# 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.

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## marked as duplicate by MJD, Dennis Gulko, Daniel Robert-Nicoud, Daniel Fischer♦, TomasOct 25 '13 at 23:31

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–  AakashM Sep 17 '13 at 9:42

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

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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 $\sqrt 3=\frac 12 (\frac ab + \frac ba)$

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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 $\sqrt{3}$.

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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 $X^4-10X^2+1$.

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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 $\sqrt{6}$. –  Steven Stadnicki Sep 17 '13 at 3:54
I like proofs of Gauss' criterion more than any proof of the irrationality of $\sqrt{n}$ for non-square $n$... :P –  Pipicito Sep 17 '13 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. –  user43208 Sep 17 '13 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 Sep 17 '13 at 14:37

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.

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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.

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