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I am giving the following propositions:

$p: a < b$

$q: \frac{a+b}{2} > a$

$r: \frac{a+b}{2} < b$

$a$ and $b$ are real numbers.

I need to show that these are equivalent by proving the following:

$(p \implies q) \land (q \implies r) \land (r \implies p)$

So they are all strong together by the and I assume I should prove each of these implications separately and then show that the conclusions are all equivalent. I don't want the whole question solved for me, but I am looking for a base.

If I look at $p \implies q$, where would I start here? Should I be using a direct proof, contraposition, or a contradiction to best solve this?

ComputerLocus
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2 Answers2

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Use a direct proof for each. Remember you can add or subtract the same thing to both sides of an inequality, or multiply or divide by a positive constant, and the inequality is preserved.

Matthew Leingang
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I'll outline a much more efficient way to structure your proof.

Suppose $a<b$. Then $2a=a+a<a+b<b+b=2b$. Since $2=1+1>0$, we have that $2^{-1}>0$, so that $a<(a+b)\cdot 2^{-1} < b$, which we may write as $a<\frac{a+b}{2}<b$, as desired.

If you absolutely must do all of the proofs by implication (really unnecessary and inefficient), then here is how you may do the first one (I trust you can do the rest): \begin{align} a < b &\Rightarrow a+a < a+b\\[0.5em] &\Rightarrow a < \frac{a+b}{2}. \end{align} The other two implications may be shown similarly, but for the inequality you want to prove...this line of attack seems like a waste of time.

Daniel W. Farlow
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  • Well I don't know if I necessarily need to do this by implication. The question does not specify. However I am pretty new to this, and doing it via implication just seemed like the easier way to do it for me, as the other way seems more complex than I fully understand. – ComputerLocus Feb 07 '15 at 02:39
  • @Fogest Ah. Then definitely use the approach I first gave. Your approach is a major pain in the butt. The only time you would use a series of implications like what you're attempting to do would be to prove a series or chain of inequalities where proving the implications actually saves some work. For example, check out this answer I gave which makes use of your line of reasoning. That was necessary there because of how many inequalities there were. Yours is small though :) – Daniel W. Farlow Feb 07 '15 at 02:42
  • Why is $a+a < a+b$ in this? (regarding the implication) – ComputerLocus Feb 07 '15 at 02:46
  • I would love to use your first approach, but I do not fully understand it, so trying to use it as an answer isn't going to work out well as I won't be able to put it down on paper in a way that makes sense. – ComputerLocus Feb 07 '15 at 02:48
  • @Fogest Why does $a<b$ imply $a+a<a+b$? Well, consider something concrete: $2<3$; thus, $2+2<2+3$. – Daniel W. Farlow Feb 07 '15 at 02:51
  • Okay ,that makes sense, thanks! Would I apply the same concepts to the other two implications, or do I need to use different reasoning? – ComputerLocus Feb 07 '15 at 02:57
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    @Fogest Yes, you can use the same reasoning. Honestly, I would consult my teacher about these things. He/she may be able to clarify some things more easily than I can by being in person. – Daniel W. Farlow Feb 07 '15 at 03:02