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

Question

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!

Answer 1

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.

Answer 2

@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?

Answer 3

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

Answer 4

A much less elegant but maybe clearer proof than those involving greatest common denominator (g.c.d.):-

If $\sqrt{2}$ is rational, the we can express it as a ratio of two integers, $A$ and $B$.

So:

$$ \sqrt{2} = \frac {A}{B} $$

where $$ A, B \in \mathbb{N} $$

$$ \implies A = \sqrt{2}B $$

Since $ A^2 = 2B^2$ we can see that $A^2$ is even, i.e.

$$ A^2 \in \mathbb{E} $$

Each integer can be written as the product of a unique set of prime factors to various powers. So any integer that is the square of a smaller integer has each of its prime factors at least twice. So the integer square root of a number that is even will itself also be even. (And indeed if the integer square is divisible by $N$ then so will the starting integer be divisible by $N$.)

$$\implies A \in \mathbb{E} $$

Writing $A$ as some even number $2K$, we see that

$$ B^2 = \frac{A^2}{2} = \frac{(2K)^2}{2} = 2K^2 $$

$$ \implies B^2 \in \mathbb{E} $$

$$ \implies B \in \mathbb{E} $$

But from our earliest definition, we have: $$A^2 = 2B^2$$

and, with $B^2$ resolved into pairs of prime factors, this would imply that $A^2$ has one unpaired prime factor, i.e. $2$.

Yet, from our reasoning on squared integers, we know that this cannot be so since squared integers may only have pairs of prime factors.

Hence, since our initial assertion that $\sqrt{2}$ is rational has led us to an erroneous deduction, we can conclude that $\sqrt{2}$ is not a rational number.

It is clear from the above reasoning that neither is any $\sqrt{K} $ rational, where $K \in \mathbb{N}$ and $\sqrt{K} \notin \mathbb{N}$

The latter generality covers $\sqrt{338}$ or $13 \sqrt{2}$.

It also covers the more obvious situation where $K$ is a prime number, i.e. $K$ has no factors (unless you regard $1$ as a factor) let alone an integer root.