrt{11}$ irrational would require
checking whether $11$ is close to a perfect square
assuming $\sqrt{11}=a/b$ coprime, then contradicting that
reusing the $\sqrt{2}$ proof’s numbers directly, without change
computing $\sqrt{11}$ to enough decimal places
Check your answer
checking whether $11$ is close to a perfect square — Closeness to a perfect square plays no role — the method is the same contradiction argument, run with the prime $11$.
✓ assuming $\sqrt{11}=a/b$ coprime, then contradicting that — (B) The same template runs through unchanged, with $11$ in place of $3$.
reusing the $\sqrt{2}$ proof’s numbers directly, without change — The STRUCTURE is reused, but the working numbers change to match $11$ — the $\sqrt{2}$ proof’s specific numbers do not carry over unchanged.
computing $\sqrt{11}$ to enough decimal places — A decimal check was already ruled out as invalid — the method instead runs the same contradiction argument with $11$.
assume, to the contrary, $\sqrt{5} = a/b$ with $a$, $b$ coprime This is proof by contradiction. It reuses the same method with the prime $5$.
$a^2 = 5 b^2$ Square both sides. Then clear the denominator.
$5$ divides $a^2$, so $5$ divides $a$ The same lemma again. Apply it with $p = 5$.
$a = 5 c$, so $25 c^2 = 5 b^2$, so $b^2 = 5 c^2$ Substitute $a = 5 c$. $(5 c)^2 = 25 c^2$, so $25 c^2 = 5 b^2$. Divide both sides by $5$.
$5$ divides $b^2$, so $5$ divides $b$ The same lemma again.
$5$ divides both $a$ and $b$ But $a$ and $b$ were assumed coprime. This is the same contradiction, this time with the prime $5$.
$\sqrt{5}$ is irrational The assumption fails. So its opposite holds.
$\sqrt{5} \approx 2.2360679$ Check it for yourself. This decimal never settles into a repeating block either. That is the same signature you just saw for $\sqrt{3}$.
Check yourself
In the $\sqrt{5}$ proof, squaring $a/b$ gives $a^2 = 5b^2$. The next step concludes $5$ divides $a$ because
$a^2$ is always a multiple of $5$ when $b$ is an integer
$5b^2$ is even, so $a$ must be even too
$a$ was already assumed to be a multiple of $5$ from the start
$5$ is prime and $5$ divides $a^2$
Check your answer
$a^2$ is always a multiple of $5$ when $b$ is an integer — That $a^2=5b^2$ is a multiple of $5$ is the equation itself — getting from there to “$5$ divides $a$” is what the lemma does.
$5b^2$ is even, so $a$ must be even too — The argument is about the prime $5$, not about evenness — $5b^2$ being a multiple of $5$ has nothing to do with parity.
$a$ was already assumed to be a multiple of $5$ from the start — Nothing in the setup assumes $a$ is a multiple of $5$ — that is exactly what this step derives, using the lemma.
✓ $5$ is prime and $5$ divides $a^2$ — (D) Exactly the same lemma used for $\sqrt{2}$ and $\sqrt{3}$ applies again, now with $p=5$.
A student writes $a = 5c$ and then claims $25c^2 = 5b^2$ directly gives $c^2 = 5b^2$. What is the error?
there is no error; both forms are equivalent
the error is in writing $a = 5c$ in the first place
dividing by $5$ gives $b^2=5c^2$, not $c^2=5b^2$
the equation should have been divided by $25$, not $5$
Check your answer
there is no error; both forms are equivalent — $b^2=5c^2$ and $c^2=5b^2$ are not the same statement — only the first follows correctly from dividing by $5$.
the error is in writing $a = 5c$ in the first place — Writing $a=5c$ is exactly what “$5$ divides $a$” means — the actual slip happens one step later, in the division.
