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
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
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).