I know $\sqrt2$ and $\sqrt3$ are irrational numbers. However, how can I prove that $\sqrt2$ + $\sqrt3$ is an irrational number? Thank you.
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The square of it is irrational. – Eman Yalpsid Oct 02 '16 at 12:59
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4And there's this one too. – Oct 02 '16 at 13:14
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If it were rational, its square would also be rational, so $2+3+2\sqrt{6}$ would be rational and hence $\sqrt{6}$ would be rational.
Nacho Darago
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$$\sqrt2+\sqrt3=\frac pq\implies 2=\left(\frac pq-\sqrt 3\right)^2\implies\frac{3+\frac{p^2}{q^2}-2}{2 \frac pq}=\sqrt 3$$
Can this be?
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If $u=\sqrt 2+\sqrt 3$, we have \begin{align*}u-\sqrt 2=\sqrt 3&\implies u^2+2-2\sqrt 2u=3\iff u^2-1=2\sqrt 2 u\\ &\implies (u^2-1)^2=8u^2 \iff u^4-10u^2+1=0 \end{align*} This is a polynomial with integer coefficients. By the rational roots theorem, if $u=\frac pq$ is a rational root of this polynomial, necessarily $u=\pm 1$. As one can check, none is a root. So $u$ is irrational.
Bernard
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