(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

a) Prove that [tex]\sqrt{3}, \sqrt{5}, \sqrt{6}[/tex] are irrational. Hint: To treat [tex]\sqrt{3}[/tex], for example, use the fact that every integer is of the form [tex]3n[/tex] or [tex]3n + 1[/tex] or [tex]3n + 2[/tex]. Why doesn't this proof work for [tex]\sqrt{4}[/tex]?

b) Prove that [tex]2 ^ 1/3[/tex] and [tex]3 ^ 1/3[/tex] are irrational.

2. Relevant equations

Can't think of any, really.

3. The attempt at a solution

a) I futzed around with fractions involving [tex]3n[/tex] and [tex]3n + 1[/tex] and so forth but I never really got anywhere. I figure that the proof must have something to do with 3, or [tex]\sqrt{3}^2[/tex], but I can't tell exactly what.

b) I actually got something here: I reasoned that [tex]\sqrt[3]{2} = (\sqrt[3]{2}^2)\sqrt{2}[/tex]. Since we know that [tex]\sqrt{2}[/tex] is irrational, and that (rational) x (irrational) is irrational unless one of the terms is [tex]0[/tex], and that (rational) x (rational) is rational, we can conclude that it is impossible for [tex]\sqrt[3]{2}[/tex] to be rational because if it was, then a rational number ([tex]\sqrt[3]{2}^2[/tex]) multiplied by an irrational number ([tex]\sqrt{2}[/tex]) would equal a rational number, which here cannot be possible. Is this a proof by contradiction?

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# Homework Help: Irrationality Proofs

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