Prove $\sqrt{2} + \sqrt{5}$ is irrational [duplicate]

Solution 1:

$\sqrt 2=\dfrac 12\left(\sqrt 2+\sqrt 5-\dfrac 3{\sqrt 2+\sqrt 5}\right)\notin\mathbb Q$, hence $\sqrt 2+\sqrt 5\notin \mathbb Q$

Solution 2:

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.

Solution 3:

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.

Solution 4:

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 $\sqrt5$

Solution 5:

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.