✓ dividing by $5$ gives $b^2=5c^2$, not $c^2=5b^2$ — (C) The division is right but the result was written with $b^2$ and $c^2$ in the wrong places.
the equation should have been divided by $25$, not $5$ — Dividing by $5$ is correct here — the error is not the divisor but which variable ended up squared on which side.
assume, to the contrary, $3 + 2 \sqrt{5} = r$ for some rational $r$ This is proof by contradiction. Assume the opposite of what we want to prove.
$2 \sqrt{5} = r - 3$ Rearrange. A difference of two rationals is rational.
$\sqrt{5} = (r - 3)/2$ Divide by $2$. A rational number divided by a nonzero integer is still rational.
but $\sqrt{5}$ is irrational We already proved this.
contradiction, so $3 + 2 \sqrt{5}$ is irrational The assumption forces $\sqrt{5}$ to be rational. That is false. So the assumption itself must be false.
$3 + 2 \cdot 2.2360679 \approx 7.4721359$ Check it yourself. This decimal never settles into a repeating block either. That is exactly what the algebra above just proved.
The same picture works for any number built by multiplying a known irrational by a rational number and adding another.
Check yourself
To prove $3 + 2\sqrt{5}$ is irrational, the proof assumes it equals a rational $r$ and then
repeats the full coprime-contradiction argument from scratch on $3 + 2\sqrt{5}$
isolates $\sqrt{5}$ algebraically, showing it would have to be rational too
checks a few decimal values of $3 + 2\sqrt{5}$
assumes $3$ and $2\sqrt{5}$ are each irrational separately
Check your answer
repeats the full coprime-contradiction argument from scratch on $3 + 2\sqrt{5}$ — The proof reuses the already-established irrationality of $\sqrt{5}$ instead of repeating the whole coprime argument again.
✓ isolates $\sqrt{5}$ algebraically, showing it would have to be rational too — (B) The reduction turns a new claim into the already-proved irrationality of $\sqrt{5}$.
checks a few decimal values of $3 + 2\sqrt{5}$ — A decimal check was already ruled out as invalid earlier in the chapter — this proof instead reduces algebraically to $\sqrt{5}$’s known irrationality.
assumes $3$ and $2\sqrt{5}$ are each irrational separately — The proof does not assume irrationality of either piece — it assumes the SUM is rational and derives a contradiction from that.
The step “$2\sqrt{5} = r - 3$, so it is rational” relies on which fact?
a difference of two rational numbers is always rational
a difference of a rational and an irrational number is always rational
any number that can be isolated algebraically is automatically rational
$r - 3$ is rational only when $r$ is a whole number
Check your answer
✓ a difference of two rational numbers is always rational — (A) Since $r$ and $3$ are both rational, their difference must be rational too, and that rational value equals $2\sqrt{5}$.
a difference of a rational and an irrational number is always rational — A rational minus an irrational is generally irrational — the fact used here is about TWO rationals, $r$ and $3$, both rational.
any number that can be isolated algebraically is automatically rational — Isolating a term algebraically proves nothing about rationality by itself — the actual reason is that $r$ and $3$ are both rational, so their difference is too.
$r - 3$ is rational only when $r$ is a whole number — The closure fact holds for any rational $r$, whole number or not.
Proving $6 + \sqrt{2}$ irrational by this method would begin by
assuming $6$ is irrational and deriving a contradiction from that
checking whether $6 + \sqrt{2}$ is close to a whole number
reusing the exact numbers from the $3 + 2\sqrt{5}$ proof unchanged
assuming $6+\sqrt{2}=r$, then isolating $\sqrt{2}=r-6$
Check your answer
assuming $6$ is irrational and deriving a contradiction from that — $6$ is an integer and rational by definition — the term under question is $\sqrt{2}$, not $6$.
checking whether $6 + \sqrt{2}$ is close to a whole number — The method isolates $\sqrt{2}$ algebraically and reduces to its known irrationality — it does not check closeness to a whole number.
reusing the exact numbers from the $3 + 2\sqrt{5}$ proof unchanged — The STRUCTURE carries over, but $6$ and $\sqrt{2}$ replace $3$ and $2\sqrt{5}$ — the specific numbers do not stay the same.
