Proving/disproving that √7 - √2 is irrational

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.