<s>[INST] Proof that $\sqrt[3]{3}$ is irrational Proof that $\sqrt[3]{3}$ (Just need a check)
Let $\sqrt[3]{3}= \frac{a}{b}$ both a,b are integers of course. $\Rightarrow 3=\frac{a^{3}}{b^{3}}$ $\Rightarrow$ $3b^{3}=a^{3}$ $\Rightarrow$ 3b=a $\Rightarrow$ $3b^{3}=27b^{3}$ and this is a contradiction because the cubing function is a 1-1 function. Does this work? 
 [/INST] No, $3b^3=a^3$ does not imply that $a=3b$. In fact if $a=3b$, then $a^3=(3b)^3=3^3b^3=27b^3$. What $3b^3=a^3$ does imply is that $3\mid a^3$, which then further implies that $3\mid a$. Then you have $a=3m$ for some integer $m$, and $3b^3=a^3=(3m)^3=27m^3$. Divide through by $3$ to get $b^3=9m^3$. Now $3\mid b^3$; what does this tell you about $b$? If you had begun by assuming (as you certainly may) that $\frac{a}b$ was in lowest terms, how would this be a contradiction?
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