Irrational Numbers

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

#IrrationalNumbers #SquareRoot

Note by Jonathan Hsu
5 years, 9 months ago

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Comments

This can be proved by contradiction. Suppose 2+3\sqrt{2} + \sqrt{3} is rational. Then 12+3=32\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

12((2+3)+(32))=3\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., 2+3\sqrt{2} + \sqrt{3} is irrational.

Note that we can prove 3\sqrt{3} is irrational by contradiction. Suppose that 3=ab\sqrt{3} = \dfrac{a}{b} for positive coprime integers a,b.a,b. Squaring both sides gives us that 3b2=a2,3b^{2} = a^{2}, which by the Fundamental Theorem of Arithmetic implies that 3a.3|a. But if a=3ma = 3m then 3b2=9m2b2=3m2,3b^{2} = 9m^{2} \Longrightarrow b^{2} = 3m^{2}, implying that 3b.3|b. But we had assumed that a,ba,b were coprime, and thus our original assumption that 3\sqrt{3} was rational is false.

Brian Charlesworth - 5 years, 9 months ago

Note that 2,3\sqrt{2},\sqrt{3} are both algebraic integers, so 2+3\sqrt{2}+\sqrt{3} is also an algebraic integer. However, if 2+3\sqrt{2}+\sqrt{3} is rational then that means it is a rational integer, which is clearly false as 3<2+3<43 < \sqrt{2}+\sqrt{3} < 4.

Daniel Liu - 5 years, 9 months ago

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Extension: if a1+a2++an\sqrt{a_1}+\sqrt{a_2}+\cdots +\sqrt{a_n} is rational, prove that it is integer.

In particular, prove that a1,a2,,ana_1, a_2, \ldots , a_n are perfect squares.

Daniel Liu - 5 years, 9 months ago
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