So, I was recently asked by a friend about the lcm of two irrational numbers.

As far as I know, mathematically speaking, lcm is generally defined only for positive integers (and sometimes extended to negative integers and also even rationals).

But I have never heard of it being extended to irrationals. But my friend argues that lcm of irrationals is defined (in general).

He gives me an example: $\textrm{lcm}(e,2e)=2e$ where $e$ is the irrational Napier's constant.

I think he is wrong. I still want to hear opinions from others on this. What do you guys think, fellow MSE users?

  • $\begingroup$ I say he's right. Two irrationals may or may not have a common multiple, but if there is a common multiple there is a least one. I don't see any reason not to call the least common multiple the least common multiple... $\endgroup$ – David C. Ullrich Jun 23 '15 at 20:24
  • $\begingroup$ @DavidC.Ullrich, I'm speaking strictly mathematically. If you check out the exact mathematical definition of lcm given in Wolfram Mathworld, you'll see that it gives no scope for lcm of irrationals to be defined. Even W|A can't calculate lcm(e,2e). $\endgroup$ – lcm Jun 23 '15 at 20:26
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    $\begingroup$ @lcm And exactly how do you determine whether a given source is reliable? (That's "you", not "one" - how do you do this?) If you regard Mathworld as reliable you're going to get a lot of "nonstandard" views on mathematics. What's especially problematic is assuming that it's reliable in the sense of being complete - if something's not there it doesn't exist. There is no source that's reliable in that sense. Not that it's directly relevant, but look here en.wikipedia.org/wiki/… and note that what it says there is not on Mathworld. $\endgroup$ – David C. Ullrich Jun 23 '15 at 20:50
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    $\begingroup$ The concept of LCM is defined with respect to a given commutative ring (see Wikipedia for instance). So for this question to be answerable, you need to tell us which ring you consider your numbers to be elements of. $\mathbb{R}$? $\mathbb{Z}[e]$? Something else? $\endgroup$ – Hans Lundmark Jun 23 '15 at 20:52
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    $\begingroup$ @HansLundmark: i think it is fair to say that there is only likely candidate for the ring in question: $\mathbb{Z}[\xi, \eta]$ where $\xi$ and $\eta$ are the numbers whose l.c.m. is at issue. $\endgroup$ – Rob Arthan Jun 23 '15 at 22:28

The ancient Greeks knew about this. Finding a least common (integer) multiple of two positive numbers is essentially the same problem as finding a greatest common divisor (if $\zeta$ is a g.c.d of $\xi$ and $\eta$, then $(\xi\eta)/\zeta$ is an l.c.m.). The Greeks understood that greatest common divisors exist for some $\xi$ and $\eta$ but not for all $\xi$ and $\eta$. They called $\xi$ and $\eta$ commensurable if they have a greatest common divisor: we would say that $\xi$ and $\eta$ are commensurable if $\xi/\eta$ is rational.

Euclid's algorithm works perfectly well for two arbitrary positive real numbers $\xi$ and $\eta$. From a modern point of view, you carry out the arithmetic in the ring $\mathbb{Z}[\xi, \eta]$. If the algorithm terminates, it finds the greatest common divisor: the largest $\zeta$ such that $\xi$ and $\eta$ are both integer multiples of $\zeta$.

What ought to be much better known is that the Greeks studied the case when the algorithm does not terminate: Euclid's Proposition X.2 gives non-termination of the algorithm as a necessary and sufficient condition for the irrationality of $\xi/\eta$.


I would write about an l.c.m. of two irrational numbers only if either (1) I explicitly say what I mean by the term rather than assuming the reader knows, or (2) I'm writing in some context in which it is appropriate to expect the reader to think about what the term ought to mean.

The smallest number that can be written as $n\Big(4\sqrt2\Big)=m\Big(6\sqrt2\Big)$ where $n$ and $m$ are positive integers is $12\sqrt2$, so one could say that is the l.c.m. of $4\sqrt2$ and $6\sqrt2$. One can write a sensible definition of l.c.m. so that the question of what is $\operatorname{lcm}\left(4\sqrt2,6\sqrt2\right)$ makes sense, but I would heed what the paragraph above says.

Note, however, that in that sense, $\operatorname{lcm}(a,b)$ will exist only if $a/b$ is rational and positive.

  • $\begingroup$ ??? There is no smallest number that can be written as $n(4\sqrt{2})+m(6\sqrt{2})$ where $n$ and $m$ are integers. If $n$ and $m$ are required to be positive integers the smallest such number is not $12\sqrt{2}$. The smallest number that can be written as $n(4\sqrt{2})$ and can also be written as $m(6\sqrt{2})$ is $12\sqrt{2}$. That's why it's the lcm - it is the smallest common multiple. $\endgroup$ – David C. Ullrich Jun 23 '15 at 20:41
  • $\begingroup$ @DavidC.Ullrich : Sorry --- I've rephrased it. I had "$+$" where I meant "$=$". Also, I should of course have said "positive" (otherwise the value sought would be $0$). ${}\qquad{}$ $\endgroup$ – Michael Hardy Jun 23 '15 at 20:43

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