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I need to proof that $\sqrt{3} + \sqrt{2}$ is irrational, without using the fact that an irrational number plus a rational number equals irrational. also, i can't use the rational root theorem. that's why i posted a new question..

thanks for help!


marked as duplicate by Martin R, Jean-Claude Arbaut, user223391, Arpit Kansal, MathOverview Nov 6 '15 at 20:35

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  • $\begingroup$ I saw this answer, couldn't understand it well... @MartinR $\endgroup$ – tanto30 Nov 6 '15 at 20:04

Assume to get contradiction $\sqrt{3} + \sqrt{2}=\frac{p}{q}$ where $p,q\in \mathbb{N}$. $\sqrt{3} + \sqrt{2}=\frac{p}{q} \Rightarrow (\sqrt{3} + \sqrt{2})^2=(\frac{p}{q})^2 \Rightarrow 3+2\sqrt{6}+2=\frac{p^2}{q^2} \Rightarrow \sqrt{6}=\frac{1}{2}(\frac{p^2}{q^2}-5)$. So now we have $\sqrt{3}+\sqrt{2}\not \in \mathbb{Q} \iff \sqrt{6} \not \in \mathbb{Q}$. Let us show $\sqrt{6} \not \in \mathbb{Q}$. Assume $\sqrt{6}=\frac{r}{t}$ where $r,t\in \mathbb{N}$ s.t. $gcd(r,t)=1$. $\sqrt{6}=\frac{r}{t} \Rightarrow 6=\frac{r^2}{t^2} \Rightarrow r^2=6t^2 \Rightarrow 2\mid r \Rightarrow r=2m \Rightarrow 6t^2=(2m)^2=4m^2$. So we get $4 \mid 6t^2$ but we know $6=2\cdot 3$ which implies $2^2 \not \mid 6$ thus we must have $2\mid t^2 \Rightarrow 2\mid t \Rightarrow t=2n$. But now we have $gcd(r,t)=gcd(2m,2n)\geq 2 > 1 \Rightarrow \Leftarrow$


Note that $\sqrt 2=\dfrac 12\left(\sqrt 2+\sqrt 3-\dfrac 1{\sqrt 2+\sqrt 3}\right)\notin\mathbb Q$, hence $\sqrt 2+\sqrt 3\notin \mathbb Q$


Suppose $\sqrt{2} + \sqrt{3}$ is rational, then so is $(\sqrt{2} + \sqrt{3})^2 = 5 + 2 \sqrt{6}$. Hence, $\sqrt{6}$ is rational which is of course not true.

Hence we are done.

  • $\begingroup$ In the proof of $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{3}+\sqrt{2})$ you basically end up showing the desired result. I like your first proof. I am less of a fan of your sledgehammer $\endgroup$ – user45150 Nov 6 '15 at 19:52
  • $\begingroup$ @user45150 I see..I have edited my post.Regards, $\endgroup$ – Arpit Kansal Nov 6 '15 at 19:54

Assume that $\sqrt2 + \sqrt3 =$$ p \over q $$p,q \in \mathbb{Z}$. Then $\sqrt3 = $$p \over q$ $-\sqrt2$.

Squaring gives $\sqrt2 = $$p^2-q^2 \over 2pq$ which is a contradiction since $\sqrt2 \notin \mathbb{Q}$

  • $\begingroup$ This is precisely the same method the accepted answer used in the duplicate question. $\endgroup$ – user236182 Nov 6 '15 at 21:04

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