Is f(x, y) = 2^x(6^y) on NxN->N 1-to-1?

[dagr8est]

N = non-negative integers

Is f(x, y) = 2^x(6^y) on NxN->N 1-to-1? Someone told me it isn't 1-to-1. All I see is that 2^x(6^y) = 2^(x+y)(3^y) so it is possible that it is not 1-to-1. I can't find an actual example though. Any help is appreciated.

 

[stapel]

For f(x, y) to be one-to-one onto its range (which obviously isn't all of the natural numbers), you must have, for f(a, b) = f(c, d), that a = c and b = d.

. . . . .2^a 6^b = 2^c 6^d

. . . . .2^a 2^b 3^b = 2^c 2^d 3^d

. . . . .2^a+b 3^b = 2^c+d 3^d

Divide through by the right-hand side:

. . . . .2^a+b-c-d 3^b-d = 1

There is no way for powers of 2 to "cancel out" powers of 3, so the only way for this product to equal 1 is for each of the factors to equal 1.

What then can you conclude about the powers a + b - c - d and b - d?

Eliz.