# Homework Help: Help with a Real Analysis Proof

1. Feb 4, 2010

### vyro

1. The problem statement, all variables and given/known data

Prove that 2^n + 3^n is a multiple of 5 for all odd n that exist in the set of natural numbers.

2. Relevant equations

3. The attempt at a solution

Suppose the contrary perhaps and do a proof by contradiction? Perhaps induction?

edit: done, thank you. please look at second proof :)

Last edited: Feb 4, 2010
2. Feb 4, 2010

### vyro

Also, another proof:
1. The problem statement, all variables and given/known data

Prove that for each a that exists in real numbers and n that exists in natural numbers there exists a rational r such that abs(a-r) < 1/n

2. Relevant equations

3. The attempt at a solution

I believe it has something to do with the theorem on density of rationals.

3. Feb 4, 2010

### boboYO

i don't think the first question is a real analysis problem. try working in mod 5. whats 2^n mod 5 and 3^n mod 5 for odd n?

4. Feb 4, 2010

### vyro

yeah, i redid the first one using a proof by induction. thanks for the response though.

5. Feb 4, 2010

### HallsofIvy

Yes, saying that |a- r|= |r- a|< 1/n is the same as saying that -1/n< r- a< 1/n or that a- 1/n< r< a+ 1/n. Since there exist a rational number in any interval, there exist a rational number in the interval (a- 1/n, a+ 1/n).

6. Feb 4, 2010

### vyro

is that valid as a proof though, starting from the conclusion and then reaffirming it?