Problem in the proof of root 2 is irrational

Question

The standard way we prove that the square root of $2$ is irrational is the following:

Let us assume the square root of $2$ is rational and is equal to $\frac{a}{b}$ where $a$ and $b$ are co-prime.

∴$\sqrt{2} = \frac{a}{b} \Rightarrow 2=\frac{a^2}{b^2}\Rightarrow b^2 = \frac{a^2}{2}$

∴ $a^2$ is divisible by $2$

∴ $a$ is divisible by $2$

I want to stop here. How does the statement $a^2$ is divisible by $2$ imply $a$ is divisible by $2$? For example, $4$ is divisible by $4$, but that does not mean $2$ is divisible by $4$.

Without knowing any arithmetic, you can also prove easily that a number is even iff its square is even. — TheSilverDoe, Jun 27 '23 at 12:09
See Euclid's Lemma. The special case $p=2$ can easily be proven on its own. — lulu, Jun 27 '23 at 12:09
With regards to the "special case proven on its own"... if you suppose $a$ were not divisible by $2$, i.e. that $a$ is odd, then it would follow that $a^2$ were also odd. This proof is classic for a reason. Make sure you understand it well. — JMoravitz, Jun 27 '23 at 12:11
@NikitaArtemenko I have taken the liberty to rollback your changes. Replacing $\sqrt{2}$ by $√2$ and $\Rightarrow$ by $=>$ is not an improvement, quite the contrary. — Martin R, Jun 27 '23 at 12:31

CyclotomicField · Accepted Answer · 2023-06-27T12:25:43.673

1

An even square must be even. To see this note that $(2k)^2=2(2k^2)$ which is even. An odd number squared is $(2k+1)^2 = 4k^2+4k +1 = 2(2k^2+2k) + 1$ which is odd. Since all integers are either even or odd this accounts for all squares which can be verified by considering their remainder when divided by $2$.

edited Jun 27 '23 at 12:25

answered Jun 27 '23 at 12:14

CyclotomicField

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@hardmath sure, I added it. – CyclotomicField Jun 27 '23 at 12:25

Problem in the proof of root 2 is irrational

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