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

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.
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Brian Charlesworth
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1 year, 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 \(3 < \sqrt{2}+\sqrt{3} < 4\).
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Daniel Liu
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1 year, 9 months ago

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@Daniel Liu
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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
·
1 year, 9 months ago

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TopNewestThis 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 · 1 year, 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 · 1 year, 9 months ago

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In particular, prove that \(a_1, a_2, \ldots , a_n\) are perfect squares. – Daniel Liu · 1 year, 9 months ago

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