Prove $\sqrt {92}$ is an irrational number

Question

Prove $\sqrt {92}$ is an irrational number is the question. I know how to do questions like root $4$ etc., however I do not know how to go about this any help is appreciated! This is what I have so far.

In order to prove $\sqrt {92}$ is irrational, we must use proof by contradiction. Hence we must first prove it isn’t rational. In order to prove $\sqrt {92}$ is rational, we must see if it can be written as the ratio of 2 co-prime integers $a$ and $b$ such that $a/b= \sqrt {92}$ and $a, b$ are no longer reducible. By squaring both sides, we get:

$92=a^2/b^2 \\ 92b^2=a^2$

Therefore $a^2$ is an even number as anything multiplied by $92$ is an even number Therefore $a$ is even as $a \times a$ is even. Therefore $\sqrt {92}$ is irrational.

Answer 1

Your approach is a good one.

From $92b^2 = a^2$, you have that

$23|a^2 \implies 23|a \implies $

[since $a,b$ assumed coprime] $(23)$ does not divide $(b)$.

Therefore, you have that

$(23)^2$ divides $(a^2)$ but $(23)^2$ does not divide $92b^2$.

This is your contradiction.

Answer 2

Therefore a2 is an even number as anything multiplied by 92 is an even number Therefore a is even as a×a is even.

True. $a^2$ is even. But.... so what? So long as $b^2$ is odd we have no contradiction.

So you have $92b^2 = a^2$ where $a$ is even.

Let $a = 2a'$ so $a^2 = 4a'^2$.

We have $92b^2 = 4a'^2$ which means $23b^2 = a'^2$.

That's not at all a contradiction.

Yet.

But this means $23|a'^2$. Could that be an issue?

Well, $23$ is prime so if $23|a'^2$ then $23|a$.... so can you finish?

If $23|a'$ then let's let $a'= 23\alpha$.

Then $23b^2 = a'^2 = (23\alpha)^2 = 23^2 \alpha^2$.

So if we divide both sides by $23$ we have

$b^2 = 23 \alpha^2$.

So $23|b^2$. And as $23$ is prime $23|b$.

.... and !THAT! is a problem. We assumed $a,b$ were coprime. But we have $23|\alpha$ (so $23|a$) and we have $23|b$. Well, that contradicts they are coprime.

Answer 3

As $\sqrt{92}=2\sqrt{23}$, it suffices to prove for $23$, which is a prime.

Let $\sqrt{p}=\dfrac ab$ where $a,b$ are relative primes. Then $pb^2=a^2$ and $a^2$ is a multiple of $p$ and it must also be a multiple of $p^2$. Then $b^2$ is a multiple of $p$, a contradiction.