Proving a function is one-to-one

1. Nov 26, 2007

hannahs

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

Prove that the function f: Q ---> R, f(a/b) = (2^a)(3^b) is one to one, assuming that gcd(a,b) = 1, that is the fraction a/b is reduced.

2. Relevant equations

3. The attempt at a solution

I know f is one to one if it never maps 2 different elements to the same place, i.e. f(a) does not equal f(b) whenever a does not equal b.

I have tried looking at how other problems are proved to be one to one and am attempting to do it in a similar way:

Suppose f(a/b) = f(c/d)
then (2^a)(3^b) = (2^c)(3^d)

I'm not sure what to do though. I don't think I'm doing this right. Please help me with this problem.

2. Nov 26, 2007

EnumaElish

You're on the right track. Now suppose a/b is not equal to c/d. Can (2^a)(3^b) = (2^c)(3^d) still be the case?

3. Nov 26, 2007

hannahs

Well if a/b does not equal c/d then the equality does not hold. I think that a/b must equal c/d but I'm not sure how to really write it formally as a proof. Also,I just am not sure that this problem is so simple though. What does gcd(a,b) = 1 have to do with the problem?

Nevermind, I got it now. Thanks for your help!

Last edited: Nov 27, 2007
4. Nov 27, 2007

Office_Shredder

Staff Emeritus
gcd(a,b)=1 iff the fraction is reduced (i.e. there is no common multiple of a and b that you can cancel).

Note that 2 and 3 are primes, and the fundamental theorem of arithmetic tells us something about primes being multiplied together.

As an aside, this is quite a nice way of showing the set of rationals is countable, as 2a and 3b are in fact integers

5. Nov 27, 2007

Hurkyl

Staff Emeritus
That's not well defined. For example:
6 = f(1/1) = f(-1/-1) = 1/6

6. Nov 27, 2007

Office_Shredder

Staff Emeritus
That's a good point... usually for stuff like this it's assumed the denominator is positive I think

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook