# Homework Help: Prove m + m/1

1. May 20, 2007

### disregardthat

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

Prove that if m is a positive rational number, then:

m + 1/m = integer

Only when m = 1

2. Relevant equations

Don't know any

3. The attempt at a solution

That's the problem, I don't know where to start. I have tryed a few things, but none of them works out for me.

I don't know much of proofs and such, the problem I have is always where shall I start? Do you have any tips?

EDIT: The topics name is not correct, the correct question is in the topic

2. May 20, 2007

If m is a positive rational number, then m = p/q, for some natural numbers p and q. Hence, you have p/q + q/p = integer => p/q & q/p are integers. Now, what does that mean?

3. May 20, 2007

### D H

Staff Emeritus
Just because the sum of two numbers is an integer does not mean the two numbers themselves are integers. This final step needs further justification.

4. May 20, 2007

### disregardthat

I was at that stage:

a/b + b/a = integer

But I couldn't state that this only works when m=1

Is it ok to say: a and b is integers, if a is more that 1, we would have a decimal number in b/a and an integer in a/b, if b is more than one, we would have a decimal number in a/b and an integer in b/a.

Since m must be positive, and that we can't divide by 0, m must be 1 to get an integer.

Last edited: May 20, 2007
5. May 20, 2007

### D H

Staff Emeritus
How would you express a/b+b/a as a fraction? What are the conditions that this fraction be an integer?

Now put this back in the context of the original problem, which is to prove m+1/m is an integer only if m is 1.

6. May 20, 2007

a/b + b/a is not an integer if a doesn't divide b or if b doesn't divide a (or both). So, for a/b + b/a to be an integer, a must divide b, and b must divide a.

7. May 20, 2007

### D H

Staff Emeritus
radou, you just gave out the answer. Not a good idea in the homework section ...

8. May 20, 2007

Hm, I'd let Jarle decide about that.

9. May 20, 2007

### Dick

This is all true, except that the sum of two nonintegers can be a integer. So it's not so clear and I don't think the answer is given away yet. You'll want to clear the fractions in a/b+b/a=1 and assume a and b are relatively prime.

Last edited: May 20, 2007
10. May 21, 2007

### disregardthat

I don't know how to prove it, I tryed, but I can't

Last edited: May 21, 2007
11. May 21, 2007

### Dick

I'm not sure what you mean by 'shortened'. I would write it as a^2+b^2=a*b. Is this relation possible if a and b have no common divisors?

12. May 21, 2007

### HallsofIvy

I'm a bit unclear as to whether the problem is to show that m+ 1/m is an integer or (m+1)/m is an integer only if m= 1.

If it is the first case, I don't think you need to worry about "divisors". Since the difference of two integers is an integer, if m+ 1/m= n, an integer, then 1/m= n- m is an integer. What does that tell you?

13. May 21, 2007

### D H

Staff Emeritus
Halls, it looks like you assumed that m is an integer. It is a rational.

14. May 21, 2007

### Dick

You can assume a and b have no common divisors, right? Otherwise you could just simplify the fraction. This mean one of a or b has a prime divisor that does not divide the other. Suppose p divides a but not b. How many of the terms in a^2+b^2=a*b are divisible by p? How many are not? Is there anything wrong here?

Last edited: May 22, 2007
15. May 21, 2007

### D H

Staff Emeritus
Dick, you arrived at $a/b+b/a=1$ in post #9 and then used that in posts #11 and #14. However, you do not know $a/b+b/a=1$. The given condition is that $a/b+b/a=n$, where $a, b, n \in \mathbb N$. So the problem becomes one of showing that $a^2+b^2=nab$ only has solution if $a=b=1$.

16. May 21, 2007

### Dick

Ooooops. Sorry! Same argument, though.

17. May 21, 2007

### Jimmy Snyder

That should read "only has solution if $a=b$"

Last edited: May 21, 2007
18. May 21, 2007

Nah, it should read "only has solution in the set of integers".

19. May 21, 2007

### Jimmy Snyder

a and b may both be integers and yet not be a solution. a = b is crucial. a and b were chosen from the set of integers.

20. May 21, 2007