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I am trying to prove, using the fundamental theorem of arithmetic and contradiction, that $\sqrt{2}$ is irrational. Here is my attempt.

Suppose otherwise that $\sqrt{2} \in \mathbb{Q}$, so there exists $a,b \in \mathbb{Z}$ with $b \neq 0$ such that $\sqrt{2} = \frac{a}{b}$. Without loss of generality, we can assume $a,b > 0$ since $q^2 = (-q)^2$, and that $a$ and $b$ have no common factors. Then $2 = \frac{a^2}{b^2}$, so $2b^2 = a^2$. By the fundamental theorem of arithmetic, $a$ and $b$ can be factored uniquely as a product of primes. Let $M$ be the number of prime factors, with multiplicity, of $a$ and $N$ the number of prime factors, with multiplicity, of $b$. Then $a^2$ and $b^2$ have $2M$ and $2N$ prime factors, with multiplicity. As $2$ is prime, $2b^2$ has $2N + 1$ prime factors with multiplicity, which is not even. That is, with multiplicity, $a^2$ has an even number of prime factors and $2b^2$ has an odd number of prime factors, contradicting the fundamental theorem of arithmetic.

How does this proof look?

MathNewbie
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  • From the [tag:solution-verification] tag description: "This should not be the only tag for a question." – Shaun Jan 04 '22 at 21:33
  • @Shaun My apologies. Would you classify this problem as number theory? – MathNewbie Jan 04 '22 at 21:34
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    It's alright. You're new. Elementary number theory is fine. – Shaun Jan 04 '22 at 21:35
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    The argument is fine. Note that you really only need to focus on the order to which $2$ divides both sides. $2$ divides $a^2$ an even number of times, $2$ divides $2b^2$ an odd number of times. To be sure, we need Unique Factorization to demonstrate that the "order", in this sense, is well defined. – lulu Jan 04 '22 at 21:37
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    Although the proof is fine, it’s more complicated than the standard proof. The fundamental theorem of arithmetic is a much more complicated theorem than that the root 2 is irrational – Eric Jan 04 '22 at 22:49
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    Also, if you are going to use the overpowered fundamental theorem, you could make it even simpler. In particular, $2=a^2/b^2=2^1$ has no primes to the negative power, and any prime $p$ that divides $b$ would be negative in any prime factorization, so $b^2=1$, so $a^2=2$. Since $a^2$ is either 0,1, or at least 4, it isn’t $2$ giving the desired contradiction – Eric Jan 04 '22 at 22:56
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    This is almost exactly the proof that I prefer. – Lubin Jan 04 '22 at 23:53
  • It is correct, and is one of the most common proofs, and it doesn't require high-powered unique prime factorization, as explained in this answer in the linked dupe. – Bill Dubuque Jan 05 '22 at 02:50
  • But, to be completely rigorous, one must explicitly invoke FTA (existence and uniqueness of prime factorizations) in the several places where it is implicitly used. For more on this see here and here and here. – Bill Dubuque Jan 05 '22 at 09:31

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