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This probably involves some very easy algebra, but I am stuck and would appreciate some help. Walter Rudin's Example 1.1 on page 2 of Principles of Mathematical Analysis includes the following observation:

If $p > 0$ and $q = p - \dfrac{p^2 - 2}{p + 2} = \dfrac{2p + 2}{p + 2}$ then

If $p^2 - 2 > 0$ this implies $0 < q < p$.

I can see how $q < p$, but why is $q$ (necessarily) $> 0$?

I'm sure I am missing something obvious.

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    See also http://math.stackexchange.com/questions/141774/choice-of-q-in-baby-rudins-example-1-1 for an explanation of why this $q$ was chosen.2012-10-30

2 Answers 2

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Because p is positive, so also $\,\displaystyle{q:=\frac{2p+2}{p+2}>0}$

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    Thanks, I knew it had to be something obvious.2012-05-25
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Because from your argument we have, $\displaystyle q= p-\frac{p^{2}-2}{p+2}.$ $\displaystyle \therefore q=\frac{p(p+2)-(p^{2}-2)}{p+2}.$ $\displaystyle \Rightarrow q= \frac{2p+2}{p+2}>0.(\because p>0).$

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    So basically, your explanation is that "the result follows from the hypotheses" ??2012-05-25