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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.

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Apologies for the misplaced edit (which was supposed to be to my answer). Phone screens are too small to see some things... – Bill Dubuque Nov 23 '12 at 16:22
@Bill: No problem; it led me to catch a typo. – Brian M. Scott Nov 23 '12 at 16:28

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).

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