Analysis Proof: Looking for Help with Rudin's Book

[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

 

[micromass]

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 going to 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.

 

[bedi]

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

 

[jgens]

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

 

[bedi]

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

 

[jgens]

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

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

 

[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 irrationalProof:
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.