Proving/disproving that √7 - √2 is irrational It's been proven that √7 and √2 are irrational.
However, I am not sure how to go about proving that √7 - √2. Is it an acceptable proof to just solve the equation which would prove/disprove the equation or as should the proof be done as a contrapositive, similar to how √7 and √2 are proven to be irrational. 
What would a valid proof/disproof of irrationality look like in this case?
 A: Hint:  suppose $\,\sqrt{7}-\sqrt{2}\,$ were rational, then so would be $\,\dfrac{5}{\sqrt{7}-\sqrt{2}}=\sqrt{7}+\sqrt{2}\,$, then so would be their difference $\,2 \sqrt{2}\,$.

[ EDIT ] Following up on the previous comment: let $\,x=\sqrt{7}-\sqrt{2}\,$, then $\,x^2=9-2\sqrt{14}\,$, then $(x^2-9)^2=4 \cdot 14 \iff x^4 - 18 x^2 + 25 = 0\,$. But the latter equation has no rational roots, since by the rational root theorem the only such roots could be $\,\pm1, \pm5, \pm25\,$ which none work.
A: Suppose
$\sqrt{a}-\sqrt{b}
= r
$
is rational.
Squaring this,
$a+b-2\sqrt{ab} = r^2$,
so
$\sqrt{ab}$ is rational.
If $\sqrt{ab}$ is irrational,
this can not hold.
Therefore,
if $\sqrt{ab}$ is irrational,
so is
$\sqrt{a}-\sqrt{b}$.
Since $\sqrt{14}$ is irrational,
so is
$\sqrt{7}-\sqrt{2}$.
Note that
this works for
$\sqrt{a}+\sqrt{b}$
also.
Note 2:
There are many proofs here that
if $n$ is not a perfect square
then $\sqrt{n}$
is irrational.
A: Suppose $\sqrt{7}-\sqrt{2}$ were rational; that is, suppose 
$$\sqrt{7}-\sqrt{2}=\frac{a}{b},
$$ 
where $\text{gcd}(a,b)=1$. 
Multiply both sides of the equation by $\sqrt{7}+\sqrt{2}$ to obtain 
$$
5=7-2=(\sqrt{7}-\sqrt{2})(\sqrt{7}+\sqrt{2}) = \frac{a}{b}(\sqrt{7}+\sqrt{2}).  
$$
Since $\frac{5b}{a}\in\mathbb{Q}$, $\sqrt{7}+\sqrt{2}$ is also a rational number. 
Since the sum of two rational numbers is rational, 
$$
(\sqrt{7}-\sqrt{2}) + (\sqrt{7}+\sqrt{2}) = 2\sqrt{7} 
$$ 
is rational. So $\sqrt{7}$ is rational. This is a contradiction. 
