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# Irrational Numbers

Prove that the $$\sqrt{2} + \sqrt{3}$$ is irrational.

Note by Jonathan Hsu
2 years, 5 months ago

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This can be proved by contradiction. Suppose $$\sqrt{2} + \sqrt{3}$$ is rational. Then $$\dfrac{1}{\sqrt{2} + \sqrt{3}} = \sqrt{3} - \sqrt{2}$$ is rational as well. Now since the sum of rational numbers is always rational, we would then have that

$$\dfrac{1}{2}((\sqrt{2} + \sqrt{3}) + (\sqrt{3} - \sqrt{2})) = \sqrt{3}$$

is rational, which is not the case. So our original assumption must be false, i.e., $$\sqrt{2} + \sqrt{3}$$ is irrational.

Note that we can prove $$\sqrt{3}$$ is irrational by contradiction. Suppose that $$\sqrt{3} = \dfrac{a}{b}$$ for positive coprime integers $$a,b.$$ Squaring both sides gives us that $$3b^{2} = a^{2},$$ which by the Fundamental Theorem of Arithmetic implies that $$3|a.$$ But if $$a = 3m$$ then $$3b^{2} = 9m^{2} \Longrightarrow b^{2} = 3m^{2},$$ implying that $$3|b.$$ But we had assumed that $$a,b$$ were coprime, and thus our original assumption that $$\sqrt{3}$$ was rational is false.

- 2 years, 5 months ago

Note that $$\sqrt{2},\sqrt{3}$$ are both algebraic integers, so $$\sqrt{2}+\sqrt{3}$$ is also an algebraic integer. However, if $$\sqrt{2}+\sqrt{3}$$ is rational then that means it is a rational integer, which is clearly false as $$3 < \sqrt{2}+\sqrt{3} < 4$$.

- 2 years, 5 months ago

Extension: if $$\sqrt{a_1}+\sqrt{a_2}+\cdots +\sqrt{a_n}$$ is rational, prove that it is integer.

In particular, prove that $$a_1, a_2, \ldots , a_n$$ are perfect squares.

- 2 years, 5 months ago