✓ assuming $6+\sqrt{2}=r$, then isolating $\sqrt{2}=r-6$ — (D) The same reduce-to-a-known-irrational template applies, isolating $\sqrt{2}$ this time.
A grid of twelve discs comes from three choices for the 2s, two for the 3, and two for the 5.
FTA: every number breaks into primes in one way. $1400 = 2^3 \cdot 5^2 \cdot 7$. No other set of primes works.
HCF AND LCM: read them off two factorisations. For $8$ and $12$: $\text{HCF} = 4$ and $\text{LCM} = 24$.
PRODUCT RELATION: $\text{HCF}(a, b) \cdot \text{LCM}(a, b) = a \cdot b$. Check it: $4 \cdot 24 = 96$, and $8 \cdot 12 = 96$ too.
PRIME-SQUARE LEMMA: if a prime $p$ divides $a^2$, then $p$ divides $a$. For $a = 6$, $p = 2$ divides both $36$ and $6$.
CONTRADICTION TECHNIQUE: assume $\sqrt{2} = a/b$ in lowest terms. The argument forces $2$ to divide both $a$ and $b$. That is impossible, since coprime numbers share no common factor.
You have used every one of these five rules already, in the pages above. We built each one from ONE tool: a number’s prime factorisation never changes. So an assumption that forces a fraction’s numerator and denominator to keep sharing a factor forever must be false. This is what proves $\sqrt{2}$, $\sqrt{3}$ and $\sqrt{5}$ irrational. Simple algebra then extends it to numbers such as $3 + 2 \sqrt{5}$. Look back at the five rules above. Notice how often the same numbers turn up: $8$, $12$ and $6$ appear in more than one rule.
And the question this chapter opened with. How many tray sizes split $60$ exactly? Every tray size has to be built out of $60$’s own primes, $2$, $2$, $3$ and $5$, and nothing else — that is the Fundamental Theorem again. Take any combination of them and you get $1$, $2$, $3$, $4$, $5$, $6$, $10$, $12$, $15$, $20$, $30$, $60$. Twelve sizes. There is no thirteenth, because a thirteenth would need a prime that $60$ does not have.
Check yourself
Every proof in this chapter — for $\sqrt{2}$, $\sqrt{3}$, $\sqrt{5}$ and $3+2\sqrt{5}$ — ultimately traces back to
a separate rule for each individual number proved irrational
the specific decimal expansions of each square root
the Fundamental Theorem of Arithmetic’s uniqueness
a general rule that all square roots are irrational
Check your answer
a separate rule for each individual number proved irrational — The chapter builds every proof from the SAME uniqueness tool — no number needs a rule of its own.
the specific decimal expansions of each square root — Decimal expansions were ruled out earlier as a valid method — every proof here instead uses prime factorisation’s uniqueness.
✓ the Fundamental Theorem of Arithmetic’s uniqueness — (C) One single tool, uniqueness, powers every contradiction and reduction in the chapter.
a general rule that all square roots are irrational — $\sqrt{4}=2$ is rational, so not all square roots are irrational — the chapter’s tool applies specifically through the prime-divides-square lemma.
Why does an assumption that “a fraction’s numerator and denominator keep sharing a factor forever” have to be false?
because fractions are not allowed to share any factors, by definition
because the assumption is only false for even numbers
because a number’s prime factorisation is fixed and finite
because every fraction is already in lowest terms automatically
Check your answer
because fractions are not allowed to share any factors, by definition — Fractions in general MAY share factors — it is lowest terms, a choice made at the start of the proof, that rules it out there.
because the assumption is only false for even numbers — The argument holds for any integer’s factorisation, not only even numbers.
✓ because a number’s prime factorisation is fixed and finite — (C) Uniqueness fixes a finite set of primes for any integer, which rules out an endless chain of shared factors.
because every fraction is already in lowest terms automatically — Lowest terms is a form a fraction is deliberately WRITTEN in for the proof, not a property every fraction has automatically.
The reduce-to-a-known-irrational technique (used for $3+2\sqrt{5}$) could next be applied to show which number is irrational, WITHOUT a fresh contradiction proof?
