Proving that $\sqrt{2}+\sqrt{3}$ is irrational

Question

This is from Spivak.

Prove that $\sqrt{2}+\sqrt{3}$ is irrational.

So far, I have this:

If $\sqrt{2}+\sqrt{3}$ is rational, then it can be written as $\frac{p}{q}$ with integral $p, q$ and in lowest terms.

$$\sqrt{2}+\sqrt{3}=\frac{p}{q}$$ $$2\sqrt{6}+5 =\frac{p^2}{q^2}$$ $$(2\sqrt{6}+5)q^2=p^2$$

And that's about where I get stuck. In a similar question (prove that $\sqrt{2}+\sqrt{6}$ is irrational,) I was able to show that both $p,q$ had to be even which is impossible. I obviously can't apply this trick here. Any hints?

Answer 1

$(\sqrt3+\sqrt2)(\sqrt3-\sqrt2)=3-2=1$ is rational, so if the first factor were rational, so would be the second, and their difference $2\sqrt2$. But surely you know that the latter isn't rational.

Answer 2

From your very last step: $$\sqrt{6} = \frac{p^2 - 5q^2}{2q^2}$$ This a rational number- a contradiction, hence the answer.

Answer 3

Ideas:

$$\sqrt2+\sqrt3=\frac pq\implies 3=\frac{p^2}{q^2}-2\frac pq\sqrt2+2\implies$$

$$\sqrt2=\left(\frac{p^2}{q^2}-1\right)\frac q{2p}\in\Bbb Q\;\ldots$$

Answer 4

If $2\sqrt{6} + 5 = p^2/q^2$, then $\displaystyle \sqrt{6} = \frac 1 2 \left(\frac{p^2}{q^2} - 5\right)$, so $\sqrt{6}$ is rational.

Can you take it from there?

Answer 5

Hint $\ \sqrt2 +\sqrt3 = r\in\Bbb Q\,\overset{\large\rm square}\Rightarrow\ 5+2\sqrt 6 = r^2 \,\Rightarrow\ \sqrt 6 = (r^2-5)/2 \in \Bbb Q,\,$ contradiction,

since $\,\sqrt{6} = a/b\,\Rightarrow\, 6b^2 = a^2\,$ has an odd number of $2$'s on the LHS, but an even number on RHS (by uniqueness of prime factorizations).

There are also many other ways, e.g. using the Rational Root Test on its minimal polynomial, or using Bezout, etc see prior posts here.

Answer 6

Let $x=\sqrt2+\sqrt3$. Then this is a solution to the equation $0=x^4-10x^2+1$. By rational root theorem, this equation has no rational solution. Therefore, $x$ must be irrational.

Answer 7

We can solve this problem by contradiction. Suppose that it is rational, so it is a non-zero rational number such as $\frac{a}{b}$.

$(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})=1\rightarrow \sqrt{3}-\sqrt{2}=\frac{1}{\sqrt{3}+\sqrt{2}}=\frac{b}{a}$

So we conclude that $\sqrt{3}-\sqrt{2}$ is a rational number as well.

$\sqrt{3}-\sqrt{2}=\frac{b}{a}\rightarrow 5-2\sqrt{6}=(\frac{b}{a})^2$

$\sqrt{3}+\sqrt{2}=\frac{a}{b}\rightarrow 5+2\sqrt{6}=(\frac{a}{b})^2$

Subtract the former equation from the latter one, we conclude $4\sqrt{6}=(\frac{a}{b})^2-(\frac{b}{a})^2$

From above equation we conclude $\sqrt{6}$ is rational, but we can prove the irrationality of $\sqrt{6}$ quite simple(It has been brought in following) which shows contradiction and hence $\sqrt{2}+\sqrt{3}$ is irrational.

proof for irrationality of $\sqrt{6}$

If it is rational then it is equal to $\frac{p}{q}$ such that $p$ and $q$ are relatively prime natural numbers. then $p^2=6q^2\rightarrow q^2|p^2\rightarrow$ $p$ and $q$ are relatively prime if and only if $q=1$ which implies 6 is a square integer which is a contradiction.