0
$\begingroup$

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.

$\endgroup$
  • 5
    $\begingroup$ Simplification: $\sqrt{18}=3\sqrt{2}$, i.e. the former is irrational iff $\sqrt{2}$ is irrational. $\endgroup$ – A.Γ. Nov 6 '16 at 1:14
  • 2
    $\begingroup$ Strange edit history. It seems that kunjimamu is actually two different people. $\endgroup$ – TonyK Nov 6 '16 at 22:15
1
$\begingroup$

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.

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

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.

$\endgroup$
0
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.