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Prove the following function is one to one function $$f : \mathbb N\times \mathbb N \to \mathbb N ,\quad f( i , j ) = 2^i3^j$$

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closed as off-topic by egreg, user1337, TZakrevskiy, Stefan Hamcke, anorton Dec 18 '13 at 19:45

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What have you tried for this problem? We need more context to see where you are getting stuck. –  Brian Scholl Dec 18 '13 at 18:10
tried many times but no luck :\ –  user2862699 Dec 18 '13 at 18:12
@user2862699 What did you try in those attempts? –  T. Bongers Dec 18 '13 at 18:12
I've put values for i and j and try substituting –  user2862699 Dec 18 '13 at 18:15

3 Answers 3


$$2^i3^j=2^n3^m\iff 2^{i-n}=3^{m-j}$$

Suppose $\;i\ge n\;$ , then the left side in the last equality above is an integer. Use now the Fundamental Theorem of Arithmetic....

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Do we really even need the Fundamental Theorem of Arithmetic for this? Couldn't we just argue by divisibility? –  Brian Scholl Dec 18 '13 at 18:13
Formally, @BrianScholl? I'm not sure...yet the FTA is rather basic and sometimes it is even covered in high school, so I don't think it is a heavy weapon. But perhaps there's another way to do it formally . –  DonAntonio Dec 18 '13 at 18:15
@BrianScholl: We certainly need something to assure us that $3^{m-j}$ doesn't have an alternate factorization involving a bunch of $2$'s. –  Austin Mohr Dec 18 '13 at 18:16
Precisely my point, @AustinMohr . Thanks. –  DonAntonio Dec 18 '13 at 18:17
still didn't get it :\ –  user2862699 Dec 18 '13 at 18:26

Hint $\ f$ is $1$-$1$ iff $f(a,b) = f(c,d) \Rightarrow (a,b)=(c,d),\ $ i.e. $\ 2^a 3^b = 2^c 3^d\Rightarrow a=c, b=d$

This is an immediate consequence of the uniqueness of prime factorizations. If you can't use that then prove this: if $ m,n$ are odd and $\,2^a m = 2^c n\,$ then $a = c\ $ (hint: cancel $2$'s then use parity). Applying this above, we can cancel $\,2^a = 2^c$ to obtain $\,3^b = 3^d,\,$ so $\,b = d$.

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$$ F(i_1,j_1)= F(i_2,j_2) \implies 2^{i_1}3^{j_1} = 2^{i_2}3^{j_2} \implies \dots \implies i_1=i_2,\,j_1=j_2 .$$

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This is what the OP has to prove, and I can't see how this helps him to do so, even by hint... –  DonAntonio Dec 18 '13 at 18:16
@DonAntonio: the definition of one to one $F(X_1)=F(X_2)\implies X_1=X_2$. –  Mhenni Benghorbal Dec 18 '13 at 18:18
I don' think that was the OP's problem, @Mhenni but how to prove that. –  DonAntonio Dec 18 '13 at 18:19

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