# Proof $($by contradiction$)$

can anyone please explain through these? If so, I would really appreciate it. I think one, if not both, are proof by contradiction.

1) Suppose that m and n are negative integers with $$m > n$$. Prove that $$\sqrt{(m^2 + n^2)} \neq −(m + n)$$.

2) Suppose that a and b are rational numbers and $$x^2 −ax+b = 0$$ has two distinct real solutions. Prove that one solution is irrational if and only if the other solution is irrational.

Note; 2 is a Contrapositive as we have not been taught Vieta's as of yet.

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• For $2$ , try to use Vieta's Formula Nov 29, 2019 at 13:28

1) $$m,n<0\implies 2mn>0\implies m^2+2mn+n^2>m^2+n^2\implies -(m+n)>\sqrt{m^2+n^2}.$$
2) By Vieta, if $$x_0$$ is a root, the other is $$x_1=a-x_0$$. With $$a$$ rational, $$x_0,x_1$$ are of the same type.
• @Wheatbread: 1) for negatives, $\sqrt{x^2}=-x$; 2) no, this is a straight proof.