$7\sqrt{5}$, reduced to the known irrationality of $\sqrt{5}$
$\sqrt{25}$, since it involves a square root
$\sqrt{2} + \sqrt{3}$, by adding the two known irrationals directly
$10$, since it is a multiple of $5$
Check your answer
✓ $7\sqrt{5}$, reduced to the known irrationality of $\sqrt{5}$ — (A) $7\sqrt{5}$ reduces to $\sqrt{5}$ the same way $3+2\sqrt{5}$ did, needing no new contradiction proof.
$\sqrt{25}$, since it involves a square root — $\sqrt{25} = 5$, a rational whole number — having a square root sign does not make a number irrational.
$\sqrt{2} + \sqrt{3}$, by adding the two known irrationals directly — Adding two irrationals is not shown to reduce this way in the chapter — the technique reduces a NEW expression to ONE already-proved irrational, not a sum of two.
$10$, since it is a multiple of $5$ — $10$ is an ordinary rational integer — being a multiple of $5$ does not connect it to $\sqrt{5}$’s irrationality.
Check yourself: the whole chapter
Every tool in this chapter — HCF, LCM, digit endings, and square-root proofs — comes from one fact. What is that fact?
Every prime number is itself irrational, so it can prove other numbers irrational too. Being irrational is treated as the useful property here.
Every integer greater than $1$ factors into primes in exactly one way, apart from the order of the factors.
There are infinitely many prime numbers, so a proof can always find a new one to use.
Every composite number is bigger than every prime number that divides it.
Check your answer
Every prime number is itself irrational, so it can prove other numbers irrational too. Being irrational is treated as the useful property here. — Primes are ordinary whole numbers, not irrational numbers — what makes the chapter’s proofs work is that every number’s prime factorisation is fixed, not any property of the primes being irrational.
✓ Every integer greater than $1$ factors into primes in exactly one way, apart from the order of the factors. — (B) A number’s prime factorisation is fixed — the same primes, the same number of times, however it is worked out. That single fact is the tool behind every method in this chapter.
There are infinitely many prime numbers, so a proof can always find a new one to use. — How many primes exist plays no role in these proofs — what matters is that a number’s OWN prime factorisation cannot be written a second, different way.
Every composite number is bigger than every prime number that divides it. — This comparison is true but irrelevant — it has nothing to do with why unique factorisation lets HCF, LCM, and irrationality proofs all work.
According to the Fundamental Theorem of Arithmetic, $1400 = 2^3 \cdot 5^2 \cdot 7$. What does this theorem guarantee about $1400$?
$1400$ can also be written as a product of primes in a completely different set, such as using $11$ or $13$ instead.
$1400$ has more than one correct prime factorisation, and either one may be used.
No other set of primes, in any combination, multiplies to give $1400$ — this is the only prime factorisation it has.
$1400$ itself must be a prime number, since its factorisation uses only primes.
Check your answer
$1400$ can also be written as a product of primes in a completely different set, such as using $11$ or $13$ instead. — The theorem’s whole point is that no second, different set of primes can multiply to $1400$ — $2^3 \cdot 5^2 \cdot 7$ is the only one.
$1400$ has more than one correct prime factorisation, and either one may be used. — A composite number never has more than one prime factorisation — the theorem guarantees exactly one, not several to choose from.
✓ No other set of primes, in any combination, multiplies to give $1400$ — this is the only prime factorisation it has. — (C) The Fundamental Theorem of Arithmetic guarantees that a composite number’s prime factorisation is unique — apart from the order the primes are written in, no other set of primes multiplies to the same number.
$1400$ itself must be a prime number, since its factorisation uses only primes. — Being factorable into primes is what makes $1400$ composite, not prime — a prime number has no smaller prime factors at all.
$60 = 2^2 \cdot 3 \cdot 5$ and $84 = 2^2 \cdot 3 \cdot 7$. Using their prime factorisations, what is $\text{HCF}(60, 84)$?
