Take the 2-minute tour ×  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?  share|improve this question Why is
that $3b = a$? –  Makoto Kato Sep 27 '12 at 5:10 add comment  4
Answers  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?  share|improve this answer add comment  Let $n
\ge 2$ be an integer. Let $m \ge 2$ be a square fre