Prove by the contrapositive method, that if $p$ and $q$ are positive real numbers with the property that $\sqrt{pq}$ is not equal to $(p+q)/2$, then $p$ is not equal to $q$.

The proof is easy enough but I am not sure how to structure it logically. To use the contrapositive method instead of proving $A \implies B$, I try to prove $\neg B \implies \neg A$.

So I should work from the fact that $p = q$ and reach $\sqrt{pq} = (p+q)/2$.

However the only way I am able to do this is by algebraically manipulating the conclusion $$\begin{align*} \sqrt{pq} &= (p+q)/2\\ 4pq &= (p+q)^2\\ 4pq &= p^2 + 2pq+q^2\\ 0 &= (p-q)^2 \end{align*}$$ therefore $p = q$.

But I should have been working from this fact and reached the conclusion, though in reality I worked from the conclusion and reached my assumption. My question is whether this approach is mathematically rigorous and if not how should I have proceeded. Thanks.

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    $\begingroup$ In a proof by contraposition, you should not talk about "contradiction". $\endgroup$ – Frunobulax Aug 17 '14 at 12:23

We are trying to prove that

if $\sqrt{pq} \ne (p + q)/2$, then $p \ne q$ (where $p, q$ are positive)

by proving the contrapositive. This means that we must prove the equivalent statement

if $p = q$, then $\sqrt{pq} = (p + q)/2$.

So suppose that $p = q$. We then have $$\begin{align*} \sqrt{pq} &= \sqrt{p^2}\\ &= p\\ &= (p + p)/2\\ &= (p + q)/2 \end{align*}$$ and the statement is proved.

Note that we did not presuppose that $\sqrt{pq} = (p+q)/2$. What you have proven is in fact the inverse of the original statement,

if $\sqrt{pq} = (p + q)/2$, then $p = q$.


Because $p=q$ we have $pq=q^2$ and thus $\sqrt{pq}=q$ as the numbers are positive. Likewise, we get $\sqrt{pq}=p$. If we add these two equations, we have $2\sqrt{pq}=p+q$. Now divide by $2$.

  • $\begingroup$ Thanks, does that mean that my version would not be considered rigorous and has to be rewritten in your way? $\endgroup$ – pseudomarvin Aug 17 '14 at 12:31
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    $\begingroup$ Sorry, forgot to answer that: If you explicitly say that all your manipulations above can be reversed, i.e. that the first and the last line are equivalent, then your proof is certainly correct. However, it is probably not what the person who asked the question had in mind when they specifically mentioned using the contrapositive method. $\endgroup$ – Frunobulax Aug 17 '14 at 12:33
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    $\begingroup$ Or, do it the other way around: If $p=q$, then $p-q=0$. This implies $(p-q)^2=0$. Now work your way backwards through your equations above. Then you've proved that the conclusion you want to have follows from $p=q$. $\endgroup$ – Frunobulax Aug 17 '14 at 12:36
  • $\begingroup$ Okay I think get it, thank you for your time. $\endgroup$ – pseudomarvin Aug 17 '14 at 12:38
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    $\begingroup$ While I think the posted answers are more representative of how one is "supposed" to prove the stated fact, I think your sequence of equations (in the question) is quite clever, especially since they would enable to you write "if and only if" instead of merely "implies". $\endgroup$ – David K Aug 17 '14 at 13:10

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