$2^2 \cdot 3 \cdot 5 \cdot 7 = 420$
$5 \cdot 7 = 35$
$60 \cdot 84 = 5040$
$2^2 \cdot 3 = 12$
Check your answer
$2^2 \cdot 3 \cdot 5 \cdot 7 = 420$ — $2^2 \cdot 3 \cdot 5 \cdot 7 = 420$ is $\text{LCM}(60, 84)$, the largest power of every prime appearing in either number — HCF instead wants the smallest shared power.
$5 \cdot 7 = 35$ — $5$ and $7$ are exactly the primes that are NOT shared between $60$ and $84$ — the HCF only uses primes both numbers have in common.
$60 \cdot 84 = 5040$ — $60 \cdot 84 = 5040$ multiplies the two original numbers directly; the HCF comes from comparing their prime factorisations, not from multiplying the numbers themselves.
✓ $2^2 \cdot 3 = 12$ — (D) $60$ and $84$ share the primes $2$ (to the power $2$ in both) and $3$ (to the power $1$ in both); taking the smallest shared power of each gives $\text{HCF} = 2^2 \cdot 3 = 12$.
For two positive integers $a$ and $b$, $\text{HCF}(a, b) \cdot \text{LCM}(a, b) = a \cdot b$. If $\text{HCF}(a, b) = 4$ and $a \cdot b = 96$, what is $\text{LCM}(a, b)$?
$4$
$24$
$96$
$384$
Check your answer
$4$ — $4$ is the HCF itself, not the LCM — the relation still needs $a \cdot b$ divided by the HCF to find the LCM.
✓ $24$ — (B) The relation $\text{HCF}(a, b) \cdot \text{LCM}(a, b) = a \cdot b$ gives $\text{LCM}(a, b) = 96 / 4 = 24$.
$96$ — $96$ is $a \cdot b$ itself; the relation needs this divided by the HCF, $4$, to isolate the LCM.
$384$ — Multiplying $4$ and $96$ reverses the relation — the HCF and the product are given, so the LCM comes from dividing, not multiplying.
In the standard proof that $\sqrt{2}$ is irrational, the argument assumes $\sqrt{2} = a/b$ with $a$ and $b$ coprime, and eventually shows that $2$ divides both $a$ and $b$. Why does this finish the proof?
Because $a$ and $b$ being divisible by $2$ means $\sqrt{2} = a/b$ must be a whole number.
Because dividing $a$ and $b$ by $2$ gives a simpler fraction equal to $\sqrt{2}$. This looks like ordinary fraction simplification.
Because it contradicts the assumption that $a$ and $b$ share no common factor other than $1$.
Because $2$ dividing both $a$ and $b$ shows that $2$ is not actually a prime number.
Check your answer
Because $a$ and $b$ being divisible by $2$ means $\sqrt{2} = a/b$ must be a whole number. — $a$ and $b$ both being divisible by $2$ says nothing about the value of $a/b$ — it is not evidence that $\sqrt{2}$ is a whole number.
Because dividing $a$ and $b$ by $2$ gives a simpler fraction equal to $\sqrt{2}$. This looks like ordinary fraction simplification. — The proof is not looking for a simpler fraction — finding that $a$ and $b$ share the factor $2$ is exactly the impossible outcome the proof is built to reach.
✓ Because it contradicts the assumption that $a$ and $b$ share no common factor other than $1$. — (C) The proof began by assuming $a$ and $b$ are coprime, sharing no common factor other than $1$. Showing that $2$ divides both of them breaks that assumption — a contradiction that makes the original assumption, that $\sqrt{2}$ is rational, false.
Because $2$ dividing both $a$ and $b$ shows that $2$ is not actually a prime number. — $2$ being prime was used earlier to conclude that $2$ dividing $a^2$ means $2$ divides $a$; the proof never doubts this, and nothing here calls it into question.
The same reading works for any two lengths a problem gives you: factor both, and the HCF is the biggest tile that leaves nothing over.
