Prove that the least common multiple of two non zero integers is unique. Need to know how to prove the theorem. Thanks.

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    $\begingroup$ Any attempts to this you would like to share ? $\endgroup$ – Aseem Dua Nov 9 '12 at 15:55
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    $\begingroup$ And I need to know what you tried. And many other things ;-) $\endgroup$ – martini Nov 9 '12 at 15:56

There is a theorem which follows from induction called the Well Ordering Principle, which says that any non-empty set of positive integers has a least element. It is easy to show that a least element is unique.

Given two non-zero integers, $m,n$, the set of positive common multiples is non-empty since it contains $|mn|$. So the least common multiple must exist and be unique.


The proof depends on your definition of lcm. If it is defined as the least element of the set of common multiples then the uniqueness is clear, since if $\rm\:m,n\:$ are both lcms then by the leastness of both $\rm\: m\le n,\, n\le m\:\Rightarrow\: m = n.$ Note that the set of common multiples of any finite set is always nonempty, since the product of the elements of the set is a common multiple of each element.

If, instead, you use the universal definition $\rm\: a,b\mid c\iff lcm(a,b)\mid c\:$ then the uniqueness follows similarly using $\rm\:m\mid n,\, n\mid m,\:$ so $\rm\:m = n\:$ if lcms are normalized to be positive (in general domains there may be no way to choose such a normalization, so we can only say that $\rm\:m,n\:$ are associates, i.e. they agree up to a unit factor).

The universal definition is preferred since not only does it work in any domain, but it helps to greatly simply diverse proofs involving gcds and lcms (e.g. here). It is a prototyical example of a point that I often emphasize: uniqueness theorems provide powerful tools for proving equalities.


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