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Assuming that $\sqrt{2}=\frac{p}{q}$ where $p,q\in\mathbb{N}$.

That implies that $p^2=2q^2$, and since $p$ is a natural number then the right hand side $2q^2$ must be a square of natural number and hence $2=n^2$ where $n\in\mathbb{N}$.

Since no natural number satisfies the equation $2=n^2$, then $\sqrt{2}$ can't be rational number.

Is this a valid proof?

Mohamed Mostafa
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    No, I don't think there is an easier proof. This standard proof is the easiest. – Dietrich Burde Jun 05 '24 at 16:24
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    "And hence $2 = n^2$ where $n \in \mathbb N$" - why? – mihaild Jun 05 '24 at 16:27
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    I have always thought about it as the left hand side having an even number of factors of $2$ (because of the square), and the right side having an odd number of factors of $2$. Probably my favourite proof because it generalises very easily to roots of any non-square number – George Coote Jun 05 '24 at 16:27
  • @DietrichBurde no it doesn't actually answers my question. – Mohamed Mostafa Jun 05 '24 at 16:28
  • @mihaild because $p^2$ is natural number, then it could be written as a product of squares of natural numbers – Mohamed Mostafa Jun 05 '24 at 16:31
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    You need to reference something for why $2=n^2.$ Presumably, you are using unique factorization, but if you wrote that step out, it would not be a "simpler proof." – Thomas Andrews Jun 05 '24 at 16:32
  • @ThomasAndrews So it's a valid proof even if not "simpler" ? – Mohamed Mostafa Jun 05 '24 at 16:33
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    as @mihaild said, your $2=n^2$ where $n \in \mathbb{N} $ part needs work. You cannot leave things out of a proof to make it "simpler". For example, if you're using the Fundamental Theorem of Arithmetic you have to explicitly write that. – AllCatsAreGrey Jun 05 '24 at 17:05
  • "Because $p^2$ is natural number, then it could be written as a product of squares of natural numbers." Sure it can, for example $p^2 = 1^2\cdot p^2$ is one way to do it. But how you know that you can write $p^2 = n^2q^2$ for some $n$? – Ennar Jun 05 '24 at 17:21
  • This new proof is incomplete. – Peter Jun 06 '24 at 18:01

2 Answers2

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This is the step that I think is invalid (without further justification at least):

You know that $2q^2$ is the square of an integer, and clearly $q^2$ is the square of an integer. In your next line, you claim this implies that $2$ is the square of an integer. But as far as I can tell, all we can really say from this information is that $2$ is the square of a rational number. (Indeed, this simply recovers the initial assumption $2=(\frac pq)^2$.)

Greg Martin
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  • I think OP is just using Fundamental Theorem of Arithmetic, no? Since $q^2,p^2\in\mathbb{Z}$? – J.G.131 Jun 05 '24 at 16:35
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    When writing a proof, we need to communicate what we're using. – Greg Martin Jun 05 '24 at 16:36
  • I see. I suppose had he specified that the Fundamental Theorem of Arithmetic is why he was able to reach the conclusion, then it would be valid. – J.G.131 Jun 05 '24 at 16:38
  • Not just saying that they used the fundamental theorem of arithmetic, but also how the fundamental theorem of arithmetic implies the conclusion. – Greg Martin Jun 05 '24 at 16:40
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Your proof is invalid. The argument that $2=n^2,\ n\in\mathbb N$ is wrong. Even $2=\left(\frac{a}{b}\right)^2$ can make $2q^2$ a perfect square ($p^2$). Now, you have to prove the ratio $\frac ab$ is not possible. Thus, the simplest proof of the irrationality of $\sqrt2$ is the standard one. You can learn about more proofs here.

Here, I would like to prove $\sqrt2$'s irrationality (in a different way) as follows:
By the rational root theorem, every rational root of the polynomial $n^2-2$ would have a numerator that divides $2$ and a denominator that divides $1$, i.e., the only possible rational roots: $n=\pm1,\pm2$. But $(\pm1)^2-2\neq0\neq(\pm2)^2-2$. So, its root $n=\sqrt2$ is not a rational number.

Pustam Raut
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