# Prove that $\sqrt 2 +\sqrt 3$ is irrational. [duplicate]

Please prove that $\sqrt 2 + \sqrt 3$ is irrational.

One of the proofs I've seen goes:

If $\sqrt 2 +\sqrt 3$ is rational, then consider $(\sqrt 3 +\sqrt 2)(\sqrt 3 -\sqrt 2)=1$, which implies that $\sqrt 3 − \sqrt 2$ is rational. Hence, $\sqrt 3$ would be rational. It is impossible. So $\sqrt 2 +\sqrt 3$ is irrational.

Now how do we know that if $\sqrt 3 -\sqrt 2$ is rational, then $\sqrt 3$ should be rational?

Thank you.

• PLEASE learn to use Mathjax to format your posts. Commented Sep 17, 2013 at 3:39
• I am new to mathematics.I even don't know what is Mathjax.I will be truly obliged.Please help me to learn Mathjax or please at least give me some good link for it. Commented Sep 17, 2013 at 9:34
• Commented Sep 17, 2013 at 9:42
• Prove that $\sqrt 2 + \sqrt 3$ is irrational Commented Sep 17, 2013 at 14:09

As the rationals are closed under addition, if you know $\sqrt 2 + \sqrt 3$ is rational and that $\sqrt 3 - \sqrt 2$ is rational, their sum $2 \sqrt 3$ is rational, then divide by $2$

Added: we can even make it explicit. If $\sqrt 2+\sqrt 3=\frac ab, \sqrt 3-\sqrt 2=\frac ba$ and $\sqrt 3=\frac 12 (\frac ab + \frac ba)$

It's badly phrased. Its phrasing makes it appear that it's saying that if $\sqrt{3}-\sqrt{2}$ is rational, then $\sqrt{3}$ must be rational. But it actually means that if BOTH $\sqrt{3}-\sqrt{2}$ and $\sqrt{3}+\sqrt{2}$ are rational, then so is $\sqrt{3}$. That's because if they're both rational, then their sum is rational. Their sum is $2\sqrt{3}$. It's easy to see that if that's rational, then so is $\sqrt{3}$.

if $$a,b$$ are rational, so is $$a+b$$...

As $$\sqrt{3}-\sqrt{2}$$ and $$\sqrt{3}+\sqrt{2}$$ are rational, so is $$\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}=2\sqrt{3}$$

if $$a,ab$$ are rational, so is $$b$$...

As $$2,2\sqrt{3}$$ are rational so is $$\sqrt{3}$$...

here we have used two statements

if $$a,b$$ are rational, so is $$a+b$$...

if $$a,ab$$ are rational, so is $$b$$...

convince your self that these results can be seen easily.. If not,

for first ststement:

As you can see if $$a=\frac{p}{q},b=\frac{r}{s}$$ then $$a+b=\frac{ps+qr}{qs}$$

and for second statement

suppose $$a=\frac{p}{q}$$ and $$ab=\frac{r}{s}$$ then $$b=\frac{qr}{ps}$$

P.S : I like your idea $$(\sqrt 3 +\sqrt 2)(\sqrt 3 -\sqrt 2)=1$$ to prove irrationality... :)That is the reason I have tried to help you... all the best

This is a solution in the style of "mathematics made difficult".

I'm assuming you know that $\sqrt{2}$ is irrational and of degree $2$ over $\mathbb{Q}$. It is easy to show that $\mathbb{Q}(\sqrt{2},\sqrt{3})= \mathbb{Q}(\sqrt{3}+\sqrt{2})$. If $\sqrt 3 + \sqrt 2$ is rational, then we would have $[\mathbb Q(\sqrt 3 + \sqrt 2): \mathbb Q]=1$, but $$1 = [\mathbb Q(\sqrt 3 + \sqrt 2): \mathbb Q]=[\mathbb Q (\sqrt 3, \sqrt 2):\mathbb Q(\sqrt 2)][\mathbb Q(\sqrt 2): \mathbb Q] \geq 2,$$ a contradiction.

Put $r = \sqrt2 + \sqrt3$. After squaring both sides and substracting five from both sides you have $r^2 - 5 = 2\sqrt2\sqrt3$. Square again to finally obtain $r^4-10r^2+1=0$. Now you know two things:

1. $\sqrt2 + \sqrt3$ is a root of the polynomial $X^4-10X^2+1$.
2. By Gauss' criterion the only possible rational roots of $X^4-10X^2+1$ are $1$ or $-1$

With this you obtain that $\sqrt2 + \sqrt3$ is a nonrational root of $X^4-10X^2+1$.

• Gauss's criterion is overkill here; you could just go with 'if $r$ is rational then $s=\frac12(r^2-5)=\sqrt{6}$ is rational'; then just apply your favorite classical proof of the irrationality of $\sqrt{n}$ for non-square integers $n$ to $\sqrt{6}$. Commented Sep 17, 2013 at 3:54
• I like proofs of Gauss' criterion more than any proof of the irrationality of $\sqrt{n}$ for non-square $n$... :P Commented Sep 17, 2013 at 4:33
• I agree with Pipicito. Gauss's criterion is both easy to prove and a powerful tool to dispose of such problems, rather than disposing them case-by-case, in ad hoc fashion. Commented Sep 17, 2013 at 14:26
• Whether or not Gauss' criterion is overkill, it is still another method that others may not have seen (myself included). So I, for one, appreciated the answer.
– user59083
Commented Sep 17, 2013 at 14:37

An alternate approach once you reach $f(x) = x^4-10x^2+1$ is to note that it's irreducible $\mod 5$, hence it's irreducible over the integers and by Gauss's lemma irreducible over the rationals. It follows all roots are irrational.