Solution

Prove that $\sqrt{2}+\sqrt{5}$ is irrational.


We must first show that $\sqrt{2}$ is irrational.

To do this, argue indirectly and assume $\sqrt{2}$ is rational.

But then, we can find integers $m$ and $n$ where $$\sqrt{2} = \frac{m}{n} \quad \textrm{(which is in lowest terms)}$$ Squaring both sides, we find $$\begin{align}2 = \frac{m^2}{n^2} &\Rightarrow 2n^2 = m^2\\ &\Rightarrow 2 \mid m^2\\ &\Rightarrow 2 \mid m\\ &\Rightarrow m = 2k_1 \quad \textrm{for some integer } k_1\end{align}$$ Now, substituting our new expression for $m$ into a previous equation, we find $$\begin{align} 2n^2 = m^2 &\Rightarrow 2n^2 = (2k_1)^2\\ &\Rightarrow 2n^2 = 4{k_1}^2\\ &\Rightarrow n^2 = 2{k_1}^2\\ &\Rightarrow 2 \mid n^2\\ &\Rightarrow 2 \mid n\end{align}$$ But this contradicts the assumption that we could find such a fraction $m/n$ in lowest terms (2 divides both $m$ and $n$).

Hence, our assumption must be rejected, and the opposite must be true: $\sqrt{2}$ is irrational.

Now, we proceed to the main argument. Again, we argue indirectly...

Assume that $\sqrt{2}+\sqrt{5}$ is rational.

But then, we can find new integers $m$ and $n$ where $$\sqrt{2} +\sqrt{5} = \frac{m}{n} \quad \textrm{(which is in lowest terms)}$$

Suppose we eliminate one of the radicals in the above equation, by squaring both sides... $$\begin{align}n\sqrt{2} + n\sqrt{5} &= m\\ n\sqrt{5} &= m - n\sqrt{2}\\ 5n^2 &= (m - n\sqrt{2})^2\\ 5n^2 &= m^2 - 2mn\sqrt{2} + 2n^2\\ \end{align}$$ Now, "solving for $\sqrt{2}$", we find:$$\begin{align}5n^2 + 2mn\sqrt{2} &= m^2 + 2n^2\\ 2mn\sqrt{2} &= m^2 - 3n^2\\\\ \sqrt{2} &= \frac{m^2-3n^2}{2mn} \end{align}$$ But noticing that both the numerator and denominator of the above fraction must be integers, we have just shown $\sqrt{2}$ is rational.

We know that this can't possibly be the case, given our earlier result. As such, our assumption that $\sqrt{2}+\sqrt{5}$ is rational must be rejected, and the opposite must be true: $\sqrt{2}+\sqrt{5}$ is irrational.

QED.