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

I know that $\sqrt{2}$ is irrational by contradiction method of letting square root of $2$ equal to $a/b$ were $a$ and $b$ are integers expressed in their lowest term with $b$ not equal to zero.

user373477
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  • If $ \sqrt{5} + \sqrt{2} \in \mathbb{Q} $, then $ \sqrt{5} - \sqrt{2} = \dfrac{5 - 2}{\sqrt{5} + \sqrt{2}} \in \mathbb{Q} $ as well. Hence, $ 2 \sqrt{2} = (\sqrt{5} + \sqrt{2}) - (\sqrt{5} - \sqrt{2}) \in \mathbb{Q} $, which contradicts what you know about $ \sqrt{2} $. – Transcendental Sep 29 '16 at 10:18

2 Answers2

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$$(\sqrt2+\sqrt5)^2=7+2\sqrt{10}$$

By the same argument as for $\sqrt2$, you show that $\sqrt{10}$ is irrational.

As the square root of an irrational cannot be rational, the claim follows.

  • Notice that the question title was changed. I am answering for $\sqrt2+\sqrt5$. The case of $2+\sqrt5$ is obvious and implies that of $\sqrt{2+\sqrt5}$. –  Sep 29 '16 at 09:58
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Divide your proof into two parts. First show that $\sqrt{5}$ is irrational. Then show that the sum of a rational number (2 in this case) and and an irrational number ($\sqrt{5} $ in this case) cannot be rational. Then use the fact that the square root of an irrational number is irrational.

Since you already proved that $\sqrt{2} $ is irrational, the proof that $\sqrt{5}$ is irrational is similar.Now we show that if $q$ is rational and $r $ is irrational then $r+q $ is irrational.

Suppose that $q+r$ was rational, then $r+q=\frac{m}{n} $ where $m,n\in\mathbb {Z}$ and $gcd (m,n)=1$. Note that since $q$ is rational then it also can be written as $\frac{a}{b}$ where $a,b\in\mathbb {Z}$ and $gcd (a,b)=1$. Then we get $$\frac {m }{n}=q+r=\frac {a}{b}+r$$ and thus $$r=\frac{bm-an}{bn} $$ This means that $r$ is rational (prove this) contradicting our assumption that $r$ is irrational.

Now since $2+\sqrt {5} $ is irrational, then its square root is irrational.

UserA
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