Proving m + 1/m = Integer When m = 1

[disregardthat]

Homework Statement

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

m + 1/m = integer

Only when m = 1

Homework Equations

Don't know any

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

 

[radou]

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?

 

[D H]

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.

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.

 

[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.

 

[D H]

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

Edited to add:

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.

 

[radou]

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.

 

[D H]

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

 

[radou]

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

Hm, I'd let Jarle decide about that.

 

[Dick]

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.

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.

 

[disregardthat]

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

 

[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?

 

[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?

 

[D H]

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?

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

 

[Dick]

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

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?

 

[D H]

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 $a=b=1$.

 

[Dick]

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 $a=b=1$.

Ooooops. Sorry! Same argument, though.

 

[Jimmy Snyder]

only has solution if $a=b=1$.

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

 

[radou]

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

Boy, what a thread. :tongue2:

 

[Jimmy Snyder]

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

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.

 

[radou]

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.

I was referring to n, but nevermind.

 

[D H]

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

I was assuming that any common factors in a and b have been eliminated.

 

[truewt]

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 $a=b=1$.

$a^2+b^2=nab$ only has solution if $a=b=1$

How do you go about proving this statement?

There are 3 unknowns, and I do know a and b must be integers, so what arguments are required to prove $a=b$

 

[D H]

Solve $a^2+b^2=nab$ for $a$ in terms of $b, n \in \mathbb N$. The equation is quadratic and must yield a positive integral solution. There aren't many values of $n$ that do that.

 

[truewt]

Ah so it's that simple by using quadratic equations.. Sorry have been too out of touch on mathematics, trying to refresh now before entering university.. which is in like 2 years later

 

[Dick]

Solve $a^2+b^2=nab$ for $a$ in terms of $b, n \in \mathbb N$. The equation is quadratic and must yield a positive integral solution. There aren't many values of $n$ that do that.

I really don't see how you would do it that way. Or why? Just use that fact you can take a,b to relatively prime if the problem has a nontrivial solution and arrive at the obvious contradiction.

 

[truewt]

What do you mean Dick? Taking a,b to be relatively prime (as in a and b are co-prime or something?) As in they do not have a common prime factor?

And how do you solve? Why would getting a non-trivial solution result in a contradiction?

If you can't post the solutions here, mind PM-ing me? Thanks.

 

[Dick]

Coprime. Exactly. And don't try to solve it. Just prove it can't be solved except in very special circumstances. As Jarle has undoubtedly gone on to bigger and better problems, let's just spell it out:

i) a/b+b/a=n is equivalent to a^2+b^2=abn.

ii) We can assume the fraction a/b is in lowest terms, so a and b are coprime. The case where one or both of them is 1 is left as an exercise.

iii) Hence there is a prime p that divides a but does not divide b.

iv) p divides a^2 and abn (since it divides a). But b^2=abn-a^2. Hence p divides b since it divides the RHS. Contradiction. Hence there is no such fraction.

 

[truewt]

wow thanks dick :) That's really cool stuff. Didn't thought of that..

BTW Co-prime = no common prime factors right? I hope i didn't remember it incorrectly..

 

[Dick]

wow thanks dick :) That's really cool stuff. Didn't thought of that..

BTW Co-prime = no common prime factors right? I hope i didn't remember it incorrectly..

You remembered it correctly.

 

[disregardthat]

Hey, I am reading this thread two times a day, I am not gone.

I wonder what you mean by this: iv)

p divides a^2 and abn (since it divides a). But b^2=abn-a^2. Hence p divides b since it divides the RHS. Contradiction. Hence there is no such fraction.

What did you show by saying that?

 

[Dick]

I showed that my premise, that there is a lowest terms fraction satisfying a/b+b/a=n leads to the conclusion that fraction is, in fact, not in lowest terms. This is an impossible situation. Hence I must have been wrong in assuming such a fraction exists.

 

[truewt]

I think the initial assumption that $$\frac {a}{b}$$ is in the lowest terms, hence they are co-prime (or rather no common factors), but you've proven that there exists a prime factor p that divides both of them, hence they are not co-prime which is a contradiction.

Anyway why do we even use co-prime when "no common factor" suffices?

EDIT: I can't seem to get the latex to work :(

 

[Dick]

I usually say relatively prime when "no common factor" suffices. Just force of habit, I guess.

 

[disregardthat]

What is RHS? You mentioned it in your proof.

 

[Dick]

RHS='right hand side' (of the equation).