You may not write another proof after Class 10. But primes, HCF, LCM and square roots stay with you. You will meet them in shops, on floors and on your phone. Here are eight of those places.
Tiling a verandah. Your verandah is $330$ cm by $210$ cm. You want the biggest square tile that needs no cutting. The maths: the side of that tile is the HCF of the two lengths. $330 = 2 \cdot 3 \cdot 5 \cdot 11$ and $210 = 2 \cdot 3 \cdot 5 \cdot 7$. So the HCF is $30$ cm, and you need $11 \times 7 = 77$ tiles.
The HCF answers the other question too: not how big each equal part is, but how many equal parts two amounts will make.
Packing sweet boxes for Diwali. Your shop has $48$ laddoos and $36$ barfis. Every box must be the same, with nothing left over. The maths: the most boxes you can make is the HCF of $48$ and $36$. $48 = 2^4 \cdot 3$ and $36 = 2^2 \cdot 3^2$. So the HCF is $2^2 \cdot 3 = 12$ boxes, each with $4$ laddoos and $3$ barfis.
Buying for a party. Paper plates come in packs of $8$. Cups come in packs of $12$. You want the same number of each, with none spare. The maths: the smallest such number is the LCM of $8$ and $12$. $8 = 2^3$ and $12 = 2^2 \cdot 3$. So the LCM is $2^3 \cdot 3 = 24$: three packs of plates and two packs of cups.
Halving keeps the shape only when the long side is root two times the short one, and no whole number of millimetres hits that exactly.
A square plot with 10 m sides has a diagonal fence of 10 root 2 m, about 14.14 m.
Paying two plans. Your phone plan renews every $28$ days. Your streaming plan renews every $30$ days. Both renewed today. The maths: both fall due on the same day again after the LCM of $28$ and $30$ days. $28 = 2^2 \cdot 7$ and $30 = 2 \cdot 3 \cdot 5$. So the LCM is $2^2 \cdot 3 \cdot 5 \cdot 7 = 420$ days, a little over a year.
Jogging with a friend. You run one lap of the park in $6$ minutes. Your friend takes $8$ minutes. You start together. The maths: you cross the start line together again after the LCM of $6$ and $8$ minutes. $6 = 2 \cdot 3$ and $8 = 2^3$. So the LCM is $2^3 \cdot 3 = 24$ minutes. That is your fourth lap and your friend’s third.
Paying online. Many banking and shopping sites lock your card details with one huge number. That number is two giant primes multiplied together. The maths: every whole number above $1$ has exactly one prime factorisation. For a huge number, finding it is very slow. Multiplying is quick: $7 \cdot 11 = 77$. Undoing it for a number over $600$ digits long would take far longer than a lifetime.
Choosing a paper size. Fold an A4 sheet in half and you get A5. It has the same shape as A4, only smaller. The maths: that works only when the long side is $\sqrt{2}$ times the short side. And $\sqrt{2}$ is an irrational number. A4 is $297$ mm by $210$ mm, and $297 / 210 \approx 1.414$. No sheet in whole millimetres can hit $\sqrt{2}$ exactly.
Fencing a square plot. Your plot is a square, $10$ m on each side. You want a fence along its diagonal. The maths: the diagonal is $10 \sqrt{2}$ m. Since $\sqrt{2}$ is irrational, its decimal never ends. $10^2 + 10^2 = 200$, so the diagonal is $\sqrt{200} = 10 \sqrt{2} \approx 14.14$ m.
Your turn. You are setting out chairs for a wedding. There are $42$ chairs for the bride’s guests and $30$ for the groom’s. Every row must be the same length, and no row mixes the two sides. What is the longest row you can make? Answer: $42 = 2 \cdot 3 \cdot 7$ and $30 = 2 \cdot 3 \cdot 5$, so the HCF is $6$. That gives $7$ rows for the bride’s side and $5$ for the groom’s.
practice Find the prime factorisation of $594$. (Worked in full below — read it, then do the next two the same way.)
