An analysis proof(?)

1. Aug 9, 2012

bedi

I'm going to be a math major soon, so I'm trying to learn analysis, proofs and some set theory too. Recently I started to read Rudin's analysis book and there is a question that I tried to solve(prove?). I think my proof is wrong or not complete, could you correct me please? http://pdfcast.org/pdf/proof-5 [Broken]

Last edited by a moderator: May 6, 2017
2. Aug 9, 2012

micromass

Staff Emeritus
I think there's a type. You typed that if x and r are rational, then x+r is irrational. This is of course false. Did you mean to say that if r is rational (and nonzero) and x is irrational, then x+r is irrational??

Let's assume you meant that.

I'm not gonna say your proof is wrong, but it's very weird. First of all, you spend your entire time proving that x+r is in $\mathbb{R}$. I really don't see the point of this.

And in the last line, you say "But since $x\notin \mathbb{Q}$, $x+r\notin\mathbb{Q}$ also". But this sentence is exactly what they asked you to prove!!! You should elaborate a bit on why this is true.

3. Aug 9, 2012

bedi

Ah, yes it's a typo. Thank you very much...

4. Aug 9, 2012

jgens

One quick comment: For set minus, you should write either $\mathbb{R} \setminus \mathbb{Q}$ or $\mathbb{R}-\mathbb{Q}$ instead of $\mathbb{R}/\mathbb{Q}$.

5. Aug 9, 2012

bedi

Yes I know but I'm new to latex and couldn't figure out how to use math tools yet

6. Aug 9, 2012

jgens

I see. Just use the "\setminus" command to get the slash facing the proper way.

7. Aug 9, 2012

mathphys2

You could use the contrapositive. The contrapositive of "A implies B" is "not B implies not A". The contrapositive is logically equivalent to the original statement.
Statement A:
R is rational AND X is irrational

Statement B:
RX is irrational

Proof:
Suppose not B. So RX is rational. Assume (towards contradiction) that A is true. R must be rational also.
We can write:
RX=N1/N2.
R=N3/N4. (Notation: Let "N#" variables be integers.)

So,
X=RX/R
X=(N1/N2)/(N3/N4)=(N1*N4)/(N2*N3).
Thus, X is rational.

This contradicts A. Thus, A is false. We have shown Not B => Not A. Therefore, A => B.

The proof is very similar with R+X. The only real difference is in the italicized region.

Hope this helps.