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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.

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    What does "multiple" mean to you? Specifically with respect to real numbers. – 2017-01-31
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    @svjacob So the (positive) multiples of $\pi$ are precisely $\pi,2\pi,3\pi, 4\pi, \ldots$? But not for example $\pi^2?$. And $4\pi+2$ divided by $\pi$ is $4$ with remainder $2$? And $\pi^2$ divided by $\pi$ is $3$ with remainder $0.1415926\ldots$? – 2017-01-31
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    @svjacob OK. And what does "divide" mean to you, specifically with regards to real numbers? – 2017-01-31
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    @GitGud Deleted my previous comment, here is what I mean: common multiple should mean that dividing by a, should result in b with no remainder, or dividing by b should result in a with no remainder. Or even the same remainder, not necessarily zero remainder. – 2017-01-31

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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!)

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    this seems to be what I was looking for, the part about irrational numbers intrigues me to ask further. Can we find a "c" that is not an integer? Is it possible to represent a frequency at sqrt(2)? Could I have one function run every sqrt(2) seconds and another every 1 second? Or is it impossible to represent a time as an irrational number? – 2017-01-31
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    @svjacob It's not that $c$ itself is an integer, it's that the *ratios* $c\over a$ and $c\over b$ are integers. It is impossible in general to do this: if $a$ is rational and $b$ is irrational, then there is no positive $c$ which is an integer multiple of $a$ and an integer multiple of $b$. Meanwhile, you ask: " Could I have one function run every sqrt(2) seconds and another every 1 second?" You certainly can, but they won't coincide (= run at the same time) more than once. – 2017-01-31
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    hmm thanks. You mention that they won't run at the same time more than "once". So if a is rational, b is irrational, does that mean it is possible to have an irrational number c such that "c/a" and "c/b" is the same irrational value? But it would only happen once? – 2017-01-31
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    @svjacob I was referring to the fact that $c=0$ provides integer ratios, always - this is just reflecting the fact that if I begin two things at the same time, they'll coincide at least at time $t=0$. But if $a$ is rational and $b$ is irrational, they'll never coincide again. Re: ""c/a" and "c/b" is the same irrational value", that will absolutely never happen - if $c\not=0$ and ${c\over a}={c\over b}$, then $a=b$! – 2017-01-31
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    Ah! Makes sense! Thank you, now I just gotta find out if it's possible to program irrational numbers. Which I doubt is possible. – 2017-01-31
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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 $\pi$.)

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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.