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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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    $\begingroup$ Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. $\endgroup$ Nov 29, 2019 at 13:26
  • $\begingroup$ For $2$ , try to use Vieta's Formula $\endgroup$ Nov 29, 2019 at 13:28

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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.

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  • $\begingroup$ 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 $\endgroup$ Nov 29, 2019 at 13:50
  • $\begingroup$ @Wheatbread: 1) for negatives, $\sqrt{x^2}=-x$; 2) no, this is a straight proof. $\endgroup$
    – user65203
    Nov 29, 2019 at 13:59

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