# Please verify my proof: $\sqrt{18}$ is irrational.

The following is one of my exam questions and I would like to know whether my proof is correct or not.

Prove $\sqrt{18}$ is irrational using proof by contradiction.

Following is my proof:

Assume $\sqrt{18}$ is rational. Then there exists $a,b$ in integers such that $a/b$ is written in the lowest terms and $\sqrt{18}$ $=a/b$, $b ≠ 0$. For $a/b$ to be written in the lowest term a or b or both have to be odd.
Square both sides: $∴$ $a^2/b^2$ $=18b^2$ $=>$ $a^2 = 18b^2$.
Therefore $a$ is even since an even number squared gives an even number.
Let $a = 2n$, $n$ $ϵ$ $ℤ$ $=> a^2 = 4n^2$. So $4n^2/b^2 = 18$.
$b^2$ must be an even number since only an even number $(4n^2)$ divided by an even number $(b^2)$ will produce an even number $(18)$.
$∴$ $a$ and $b$ are even, this contradicts at least one being odd. Therefore, $\sqrt{18}$ is irrational.

Please give me some feedback. Thank you.

• Simplification: $\sqrt{18}=3\sqrt{2}$, i.e. the former is irrational iff $\sqrt{2}$ is irrational. – A.Γ. Nov 6 '16 at 1:14
• Strange edit history. It seems that kunjimamu is actually two different people. – TonyK Nov 6 '16 at 22:15

What you stated is correct, because if $4n^2=18b^2$ then you can only get one factor of $2$ from from $18$ so you need another from $b^2$ so $b$ is even (all these work because $2$ is a prime). But again your reasoning isn't correct, because if $b=1$ then $4n^2/1$ is even but $b$ is not even.

The required contradiction is correctly stated, but its deduction does not follow from valid reasonings, unfortunately.

• The OP didn't say that only both $a$ and $b$ were odd, they said one or the other or both. – Ian Miller Nov 6 '16 at 1:14
• @IanMiller I think I saw it was stated both $a,b$ are odd, but it was edited. – user160738 Nov 6 '16 at 1:15
• I don't see an edit flagged on the question. – Ian Miller Nov 6 '16 at 1:16
• @IanMiller Okay I deleted that part, sorry about that. But rest of what I said is still true I think – user160738 Nov 6 '16 at 1:18

Your first part is right in deducing that $a$ but be even.

However arguing that an even must be divided by an even to give an even is incorrect. Consider 6 divided by 3. An even divided by an odd can give an even.

Instead you could have finished the proof as follows:

$$\frac{4n^2}{b^2}=18$$

$$4n^2=18b^2$$

$$2n^2=9b^2$$

Hence $b^2$ must be an even number as the left hand side is even. Hence $b$ is even as an only an even number squared gives an even number.

So both $a$ and $b$ are even but this contradicts the original assumption so the assumption must be incorrect.

Therefore $a$ is even since an even number squared gives an even number.

It's true that if $a$ is even, then $a^2$ is even, but that's not what you used here. We want the converse: if $a^2$ is even, how do we know that $a$ is even? (I would prove the contrapositive: if $a$ is odd, then $a^2$ is odd).

$b^2$ must be an even number since only an even number $(4n^2)$ divided by an even number $(b^2)$ will produce an even number $(18)$.

It's unclear which part of this statement the word "only" applies to. Furthermore, I would avoid dividing whenever possible. Stick with $4n^2 = 18b^2 \iff 9b^2 = 2n^2$ . Argue that $b$ must be even (because if $b$ were odd, then $b^2$ is odd, so $9b^2$ is odd, contradicting the fact that it equals $2n^2$ and thus must be even).