- #1

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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 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?

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 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?

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