Prove that a pair of irrational numbers is the solution to a quadratic polynomial. Suppose a, b are two irrational numbers such that ab is rational and a+b is rational. Then a, b are the solution to a quadratic polynomial with integer coeffecients.

Suppose $a$, $b$ are two irrational numbers such that $ab$ is rational and $a+b$ is rational. Then $a$, $b$ are the solution to a quadratic polynomial with integer coefficients.
Proof:
Because $a+b$ and $ab$ are rational,  from the definition
of a rational number we can write
$$a+b = \frac{m}{n}, \quad\mbox{and}\quad ab = \frac{p}{q},$$
where $m$, $n$, $p$ and $q$ are integers.  Now we can
write
\begin{eqnarray*}
(x-a)(x-b) &=& x^2 -(a+b)x + ab\\
&=& x^2 - \frac{m}{n} x + \frac{p}{q} \\
&=& \frac{1}{nq}\left(nq\, x^2 - mq\, x + np\right).  
\end{eqnarray*}
So we have  a quadratic 
polynomial with integer
coefficients ($nq$, $mq$ and $np$), and obviously 
the roots of this polynomial
are $a$ and $b$. Therefore 
we have the proof.
