Analysis Question - irrational and rational numbers - proof

[silimay]

This isn't really a question about homework specifically, it's more just that I don't understand part of my chapter...I am just starting Principles of Mathematical Analysis by Ruben...

Here is what I don't understand:

It is proving that $$p^2 = 2$$ is not satisfied by any rational p. And it says that if there were such a p, we could write $$p = m/n$$ where m and n are integers that are not both even. This is probably a foolish question, but I don't understand that...why can't m and n both be even?

My next question is, further on in the proof, it says, let $$A$$ be the set of all positive rationals p such that $$p^2 < 2$$ and let $$B$$ consist of all positive rationals p such that $$p^2 > 2$$, and that they are going to show that $$A$$ contains no largest number and $$B$$ contains no smallest. For every p in $$A$$ we can find a rational q in $$A$$ such that $$p < q$$ (and similar for B). But then there is this equation:

$$q = p - \frac{p^2 - 2}{p+2} = \frac{2p + 2}{p + 2}$$

I don't understand at all where this equation came from.

Thanks for any help in advance...Sorry if these are stupid questions :)

 

[Hurkyl]

It is proving that $$p^2 = 2$$ is not satisfied by any rational p. And it says that if there were such a p, we could write $$p = m/n$$ where m and n are integers that are not both even. This is probably a foolish question, but I don't understand that...why can't m and n both be even?

They can be even; he never said that they couldn't be. He was making a choice. The reason he made that particular choice is so that he can use the proof he gives.

 

[silimay]

But then they say that if m is even, n must be even, which implies that both m and n are ven. Which means it's impossible for p to be rational. But why can't both m and n be even?

 

[Hurkyl]

He chose m and n so that at least one of them was odd.

Because at least one of them is odd, they cannot both be even.

 

[silimay]

but like...why did he choose them so that at least one was odd? Couldn't there be a $$p=m/n$$ where both were even, where $$p^2 = 2$$?

Sorry that I am so slow in getting it...

 

[Hurkyl]

but like...why did he choose them so that at least one was odd?

Because he's going to invoke that fact somewhere in his proof.

One way to effect this proof is to go on to prove that they both have to be even. This contradicts the fact that they cannot both be even, thus disproving the assumption that the square root of 2 is rational.

From what you said next, it sounds like his proof is going to do something else... but I imagine that the fact that one of them is odd is going to be a crucial step.

 

[HallsofIvy]

Because any rational number can be written as a fraction reduced to lowest terms. In order to avoid questions about reducing, the proof should start "Assume that $\sqrt{2}= \frac{m}{n}$ reduced to lowest terms. Then, of course, you know that m and n can't both be even- arriving at the conclusions that m must be even and n must be even immediately gives a contradiction.

Of course, you could do this without that assumption- but then you would have to do an argument about "infinite descent". After showing that m and n are both even, cancel the 2 in the numerator and denominator to get the fraction m'/n'. Now use the same argument to show that m' and n' are both even. Cancel the 2's in numerator and denominator to get m"/n"- use the same argument to show thata both m" and n" are even, etc.

 

[silimay]

I still don't really understand how that proves it...but that is okay :(

 

[silimay]

Ahhh okay...thanks so much! :)

 

[silimay]

Do you know about the second question, where the equation comes from?

 

[Hurkyl]

Do you know about the second question, where the equation comes from?

I suspect you've either made a typo in your presentation or you've omitted an important part of his proof.

e.g. where did p and q come from?

 

[silimay]

Ah, sorry...
He writes,

For every p in A we can find a rational q in A such that $$p < q$$, and for ever p in B we can find a rational q in B such that $$q < p$$.

And then it goes straight to that equation...It says "We associate with each rational $$p>0$$ the number q = ...

 

[Hurkyl]

Ah, so he's just defining q as a function of p. You don't necessarily need to know how he decided to use that function, just what he can prove using it.

I don't yet understand why he made that choice, but it might be clear if I saw the rest of the proof, or at least the next step or two.

Notice, incidentally, that if p was a square root of 2, then q would equal p.

 

[silimay]

Ahh...okay...so it was just a bit arbitrary, but it got it proved.

Thank you so much :)

 

[Hurkyl]

By the way, once you understand the proof, it might be a good exercise to reverse engineer it. One thing you might do is to suppose that you had the idea that you wanted to do the proof in that way... how would you go about figuring out how to choose q?