Proof $($by contradiction$)$

Question

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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score 0 · Answer 1 · answered Nov 29 '19 at 13:33

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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.

answered Nov 29 '19 at 13:33

Thank you for the quick response. I understand all of the first answer, excluding where the negative comes from in the final form? I'll update the question in a second, but apparently the second one is a Contrapositive. Thank you either way though – Alex Stephens Nov 29 '19 at 13:50
@Wheatbread: 1) for negatives, $\sqrt{x^2}=-x$; 2) no, this is a straight proof. – Nov 29 '19 at 13:59

Proof $($by contradiction$)$

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