Is $\sqrt{2}+\sqrt{3}$ rational or irrational?

Question

I have proven that the sum of any two irrational numbers is not always irrational and that a rational plus an irrational is irrational, however I am not sure how to prove this specific case.

I was thinking along the lines of: $\sqrt{2}+\sqrt{3}=n$ where $n$ is an element of rationals. By squaring both sides

$$(\sqrt{2}+\sqrt{3})^2 = n^2$$ $$7 + 2\sqrt{2}\sqrt{3} = n^2$$ I have proven the sum of a rational and irrational is irrational yet it is too long and complicated to prove that root two and root three make root six. Moreover, it will be an additionally difficult to prove that two times root six is irrational so I was wondering if there was a quicker less complex way to finish this proof?

Answer 1

The novelty in this question is the request for a proof that doesn't involve knowing (or proving) the irrationality of $\sqrt6$. So here's a proof that only assumes the irrationality of $\sqrt2$ and/or $\sqrt3$.

If $\sqrt2+\sqrt3$ is rational, then so is $\sqrt3-\sqrt2$, since $(\sqrt2+\sqrt3)(\sqrt3-\sqrt2)=3-2=1$. But then so is

$$\sqrt2={(\sqrt2+\sqrt3)-(\sqrt3-\sqrt2)\over2}$$

(and also $\sqrt3=((\sqrt2+\sqrt3)+(\sqrt3-\sqrt2))/2$).

Answer 2

after squaring we get $$2\sqrt{6}=\frac{m^2}{n^2}-5$$ the right-hand side is rational, and the left-hand side irrational.

Answer 3

Assume that, $$(\sqrt{2}+\sqrt{3})=\frac{p}{q}$$ Then we have $$(\sqrt{2}+\sqrt{3})=\frac{p}{q}\in\Bbb Q \implies 5+2\sqrt{6} =(\sqrt{2}+\sqrt{3})^2 =\frac{p^2}{q^2}\in\Bbb Q\\\implies \sqrt{6} =\frac 52+\frac{p^2}{2q^2}\in\Bbb Q ~~(Impossible )$$

since $\sqrt 6\not \in\Bbb Q$

Hence $$\sqrt{2}+\sqrt{3}\not \in\Bbb Q$$

Answer 4

You can show that $\sqrt{2}+\sqrt{3}$ is a zero of the polynomial $x^4-10x^2+1$. By the Rational Root Theorem, the only possible rational roots of this polynomial are $1$ and $-1$.

So $\sqrt{2}+\sqrt{3}$ is irrational.

Answer 5

It is enough to generalize a little one proof of $\sqrt{2}\not\in\mathbb{Q}$, namely the following one.

Given a prime $p$ and a positive integer $n$, we may denote as $$\nu_p(n) = \max\{m\in\mathbb{N}: p^m|n\}.$$

Claim: if $\nu_p(n)$ is odd, then $\sqrt{n}\not\in\mathbb{Q}$.

Proof: The assumption $\sqrt{n}=\frac{a}{b}$ with $a,b\in\mathbb{N}^+$ leads to $$ a^2 = n b^2 $$ which is impossible, since $\nu_p(a^2)$ is even while $\nu_p(nb^2)=\nu_p(n)+\nu_p(b^2)$ is odd.

Corollary: if $n$ is a positive integer, $\sqrt{n}\in\mathbb{Q}$ only if $n$ is an integer square.

Corollary: $\sqrt{6}\not\in\mathbb{Q}$.

Corollary: $\sqrt{2}+\sqrt{3}\not\in\mathbb{Q}$, since the assumption $\sqrt{2}+\sqrt{3}\in\mathbb{Q}$ leads to $\sqrt{6}\in\mathbb{Q}$, contradicting the previous point.

Answer 6

Suppose $\sqrt{2}+\sqrt{3} = q$ for some $q\in \mathbb{Q}$. Then $$2+2\sqrt{6}+3 =q^2$$

and so $$\sqrt{6} = {q^2-5\over 2}$$

Since all elementary operation $(x,-,\cdot, :)$ are closed in $\mathbb{Q}$, the right side is rationa and the left is not. A contradiction.