Slight deviation of proof, would it be correct for integers? Review for exam!

1. Oct 16, 2006

mr_coffee

Hello everyone.

He told us he could of course change the parameters which he will of the proofs we have been working on so I'm testing out some cases but I want to make sure i'm doing it right.

Here is an example of a proof the boook had:
http://suprfile.com/src/1/3qaagxh/Untitled-1%20copy.jpg [Broken]

Now where he has the statement: "If a and b are rational numbers...."
I'm changing that to:
"If a and b are integers...."
and now here is my proof, i think its correct but I have to make sure.

If a and b are integers, b != 0, and r is an irrational number, then a+ br is irrational.

Suppose not. Suppose that a and b are integers, b != 0, and r is an irrational number such that a+br is rational. We must obtain a contradiction.

Since a, b are integers and a + br are rational, a+br = m/n for some integers m, n with n != 0.

Then

a + br = m/n

br = m/n - a
r = (m-an)/bn

where (m-an) and (bn) are integers since m, a, n, and b are integers, and bn is nonzero since b is nonzero. Therefore, r is rational, contradicting that r is irrational.

Thanks!

Last edited by a moderator: May 2, 2017
2. Oct 16, 2006

looks good to me. are you using the book my solow? btw, you you go to UPark?

Last edited: Oct 16, 2006
3. Oct 16, 2006

mr_coffee

I'm using Discrete Mathematics with appplications 3rd by Susanna S. Epp, and yeah i go to Penn State UPark. CSE260 is the class, hah how did ya know.

4. Oct 16, 2006

5. Oct 16, 2006

mr_coffee

yeah this place is pretty massive, well thanks for the help! I seem to post all questions but don't answer any hah. I did do thinkquest along time ago... i'm supprised that site is still up! well ttyl

6. Oct 16, 2006

HallsofIvy

Since all integers are rational numbers, once you had proven it for a, b rational, exactly the same proof must hold for a, b integer. Actually, I wouldn't have writen the rational numbers as fractions, i/j, etc. It is sufficient to use the fact that the rational numbers are closed under addition and multiplication:
Suppose a+ br, a, b rational b non-zero, is rational: a+ br= c where c is rational. Then br= c- a so r= (c-a)(1/b). Since b is a non-zero rational number 1/b is rational and so r= (c-a)(1/b) is rational- contradiction.

7. Oct 16, 2006

mr_coffee

That does seem alot nicer, thanks!