Proof that two positive real or natural numbers always have an LCM?

Question

I was working on an asynchronous program that had two functions that run at certain frequencies. In software, there are other easier ways to make sure they don't run at the same time when they do overlap.

But it got me thinking, is it possible for two positive (real or natural) numbers to not have a common multiple?

Or maybe there is a proof that shows any two real or natural positive numbers do have at least one common multiple to resolve my inquisition.

Answer 1

Any two positive integers $a, b$ have a least common multiple: this is because $ab$ is a common multiple, so the set of common multiples is nonempty, and any nonempty set of positive integers has a least element (by induction).

For reals, things get messy depending what you mean. Unlike integers, reals make division trivial: if $a, b$ are (positive) reals, then ${a\over b}$ is also a real. That is, any positive real is a positive real multiple of any other positive real, so "divisibility" in the context of the reals is trivial.

That said, I suspect that what you mean by "common multiple" of two positive reals $a, b$ is some positive real $c$ such that the ratios $c\over a$, $c\over b$ are both integers. In this case, the answer is no: such a $c$ need not exist. E.g. take $a=1$ and $b=\sqrt{2}$; since $a$ is rational but $b$ is irrational, no such $c$ can exist.

(Why not? Well, $c$ would have to be an integer multiple of $a$, hence $c$ is an integer itself; but no nonzero integer multiple of an irrational is an integer!)

Answer 2

But it got me thinking, is it possible for two positive (real or natural) numbers to not have a common multiple?

No. If two numbers don't have an LCM, then they necessary have no common multiples. But two numbers (real, natural, etc.) always have at least one common multiple, i.e., their product. (Note that for real numbers we'll need to be clear that by "multiple" we don't mean "integer multiple." So, for example, $e\pi$ in this context is a multiple of both $e$ and $\pi$.)

Answer 3

The proof looks something like this. . .

Suppose $X$ and $Y$ are two positive whole numbers. Say $X$ is the smaller.

We're looking for the smallest number they both divide into.

Well, there's certainly at least one number they both divide into, namely $XY$.

Now consider the set of all numbers smaller than $XY$ that $X$ and $Y$ both divide into.

There are only finitely many such numbers. One of them has to be the smallest number that $X$ and $Y$ both divide into.

That number is the lcm.