Prove $\sqrt{2} + \sqrt{5}$ is irrational How do you prove that $\sqrt{2} + \sqrt{5}$ is irrational?  
I tried to prove it by contradiction and got this equation:  $a^2/b^2 = \sqrt{40}$.
 A: Let  $\sqrt2+\sqrt5=a$ where $a$ is rational
$\implies\sqrt2=a-\sqrt5$
Squaring we get, $$2=a^2+5-2a\sqrt5\iff\sqrt5=\frac{a^2+3}{2a}$$ which is rational unlike $\sqrt5$ 
A: If your arrival at the equation $a^2/b^2=\sqrt{40}$ was correct, you’re done, because the equation $a^4=40b^4$ for integers $a$ and $b$ is a contradiction to the Fundamental Theorem of Arithmetic, which says that the expression of an integer as product of primes can be done in only one way. But your suspect fourth-degree equation has a number of $5$’s on the left that is divisible by $4$, but not so on the right.
A: If $\sqrt{2}+\sqrt{5}\in\mathbb{Q}$, i.e., $ \sqrt{2}+\sqrt{5}=\frac{b}{a}$ for $a,b\in\mathbb{Z} \mbox{ and } a\neq 0$ then $ 2-5=\frac{b}{a}(\sqrt{2}-\sqrt{5})$ and
$$
\sqrt{2}-\sqrt{5} =\frac{-3\cdot a}{b}
$$
Here $b$ is necessarily greater than zero by cause $\sqrt{2}+\sqrt{5}>0$.Finally we get,
$$
\sqrt{2}=\frac{1}{2}\left( (\sqrt{2}+\sqrt{5})+(\sqrt{2}-\sqrt{5}) \right)= \frac{1}{2}\left( \frac{b}{a}+\frac{-3\cdot a}{b} \right)\in\mathbb{Q}.
$$
But this is a contradiction since $\sqrt{2}$ is not rational.
A: $\sqrt 2=\dfrac 12\left(\sqrt 2+\sqrt 5-\dfrac 3{\sqrt 2+\sqrt 5}\right)\notin\mathbb Q$, hence $\sqrt 2+\sqrt 5\notin \mathbb Q$
A: Use proof by contradiction. Assume that the sum is rationial, that is $$\sqrt2 +\sqrt5 = {a\over b}$$ where $a$ and $b$ are integers with $b\neq0$. Now rewrite this as $$\sqrt5={a\over b}-\sqrt2.$$ Squaring both sides of this equation we obtain $$5={a^2\over b^2}-2\sqrt2{a\over b}+2.$$ Now, carefully solve for $\sqrt2$ and obtain $$\sqrt2={-3b\over 2a}+{a\over 2b.}$$ This implies that $\sqrt2$ is a rational number which is a contradiction. Thus $$\sqrt2+\sqrt5$$ is an irrational number.
A: You have $$(\sqrt{2} + \sqrt{5})^2 = 7 + 2\sqrt{10}.$$
The square of a  rational number is rational.  This number,
$7 + 2\sqrt{10}$,  is rational iff $\sqrt{10}$ is rational. The standard argument shows that $\sqrt{10}$ is not rational.  So we are done here.
