Why isn't enough to prove that a is an even number in this demonstration of \sqrt{2} not being rational?

Question

I was looking though a demonstration that $\sqrt{2}$ is a irrational number (it is in French) where the main points were:

• the reasoning is ad absurdum
• $\sqrt{2}$ is stated to be rational, therefore there exist numbers $a$ and $b$ which are relatively prime such that

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

From there it is derived that $a$ is even.

The demonstration then goes ahead to verify that $b$ is even as well.

Why isn't it enough to stop at the demonstration of $a$ being even? It means that it is divisible by $2$ and therefore not prime?

Is this because $a$ and $b$ must be relatively prime, so one needs to prove that $b$ is also prime (sorry I meant even, I realized my typo by the comments), and so that they have a common divider?

Answer 1

Hint:

Take, for instance, $\dfrac{2}{7}$.

Answer 2

Relative prime doesn't mean they are prime. Relative prime means they have no factors in common.

A rational number is one that can be written as $\frac ab$ where $a$ and $b$ are both integers. For example $\frac {10}{21}$.

If $a$ and $b$ have a common divisor, it can be factored out. For example $\frac {50}{105} = \frac {5*10}{5*21} =\frac {10}{21}$.

So if $\sqrt 2 $ is rational, then there exist two integers with no common factors so that $\frac ab = \sqrt 2$. There is no requirement that $a$ and $b$ be prime.

For example: $\frac {10}{21}$ is a rational number and $10$ and $21$ have no factors in common but neither of them are prime.