An equivalence.

Question

Where p is a prime number

Prove that: $$(\forall \left( a;b\right) \in \mathbb{Q} ^{2}) ;a+b\sqrt {p}=0\Leftrightarrow a=b=0$$

Notice that :$$\sqrt {p}\not\in \mathbb{Q}$$

Answer 1

Suppose that $a+b\sqrt{p}=0$ for $(a,b)\neq (0,0)$. Then $a=-b\sqrt{p}$. If $b\neq 0$, then $\sqrt{p}=-\frac{a}{b}\in \mathbb{Q}$ which is a contradiction. If $b=0$, then $a=0$ but $(a,b)\neq (0,0)$ which again is a contradiction. The other implication is trivial.