4

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.

The Chaz 2.0
  • 10,464
lumcti
  • 293
  • 1

    I can see how q < p, but why is q necessarily > 0. Because $q = (2p+2)/(p+2)$ is the ratio of two positive numbers?

     – Dilip Sarwate May 25 '12 at 13:30
  • Look at how it is defined. And note that $p$ is as well. – john w. May 25 '12 at 13:31
  • 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. – lhf Oct 30 '12 at 10:11

2 Answers2

5

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

DonAntonio
  • 211,718
  • 17
  • 136
  • 287
1

Because from your argument we have, $$\displaystyle q= p-\frac{p^{2}-2}{p+2}.$$ $$\displaystyle \therefore q=\frac{p(p+2)-(p^{2}-2)}{p+2}.$$ $$\displaystyle \Rightarrow q= \frac{2p+2}{p+2}>0.(\because p>0).$$

Kns
  • 3,165