# Proving a function is one-to-one

• Nov 26th 2007, 07:41 PM
hannahs
Proving a function is one-to-one
Problem:

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.

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.
• Nov 26th 2007, 08:14 PM
ThePerfectHacker
Quote:

Originally Posted by hannahs
Problem:

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.

.

Say \$\displaystyle f(a_1/b_1)=f(a_2/b_2)\implies 2^{a_1}3^{b_1}=2^{a_2}3^{b_2}\$ by unique factorization of the integers \$\displaystyle a_1=a_2,b_1=b_2\$.