PrepShorts · Study sheet · Class 10 Mathematics · Chapter 1, Real Numbers
Chapter 1 · Real Numbers
A prime dividing a square must divide the number itself
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4 divides 36 and 9 divides 36, yet neither divides 6. A divisor of a square is not generally a divisor of the number — the claim survives only when that divisor is prime.
The idea
Squaring a number cannot smuggle in a prime that was not already there: the square is built from exactly the same primes, each used twice as often. What turns that observation into Theorem 1.2 is uniqueness — because there is only one prime description of the square, the list of primes you get by doubling is not merely a list, it is the list, so a prime found in the square has nowhere to have come from except the original number. Everything hangs on the divisor being prime; a composite divisor can be assembled from pieces of two different copies, which is why 4 divides 36 without going anywhere near 6.
What you should be able to do
- Write the square of a number as a product of primes, given the primes of the number, and state which primes appear and how often
- Explain why no prime outside the original number's factorisation can appear in the square
- Reproduce the argument of Theorem 1.2, naming the point at which the uniqueness clause is used
- Complete the final step, from "the prime is one of those in the list" to "the prime divides the number"
- Give a counterexample showing the statement fails for a divisor that is not prime, and explain what breaks
- Identify, in a later proof, each place where Theorem 1.2 is being invoked and with which prime
Words to know
| Term | Definition in one line | First introduced |
|---|---|---|
| prime number | a whole number above 1 divisible only by itself and 1 | printed in §1.3, p. 6, in the statement of Theorem 1.2 |
| divides | goes into exactly, leaving no remainder | printed in §1.3, p. 6 |
| positive integer | a whole number greater than zero | printed in §1.3, p. 6, fixing what the theorem is about |
| prime factorisation | the number rewritten so that every factor in it is prime | printed in §1.3, p. 6, at the opening of the proof |
| not necessarily distinct | the primes in the list may repeat, and usually do | printed in §1.3, p. 6, as the condition on the list of primes |
| uniqueness | the clause of the Fundamental Theorem of Arithmetic that rules out a second, disagreeing factorisation | printed in §1.2, p. 4, and again in §1.3, p. 6, where it carries the proof |
| squaring | multiplying a number by itself | printed in §1.3, p. 7, in the proof that follows |
| index doubling | the fact that squaring turns each prime's index into twice itself | an added term; the chapter shows the effect without naming it |
| composite divisor | a divisor that is itself composite, for which the theorem's conclusion can fail | an added phrasing, used only to mark the counterexample; not printed in this chapter |
Where people slip up
- "Obviously, a divisor of the square is a divisor of the number." Only when that divisor is prime, and the fact that 4 divides 36 but not 6 is enough to show the general version is false. The theorem is not restating something evident; it is drawing a line around where the evident thing is true.
- "It works for any divisor." 4 and 9 both divide 36 while dividing neither factor. Run both counterexamples.
- "A prime found in the square must appear twice over in the number." No — take p = 3 and a = 3: 3 divides 9, and 9 does not divide 3. What the theorem gives is that p divides a; from there p² divides a² for free, and nothing more.
- "Squaring might produce a new prime factor." It cannot, and this is the single sentence the whole proof exists to secure. Show 30 → 900 and 12 → 144 and let the prime list stay put both times.
- "The proof is marked as not examinable, so I can ignore the theorem." The statement appears in the chapter's own summary, and every irrationality proof in §1.3 leans on it. Skipping the proof is allowed; skipping the theorem leaves the rest of the chapter unsupported.
- "a² has the same factorisation as a." It has the same primes, not the same factorisation. 30 and 900 differ in every index.
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Worked answers: Exercise 1.1 · Exercise 1.2
Transcript1,630 words
There is a moment in this argument where everything stops and a very small fact gets proved. It looks like a detour. It is not. Here is the moment it is built for. You want to show that the square root of 2 is not a fraction. Halfway through that argument you find yourself holding a whole number a, and you know that 2 divides a squared. And you want to say: so 2 divides a.
That single step is the thing this video is about. It is not obvious, it is not free, and it is false if you change one word in it. Start with what squaring actually does to a number. Take 30. Written out in primes, 30 is 2 times 3 times 5. Square it and you get 900. And 900 is 2 squared, times 3 squared, times 5 squared. Look at what changed and what did not.
The primes are the same three: 2, 3 and 5. Every index doubled: one became two, in all three columns. Try it again on a number whose indices are not all the same. 12 is 2 squared times 3, and 12 squared is 144. 144 is 2 to the fourth, times 3 squared. The indices went from 2 and 1 to 4 and 2. The primes did not move at all.
That is the whole content of this scene, and it is worth saying plainly. Squaring a number cannot introduce a prime that was not already there. It has nothing to introduce one from. A square is built entirely out of two copies of the number. So here is the claim on the table. Let p be a prime number, and let a be a whole number. If p divides a squared, then p divides a.
Check it on the numbers we already have. Take p equal to 3. 3 divides 900, and 3 divides 30. Fine. Take p equal to 7. 7 does not divide 900, and 7 does not divide 30. Those two failures are the same failure, which is exactly what the claim says they should be. Now, why is that not just obvious? Because the general version of it is false, and the general version is what most people are quietly assuming.
Here is the general version: if a number divides a product, it divides one of the two things being multiplied. That is wrong. 6 times 6 is 36. 4 divides 36, because 36 is 4 times 9. But 4 does not divide 6. So a divisor of a square is not automatically a divisor of the number. The claim we are making is narrower than that, and the narrowness is the whole point.
