# induction on injection function

Prove $$T:{N}^2\longrightarrow N$$ such that $$T(x,y)={2}^x {3}^y$$ is an one-to-one function

which method should i use on this proof? Induction maybe? Thanks

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No induction is needed; just check the definition of one-to-one. Suppose that $T(x,y)=T(u,v)$, and show that this implies that $x=u$ and $y=v$. If $T(x,y)=T(u,v)$, then $2^x3^y=2^u3^v$; now use the fundamental theorem of arithmetic.
One could use uniqueness of prime factorization (fundamental theorem of arithmetic), but that is an awfully big sledghammer to crack this little chestnut. Suppose $\rm\:2^i3^j = 2^I 3^J\:$ with $\rm\:i\le I.\:$ Then $\rm\:3^j = 2^{I-i}3^J\:$ so $\rm\:I = i\:$ else LHS is odd, but RHS is even. Cancelling $\rm\:2^i = 2^I\:$ yields $\rm\:3^j = 3^J\:$ so $\rm\:j = J.$
Remark $\$ In fact we can replace $2,3$ by any coprime integers $\rm\:a,b > 1\:$ (or, more generally, in any ring, any nonunit cancellable coprime elements). Hence the result essentially depends on cancellation and coprimality (not uniqueness of factorization).