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.
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.