It is not restating something evident. It is drawing a line around the place where the evident thing happens to be true. So let us build the argument properly, and watch where the word prime gets used. Write a out as a list of primes, multiplied together. Allow repeats: if a prime appears three times, write it three times. Now square that. You are multiplying the list by itself. So the square is the same list written out twice over.
Every prime that was in the list is still there, and each one now appears twice as often. Nothing new arrived, because there was nowhere for it to arrive from. That is the list of primes that multiply to give a squared. And here is the step that carries the whole thing. We are told that p divides a squared. So a squared can also be written as p, times whatever is left over.
Break that leftover into primes as well, and you have a second description of a squared as a product of primes. Now, a number has exactly one description as a product of primes. There is no second list. There never is. So the description we just built, the one that visibly contains p, and the description we built by doubling, have to be the same list. Which means p is on the doubled list.
That is the only place uniqueness gets used, and everything before it was bookkeeping. So p is one of the primes on the doubled list. But the doubled list is just the original list, written twice. So p is one of the primes of a itself. And a is those primes multiplied together. A number that is a product of primes is divisible by every one of them. So p divides a. That is the whole theorem.
And notice what we did not need. We never found out what a is. We never computed a squared. We only used the fact that a has one prime description and so does its square. Now break it on purpose, because a hypothesis you have never seen fail is a hypothesis you have no reason to believe in. Take a equal to 6, so a squared is 36. In primes, 6 is 2 times 3, and 36 is 2 squared times 3 squared.
Take the divisor 4, which is not prime. 4 divides 36 and 4 does not divide 6, and the factorisation shows exactly why. Draw 36 as two copies of 6 sitting side by side. Each copy carries one 2. 4 is two 2s. So it takes one 2 out of the left copy and one 2 out of the right copy. It is assembled out of pieces of both copies, and neither copy on its own has enough to give.
That is what a composite divisor can do, and it is the thing the theorem exists to rule out. Before you decide that 4 is somehow special, run the same thing with the other prime. 9 divides 36, since 36 is 9 times 4. 9 does not divide 6. And the accounting is identical, with 3 in place of 2. Each copy of 6 carries one 3. 9 is two 3s. It takes one from each copy.
So this is not a fact about the number 4. And in particular, being a power of a prime is not good enough. 9 is 3 squared. Every factor in it is the prime 3, and the theorem still fails. The hypothesis is not almost prime, or built from one prime. It is prime. So what is it that a prime cannot do? It cannot be split. To take a divisor out of both copies at once, you have to break it into two pieces first.
4 breaks into 2 and 2. 9 breaks into 3 and 3. Now try to break a prime. The only way to write 7 as one whole number times another is 1 times 7, or 7 times 1. That is what being prime means, and it is the only thing being used here. So one of the two pieces is the whole prime and the other is nothing at all.
Which means the prime has to come out of a single copy, whole. And a single copy is just a. So p divides a. That is the same theorem again, arrived at by looking at the picture instead of at the lists. Two routes, one fact. If they had disagreed, one of them would have been wrong. You might still suspect that 4 and 9 are unusual, and that composite divisors normally behave.
They do not. Take every whole number from 2 up to 300, square it, and list every divisor of that square above 1. That is 4684 divisors altogether. 578 of them are prime, and 4106 are composite. Of the 578 primes, the number that fail to divide the original number is zero. Not a small number. Zero. Of the 4106 composites, 3217 fail. So the theorem is not narrowly true and nearly general. It is true on one side of a line and mostly false on the other.
And 796 of the failures are powers of a single prime, which is why that near miss had to be killed by name. There is one more thing the theorem does not say, and it is worth being exact about. It does not say that p squared divides a. Take p equal to 3 and a equal to 3. 3 divides 3 squared, which is 9. But 9 does not divide 3.
And again that is not a lucky example. Across every prime below 200 paired with every number up to 500, the prime divides the square 952 times. In 734 of those, the square of the prime does not divide the number. What you are given is that p divides a. From there you get that p squared divides a squared, for free. You get nothing more, and you should not reach for more.
So, back to the moment we started with. You want to show the square root of 2 is not a fraction. Somewhere in that argument you have a whole number a, and you know 2 divides a squared, and you need 2 to divide a. That step is this theorem, used with p equal to 2. The same argument for the square root of 3 uses it with p equal to 3.
The square root of 5 uses it with p equal to 5. And it works there for one reason: 2, 3 and 5 are prime. Try to run the same proof on the square root of 4 and it collapses immediately, because 4 is not prime and the step is not available. Which is just as well. The square root of 4 is 2, and 2 is a perfectly good fraction.
So the small theorem is not a detour after all. It is the hinge every one of those proofs turns on, and the word prime is what holds it.
Where this fits
Taken from the notes each video was made from, not from the reading order — these are the ideas this one rests on and the ones that later rest on it.
Builds on
- Why a composite number has one prime factorisation and no otherClass 10 · Ch 1, Real Numbers
Comes up again in
- Assume the square root of 2 is a fraction, and watch the assumption collapseClass 10 · Ch 1, Real Numbers
- Re-running the same contradiction on other prime square rootsClass 10 · Ch 1, Real Numbers
Either side of this one
- Why the HCF-times-LCM shortcut works for two numbers but breaks for threeClass 10 · Ch 1, Real Numbers