I'm having trouble taking this test, I apologize if the question is duplicated and I appreciate any indication of content. My math base is not that good, so please be clear. If $0 < m < n$, can I prove:
$$m < n \Rightarrow m^2 < n^2.$$
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2This is false. Take $m = -1$ and $n = 0$. – balddraz Nov 13 '20 at 02:46
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Sorry, 0 < m < n. I will editing – William Teixeira Nov 13 '20 at 02:47
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Observe $n^2 - m^2 = (n-m) (n+m) \gt 0$. – Atul Anurag Sharma Nov 13 '20 at 02:50
3 Answers
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I will assume $n > m >0$ because it may not be true if $m$ or $n$ are negative. Let $n = m + \epsilon$, where $\epsilon$ is positive. Then, $n^{2} =(m + \epsilon)^{2} = m^{2} + 2\epsilon m + \epsilon^{2} > m^{2}$.
This is a way of proving that $x^{2}$ is increasing on positive $x$, which may or may not be assumable in your context.
Joshua Wang
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Under the assumption $0<m<n$, we have $m^2<n^2\Leftrightarrow 0<n^2-m^2$.
$n^2-m^2=(n-m)(n+m)$
Now $n-m>0$ by assumption and $n+m>0$. So the product $n^2-m^2>0$.
Cornman
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I had some problems to understand your reasoning, sorry. But I managed to understand after I took the test, thank you very much for your help. – William Teixeira Nov 13 '20 at 03:03