Verify my proof of the irrationality of $\sqrt{6}$.

Question

Here's my proof for the irrationality of $\sqrt{6}$:

Proof:

Let $\sqrt{6}$ be a rational number. Then

$\frac{a}{b}$ = $\sqrt{6}$ , where $a$ and $b$ are coprime.

Therefore,

$a^2 = 6 b^2$.

Here $a^2$ is divisible by both $2$ and $3$. Which means, $a$ too is divisible by both $2$ and $3$ (fundamental theorem of arithmetic).

Thus, $6b^2=(3\cdot2\cdot k)^2$ ; where $k$ is an integer.

$6b^2=36k^2$ or $b^2=6k^2$.

Here $b^2$ is divisible by both $2$ and $3$. Which means, $b$ too is divisible by both $2$ and $3$. Therefore, $a$ and $b$ have common factors, which is a contradiction. Therefore $\sqrt{6}$ is not rational.

Is the above proof correct? Is there an easier way to prove this?

Answer 1

This is correct! This is pretty much the standard way to show that a number of this form in not rational and can actually be adopted to all non-perfect squares if you're a little careful (hint: all non-perfect squares are the product of a square-free number and a perfect square).

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(these comments have been adopted into the OP) Two notational comments: The theorem that refer to is more commonly called the "fundamental theorem of arithmetic" or simply "the unique prime factorization theorem." Secondly, using a period to denote multiplication is something to be very much avoided. Use \cdot or\times to produce $\cdot$ and $\times$ respectively instead.

Answer 2

That works fine, but you don't need to do both $2$ and $3$. If:

$$a^2=6b^2$$

Then $a$ must be even, $a=2a_0$, and then: $4a_0^2=6b^2$ or $$2a_0^2=3b^2.$$

So $3b^2$ must be even, so by the fundamental theorem of arithmetic, $b$ must be even, and you are done.

Answer 3

Yes, your proof is valid. You have used a standard technique.