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Rudin √2

  • #1
I wrote a big question as regards this question but pretty much made
sense of it all apart from one bit, why divide by (p + 2) in
[itex] q \ = \ p \ - \ \frac{p^2 \ - \ 2}{p \ + \ 2}[/itex]
in the proof on page 2 of Rudin?
(Check it on the https://www.amazon.com/dp/007054235X/?tag=pfamazon01-20 using amazon's search feature).

The goal is to show that there is no rational number satisfying p² = 2. I'm
fine with the (m/n)² = 2 part, i.e. that a number is even if it's divisible by 2
& odd if it's divisible by 2 leaving 1 as a remainder but when he "examines
this situation a little more closely" I just don't see where (p + 2) comes
from, why not just Δp or something, why so specific? In fact, why
even bother dividing at all???


A := {p ∈ Q| p² < 2}
B := {p ∈ Q| p² > 2}

To show that A has no largest element there should be a q in A such that
p < q he invents this equality for q.

[itex] q \ = \ p \ - \ \frac{p^2 \ - \ 2}{p \ + \ 2}[/itex]

Now, the only logic I can see here is that if p² < 2 & we divide by (p + 2)
(for some reason):

[itex] (p^2 \ < \ 2) \ \Rightarrow \ ( \frac{1}{p \ + \ 2} ) \cdot \ (p^2) \ < \ ( \frac{1}{p \ + \ 2} ) \cdot \ (2) [/itex]

[itex]\Rightarrow \ \frac{2}{p \ + \ 2} \ < \ \frac{p^2}{p \ + \ 2} [/itex]

[itex] \Rightarrow \ ( \frac{p^2}{p \ + \ 2}) \ - \ \frac{2}{p \ + \ 2} \ < \ 0 [/itex]

[itex] \Rightarrow \ ( \frac{p^2 \ - \ 2}{p \ + \ 2})[/itex]

I just don't see why dividing matters. Would it not be better to just
not divide at all, wouldn't there still be a gap between p & q due to
the < sign?

If p² < 2 then (p² - 2) < 0 & q = p - (p² - 2) so q = 2 + (p - p²)

Then q - 2 = (p - p²) & that's really unhelpful.

No, it would not be better I guess :redface: But you see what I mean, I'd just like to
know why he divides. I can understand that he does it that way to strictly avoid the
dead end I've come up with when doing no division but I'm hoping there's some deeper
meaning as opposed to just an example of crazy genius at play :tongue: Any thoughts?
 
Last edited by a moderator:

Answers and Replies

  • #2
What Rudin did here was a slight variation of exercises 16 and 17 in chapter 3, page 81 of the same book. Anyways if you just want a quick answer, then the (p+2) denominator is somewhat necessary. For more abstraction, that 2 can be any number higher than √2 for the equation to work. You should be able to figure out why by changing the 2 to √2 in the bottom and seeing what happens.
 

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