practice Find the HCF and LCM of $66$ and $110$ by prime factorisation, and check that $\text{HCF} \cdot \text{LCM} = 66 \cdot 110$. (Factor each one the same way as above. Then take the smaller power of every prime they share for the HCF, and the larger power of every prime you see for the LCM.)
practice Find the HCF and LCM of $20$, $28$ and $35$ by prime factorisation. (Three numbers this time. The prime method does not change at all — but the $\text{HCF} \cdot \text{LCM}$ rule from the last item is for two numbers only, so it is not available here.)
practice Find the prime factorisation of 720.
practice Find the HCF and LCM of 96 and 120 by prime factorisation.
practice Find the HCF and LCM of 12, 18 and 24 by prime factorisation.
practice Two buses leave a bus stand together. One returns to the stand every 24 minutes, the other every 36 minutes. After how many minutes will they next leave together?
practice Using the Fundamental Theorem of Arithmetic, show that $7^n$ can never end in the digit 5, for any natural number $n$.
practice Show that $6 \cdot 7 \cdot 11 + 11$ is a composite number.
$20 = 2^2 \cdot 5$, $28 = 2^2 \cdot 7$, $35 = 5 \cdot 7$; no prime is in all three, so $\text{HCF} = 1$, and $\text{LCM} = 2^2 \cdot 5 \cdot 7 = 140$.
$720 = 2^4 \cdot 3^2 \cdot 5$.
$\text{HCF} = 24$, $\text{LCM} = 480$.
$\text{HCF} = 6$, $\text{LCM} = 72$.
72 minutes.
$7^n$’s only prime factor is 7, so 5 can never divide it — it can never end in 5.
$6 \cdot 7 \cdot 11 + 11 = 11 \cdot 43 = 473$, a product of two integers greater than 1.
Exercise 1.1 — further practice
practice Find the prime factorisation of $3675$.
practice Find the prime factorisation of $1050$.
practice Find the HCF and LCM of $45$ and $75$ by prime factorisation, and verify that $\text{HCF}(45, 75) \cdot \text{LCM}(45, 75) = 45 \cdot 75$.
practice Find the HCF and LCM of $84$ and $126$ by prime factorisation.
practice Find the HCF and LCM of $15$, $20$ and $45$, using prime factorisation.
practice Given that $\text{HCF}(180, 252) = 36$, find $\text{LCM}(180, 252)$.
practice Three bells ring at intervals of $20$, $24$ and $30$ minutes. If all three ring together at eight in the morning, after how many minutes will they next ring together?
practice Show that $15^n$ can never end in the digit $0$, for any natural number $n$.
practice Show that $5 \cdot 7 \cdot 11 \cdot 13 + 11 \cdot 13$ is a composite number.
practice The HCF of two positive integers is $4$ and their product is $240$. What is their LCM?
$40$
$48$
$60$
$96$
Answers
$3675 = 3 \cdot 5^2 \cdot 7^2$.
$1050 = 2 \cdot 3 \cdot 5^2 \cdot 7$.
$\text{HCF} = 15$, $\text{LCM} = 225$.
$\text{HCF} = 42$, $\text{LCM} = 252$.
$\text{HCF} = 5$, $\text{LCM} = 180$.
$\text{LCM}(180, 252) = 1260$.
$120$ minutes.
$15^n$’s only prime factors are $3$ and $5$, so $2$ never divides it — it can never end in the digit $0$.
$5 \cdot 7 \cdot 11 \cdot 13 + 11 \cdot 13 = 11 \cdot 13 \cdot (5 \cdot 7 + 1) = 143 \cdot 36 = 5148$, a product of two integers greater than 1.
practice Prove that $\sqrt{11}$ is irrational. (Worked in full below — read it, then do the next two the same way.)
practice Prove that $5 + \sqrt{11}$ is irrational. (Get $\sqrt{11}$ on its own first. Then use what item 1 has just proved, rather than starting again.)
practice Prove that $3 / \sqrt{11}$ is irrational. (One step further — $3$ divided by a nonzero rational is still rational.)
practice Prove that $\sqrt{5}$ is irrational.
practice Prove that $3 + 2 \sqrt{5}$ is irrational.
practice Prove that $1/\sqrt{2}$ is irrational.
practice Prove that $7 \sqrt{5}$ is irrational.
practice Prove that $6 + \sqrt{2}$ is irrational.
