3
$\begingroup$

I am currently a beginner at proofs and I am having trouble proving this problem...

I know that the square root of $2$ is irrational because the square root of $2$ can be expressed as $\frac{p}{q}$ and once both sides are squared it is true that both $p$ and $q$ are even which is a contradiction to the assumption that they have no common factors.

I am having trouble proving that $13$ and the square root of $2$ is irrational though and any help would be greatly appreciated! Since we are not dealing with the square root of $13$, I do not know how to start since we can not set it equal to $\frac{p}{q}$.

Thank you in advance!

$\endgroup$
6
$\begingroup$

If $13\sqrt{2}$ were rational then it would be of the form $a/b$ for $a,b$ integers ($b\neq 0$). But then $\sqrt{2}=(a/b)/13=a/(13b)$ would be rational.

$\endgroup$
3
  • $\begingroup$ Why would 13 = a/ (13b) be rational exactly? Wouldn't that be a contradiction? Sorry I am just confused with how the fact that it is rational can lead to proving it is irrational.. Thank you for your help though! $\endgroup$ – Marie Jul 15 '20 at 17:47
  • $\begingroup$ @Marie If $a/b$ is a ratio of two integers $a, b$, then $a/(13b)$ is a ratio of two integers $a, 13b$. $\endgroup$ – Rivers McForge Jul 15 '20 at 17:54
  • $\begingroup$ @RiversMcForge I have been thinking about it and it makes more sense! Thank you $\endgroup$ – Marie Jul 15 '20 at 17:56
2
$\begingroup$

@user722227 has given you exactly the right answer (so by all means, please give him the solution checkmark). But I thought I'd add a couple of general remarks on top of this:

(1) A rational number + rational number will always be rational

(2) A rational number + an irrational number will always be an irrational number

(3) A non-zero rational number multiplied by an irrational number will always be an irrational number.

(4) If you have an irrational + an irrational, or an irrational multiplied by an irrational you in fact cannot say anything general about the result.

Statements ${(2)}$ and ${(3)}$ both have very similar proofs given by @user722227. You simply do a proof by contradiction by assuming the contrapositive. I'll give the proof for general remark ${(3)}$ then you can prove in general ${(2)}$ for some practice (in your specific example, ${(3)}$ is the one that you needed for your question). So, take a rational number ${q\neq 0}$ and an irrational number ${r}$. If the result was rational, then

$${q\times r = \frac{p}{q}}$$

For coprime integers ${p,q}$; however, rearranging for $r$ would give us

$${r = \frac{p}{q}\times \frac{1}{q}}$$

${q}$ is rational, and hence ${\frac{1}{q}}$ is rational (this is why we needed ${q\neq 0}$, we cannot divide by $0$) - and the multiplication of two rational numbers is always rational. Hence we have deduced that ${r}$ is rational - a contradiction. This proves ${(3)}$, since it must then be the case that ${q\times r}$ is irrational.

An example of ${(4)}$ is ${\sqrt{2} + \left(-\sqrt{2}\right)}$. Both ${\sqrt{2}}$ and ${-\sqrt{2}}$ are irrational by ${(3)}$, but ${\sqrt{2} + \left(-\sqrt{2}\right)=0}$ which is obviously rational. Hence we can have irrational + irrational = rational. Can you come up with an example where irrational times irrational is rational?

$\endgroup$
2
$\begingroup$

Suppose $13\sqrt{2}$ to be Rational Then, $13\sqrt{2} = \frac{m}{n}$. Where, $\operatorname{gcd}(m,n)=1$ and $n \neq 0$. Then, $\sqrt{2 }= \frac{m}{13n}$. Therefore $\frac{m}{13n}$ is rational But, we know $\sqrt{2}$ is Irrational. And you are done.


In case, if you don't know whether $\sqrt{2}$ is rational or irrational. Then see:

Suppose $13\sqrt{2}$ to be Rational Then, $13\sqrt{2} = \frac{m}{n}$. Where, $\operatorname{gcd}(m,n)=1$ and $n \neq 0$. $(13\sqrt{2}n)^2=m^2$

$\Rightarrow$ $2(169n^2)=m^2$ Then $m^2$ is divisible by $2$. $\Rightarrow m$ is divisible by $2$. Let $m=2k$ for some integer $k$. Then $13\sqrt{2}n=2k$ $\Rightarrow$ $169n^2=2k^2$ $\Rightarrow$ $n^2$ is divisible by $2$ so as to $n$ is divisible by $2$. Contradiction since $\operatorname{gcd}(m,n)=1$.

$\endgroup$

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.