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The classic proof of $\sqrt{2}$ is irrational starts with the assumption that $\sqrt{2}$ can be represented as a rational number $\frac pq$ where $(p,q \in \Bbb Z , q\neq 0$, p and q are coprimes)

$$\Rightarrow \sqrt{2^2}=\frac {p^2}{q^2}$$ $$\Rightarrow 2q^2 = p^2$$ Since, $$2|p^2 \Rightarrow 2|p$$ $$\Rightarrow p= 2r$$ where, $(r \in \Bbb Z)$ $$\Rightarrow 2q^2=(2r)^2$$ $$\Rightarrow q^2 = 2r^2$$ Since, $$2|q^2 \Rightarrow 2|q$$ since p and q have a common factor of 2, this contradicts the statement that p and q are coprimes and hence our assumption that $\sqrt2$ is rational is incorrect which implies that $\sqrt2$ is irrational. While this argument is pretty reasonable,

I think it can be shortened, in the second line of the proof (i.e. $2q^2=p^2$) since p and q are coprimes they don't share any common prime factors, and since p is an integer which is divisible by 2, its square (i.e. $p^2$) must have an even number of 2s but as we can see it only has one factor of $2$ (i.e.$2q^2=p^2$) which implies that $p$ is not an integer which is a contradiction to the given statement that $p$ is an integer. Hence $\sqrt2$ is an irrational number.

Please correct me if this proof is wrong.

Bill Dubuque
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Anubhav Kumar
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    https://math.stackexchange.com/a/2970868/977780 here you will find your answer and you can verify it by your own – Sourav Ghosh Dec 13 '21 at 05:12
  • The standard proof requires you to know that if $p$ is prime and $p\mid a^n$ for some $n,$ then $p\mid a.$ Unique factorization might let you shorten the proof, but unique factorization is in some ways a fairly complicated theorem. Try writing out what unique factorization means, and notice you end up with the dreaded ellipsis “$\dots$“ in a product. In first order Peano arithmetic, it is particularly hard to write out unique factorization. It is not even remotely obvious that it can be written - it seems to require the result that “exponentiation is diophantine” to write it out. – Thomas Andrews Dec 13 '21 at 06:30
  • So, it is slightly shorter using unique factorization, but with a boatload of hidden complications. – Thomas Andrews Dec 13 '21 at 06:30
  • Your shorter form is a variant of a well-known proof - compare the parity of the powers of $2$ on both sides, using FTA (existence & uniqueness of prime factorizations), e.g. see this answer in the linked dupe. As emphasized there we don't even need the very strong FTA theorem, only the much simpler result that every natural can be written uniquely in the form $,2^i n,$ for $,n,$ odd, which has a short simple proof. – Bill Dubuque Dec 13 '21 at 08:36
  • Note that in the linked form of the proof we don't even need to assume $p$ and $q$ are coprime - the parity mismatch still works even in the non-coprime case. – Bill Dubuque Dec 13 '21 at 08:45
  • See also here for how to generalize this style of proof to show that the square-root of any nonsquare is irrational, and see here and here and here for many other irrationality proofs and generalizations. – Bill Dubuque Dec 13 '21 at 09:20

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The proof you suggest uses (two particular cases of) the fundamental theorem of arithmetic: 1) the product of two odd numbers is odd, and 2) a product of $2$ and an odd number cannot be divisible by $4$. Both facts are easy to prove without FToA, but the proof will be longer than the standard proof of the statement.

Note that 1) but not 2) is used implicitly in the standard proof too.

markvs
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