# Prove that $13\sqrt{2}$ is irrational.

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!

## 3 Answers

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.

• 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! – Marie Jul 15 '20 at 17:47
• @Marie If $a/b$ is a ratio of two integers $a, b$, then $a/(13b)$ is a ratio of two integers $a, 13b$. – Rivers McForge Jul 15 '20 at 17:54
• @RiversMcForge I have been thinking about it and it makes more sense! Thank you – Marie Jul 15 '20 at 17:56

@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?

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 $$\operatorname{gcd}(m,n)=1$$.