Prove that if $n$ is a positive integer then $\sqrt{n}+ \sqrt{2}$ is irrational

Question

Prove that if $n$ is a positive integer then $\sqrt{n}+ \sqrt{2}$ is irrational.

The sum of a rational and irrational number is always irrational, that much I know - thus, if $n$ is a perfect square, we are finished.
However, is it not possible that the sum of two irrational numbers be rational? If not, how would I prove this?

This is a homework question in my proofs course.

Answer 1

Multiply both sides by $\sqrt n - \sqrt 2$. Then $n - 2 = \frac{p}{q} ( \sqrt n - \sqrt 2 )$ so $\sqrt n - \sqrt 2$ is also rational. So we have two rational numbers whose difference (which must be rational) is $2 \sqrt 2$, meaning that $\sqrt 2$ is rational.

Answer 2

Assume that $\sqrt{n}+\sqrt{2}$ is rational, i.e., exists $a,b\in \mathbb{Z}$ such that $$ \sqrt{n}+\sqrt{2}=\frac{a}{b} $$ Then $$ n=(\frac{a}{b}-\sqrt{2})^2=(\frac{a}{b})^2-2\frac{a}{b}\sqrt{2}+2. $$ Therefore, $$ \sqrt{2}=\frac{b((\frac{a}{b})^2-n+2)}{2a}. $$ Since $b((\frac{a}{b})^2-n+2), 2a\in \mathbb{Q}$ and $\mathbb{Q}$ is field, $\sqrt{2}\in\mathbb{Q}$, but this is a contradiction!

Answer 3

Assume $\sqrt{n} + \sqrt{2} = \frac{a}{b}$ for some $a, b, n \in \mathbb{Z}$ where $n$ is not a perfect square.

You already know that the sum of a rational and an irrational is always irrational. Therefore, $\sqrt{n}$ cannot be rational for this to work. So assume $\sqrt{n}$ is irrational.

With a little rearranging the above, we get:

$$\sqrt{n} = \frac{a}{b} - \sqrt{2}$$

Squaring both sides:

$$n = \frac{a^2}{b^2} - \frac{2a}{b}\sqrt{2} + 2$$

Can you get a contradiction from here?

Answer 4

Hint:

Assume $\sqrt n + \sqrt 2$ is rational. Then there are $p, q \in \Bbb Z$ such that $$ \sqrt n + \sqrt 2 = \frac p q $$

Now try to rearrange and square both sides to look for a fairly blatant contradiction which follows from the fact the sum of a group of rational numbers is always rational.

Answer 5

Hint:

Assume $\sqrt{n}+\sqrt{2}$ is rational. Then $n+2+2\sqrt{2n}$ is rational and so $\sqrt{2n}$ is rational. Assume $\sqrt{2n}=\frac{a}{b}$ with $a,b$ co-prime. Then $2nb^2=a^2$ so $2|a^2$ hence $2|a$. Let $a=2c$, then $2nb^2=4c^2$ so $nb^2=2c^2$. If $2|b$ we have a contradiction since $a,b$ co-prime. So suppose $2|n$. You need to obtain a contradiction.

Answer 6

By a simple 1-line proof: $\,\Bbb Q(\sqrt n\! +\! \sqrt 2) = \Bbb Q(\sqrt n,\sqrt 2),\, $ so $\,\sqrt n\! +\! \sqrt 2\in\Bbb Q\,\Rightarrow\sqrt 2\in \Bbb Q\,\Rightarrow\!\Leftarrow$