Answers
Assume $\sqrt{11} = a/b$ with $a$, $b$ coprime. Squaring gives $a^2 = 11 b^2$, so $11$ divides $a^2$ and hence $a$. Writing $a = 11 c$ leads to $b^2 = 11 c^2$, so $11$ divides $b$ too — contradicting that $a$ and $b$ are coprime.
Assume $5 + \sqrt{11} = r$ is rational; then $\sqrt{11} = r - 5$ would be a difference of two rationals, so rational — but $\sqrt{11}$ is irrational, a contradiction.
Assume $3 / \sqrt{11} = r$ is rational, and $r \neq 0$; then $\sqrt{11} = 3 / r$ would be rational, since $3$ divided by a nonzero rational is rational — but $\sqrt{11}$ is irrational, a contradiction.
Assume $\sqrt{5} = a/b$ with $a$, $b$ coprime; the same contradiction as the worked example — 5 divides both $a$ and $b$.
Assume $3 + 2 \sqrt{5}$ is rational; then $\sqrt{5}$ would be rational too, which is false.
Assume $1/\sqrt{2} = r$ is rational; then $\sqrt{2} = 1/r$ would be rational, since the reciprocal of a nonzero rational is rational — but $\sqrt{2}$ is irrational.
Assume $7 \sqrt{5} = r$ is rational; then $\sqrt{5} = r/7$ would be rational — but $\sqrt{5}$ is irrational.
Assume $6 + \sqrt{2} = r$ is rational; then $\sqrt{2} = r - 6$ would be rational, as a difference of two rationals — but $\sqrt{2}$ is irrational.
Exercise 1.2 — further practice
practice Prove that $2 + \sqrt{3}$ is irrational.
practice Prove that $5 \sqrt{2}$ is irrational.
practice Prove that $1 / \sqrt{3}$ is irrational.
practice Prove that $4 - \sqrt{5}$ is irrational.
practice Prove that $2 / \sqrt{5}$ is irrational.
practice Prove that $\sqrt{7}$ is irrational.
practice Given that $\sqrt{7}$ is irrational, prove that $3 \sqrt{7}$ is irrational.
Answers
Assume $2 + \sqrt{3}$ is rational; then $\sqrt{3} = (2 + \sqrt{3}) - 2$ would be a difference of two rationals, so rational — but $\sqrt{3}$ is irrational, a contradiction.
Assume $5 \sqrt{2} = r$ is rational; then $\sqrt{2} = r / 5$ would be rational, since a nonzero rational divided by 5 is rational — but $\sqrt{2}$ is irrational, a contradiction.
Assume $1 / \sqrt{3} = r$ is rational; then $\sqrt{3} = 1 / r$ would be rational, since the reciprocal of a nonzero rational is rational — but $\sqrt{3}$ is irrational, a contradiction.
Assume $4 - \sqrt{5} = r$ is rational; then $\sqrt{5} = 4 - r$ would be a difference of two rationals, so rational — but $\sqrt{5}$ is irrational, a contradiction.
Assume $2 / \sqrt{5} = r$ is rational; then $\sqrt{5} = 2 / r$ would be rational, since 2 divided by a nonzero rational is rational — but $\sqrt{5}$ is irrational, a contradiction.
Assume $\sqrt{7} = a / b$ with $a$, $b$ coprime. Squaring gives $a^2 = 7 b^2$, so $7$ divides $a^2$ and hence $a$. Writing $a = 7 c$ leads to $b^2 = 7 c^2$, so $7$ divides $b$ too — contradicting that $a$ and $b$ are coprime.
Assume $3 \sqrt{7} = r$ is rational; then $\sqrt{7} = r / 3$ would be rational, since a nonzero rational divided by 3 is rational — but $\sqrt{7}$ is irrational, a contradiction.
