Irrational Numbers

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

Note by Jonathan Hsu
2 years, 9 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.

Brian Charlesworth - 2 years, 9 months ago

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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 \(3 < \sqrt{2}+\sqrt{3} < 4\).

Daniel Liu - 2 years, 9 months ago

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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.

Daniel Liu - 2 years, 9 months ago

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