Proving a function is one-to-one

[hannahs]

Homework Statement

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.

Homework Equations

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.

 

[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?

 

[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!

 

[Office_Shredder]

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 2^a and 3^b are in fact integers

 

[Hurkyl]

Homework Statement

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

 

[Office_Shredder]

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

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