\problemdata     
{proofs-6}           
{proofs}              
{}              
{F98, S01 PS-1} 		
{S01 Karger}        	

\begin{problem}A real number $r$ is called {\em sensible} if there
exist positive integers $a$ and $b$ such that $\sqrt{a / b} = r$.  For
example, setting $a = 2$ and $b = 1$ shows that $\sqrt{2}$ is
sensible.  Prove that $\sqrt[3]{2}$ is not sensible.  (Consider only
positive real roots in this problem)
\solution{
The proof is by contradiction.  Assume for the purpose of
contradiction that $\sqrt[3]{2}$ is sensible.  Then there exist
positive integers $a$ and $b$ such that $\sqrt{a / b} = \sqrt[3]{2}$.
Squaring both sides of this equation gives $a / b = \sqrt[3]{4}$,
which implies that $\sqrt[3]{4}$ is rational.

Since $\sqrt[3]{4}$ is rational, we can write it as a fraction $x / y$
in lowest-terms, where $x$ is an integer and $y$ is a positive
integer.  Therefore, we have:

\begin{eqnarray*}
\sqrt[3]{4} & = &   x / y \\
4       & = &   x^3 / y^3 \\
4 y^3       & = &   x^3
\end{eqnarray*}

In the last equation, the left side is even, and so the right side
must be even.  Since $x^3$ is even, $x$ itself must be even.  This
implies that the right side is actually divisible by 8, and so the
left side must also be divisible by 8.  Therefore, $y^3$ is even, and
so $y$ itself must be even.

The fact that both $x$ and $y$ are even contradicts the fact that $x /
y$ is a fraction in lowest terms.  Therefore, $\sqrt[3]{2}$ is not
sensible.  
}
\end{problem}