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Irrational numbers can not have LCM. But in one of the books I read and LCM for thr triplet $( \pi/2 , \pi , 3\pi/2)$ was calculated and the answer was $3\pi$. If we can't find LCM for irrational numbers and pi is one,who is the result possible for the problem I read.

Test Test
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    Given a collection ${a_i}$ of real numbers I suppose you can ask for the least (positive) real number $L$ such that $L$ is an integer multiple of each of the $a_i$. Of course, such an $L$ usually won't exist. In the example you give, I can't imagine why the answer wouldn't be $\pi$, however. – lulu Jun 16 '22 at 15:23
  • You can define anything any way you want. The question is what do you want to infer from that definition. – DanielV Jun 16 '22 at 15:44
  • I m very sorry. There has been some mistake in typing the question from my side the values to take LCM of were (pi/2,pi,3pi/2) which justified the answer. Again sorry for the mistake that led to confusion. – Test Test Jun 16 '22 at 17:47

2 Answers2

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Suppose for simplicity that the numbers are positive. Then one can define $\gcd(a,b)$ of real numbers provided the Eclidean algorithm for determining GCD stops atfer a finite number of steps. This is the case if $a$ and $b$ are commensurable.$\def\lcm{\operatorname{lcm}}$

This means $a$ and $b$ have a "common measure", i.e. a finite length that fits into either value without remainder.

Put differently, the quotient $a/b$ must be rational.

$$\text{Euclidean algorithm stops for } a,b \quad\iff\quad \frac ab\in\Bbb Q $$

You can then introduce LCM as $$\begin{align} \lcm(a,b) &= \frac{a\cdot b} {\gcd(a,b)} \tag 1 \\ \lcm(a,b_1,\ldots,b_n) &= \lcm(a, \lcm(b_1,\ldots,b_n)) && \text{ for } n\geqslant 2 \tag 2 \end{align}$$

As an aside, this also relates to regular continued fraction expansion of real numbers: The expansion of $r\neq 0$ has fintely many terms iff $r\in \Bbb Q$, i.e. iff $r$ and $1$ are commensurable.

In your specific case,

$$\gcd(\pi/2, \pi, 3\pi/2)=\frac \pi2$$ $$\lcm(\pi/2, \pi, 3\pi/3)=3\pi$$

emacs drives me nuts
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  • The edit in the question justifies for the answer. But if lcm cant be calculated for irrational no. how a answer comes for pi. Is there some condition about LCM not being calculatable for irrational no. that im missing. – Test Test Jun 16 '22 at 17:49
  • Great explanation though. – Test Test Jun 16 '22 at 17:50
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Yes, we can calculate the LCM of irrational numbers if and only if the ratios between all the pairs of irrational numbers are rational. Basically, LCM is the least common positive integral multiple of the given numbers. In your case, (pi/2)/(pi)=1/2,(pi/2)/(3pi/2)=1/3 and (pi)/(3pi/2)=2/3 are all rational and 3pi is the least common positive integral multiple of (pi/2),pi and (3pi/2).

Bhargav Roy
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    Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. – Community Jun 16 '22 at 16:30
  • Does that mean if the values taken for LCM are similar or relative to one value then it is possible. Had it been (2,pi) could we still get the answer. – Test Test Jun 16 '22 at 17:51
  • Yes, generally the values are commensurable i.e. rational multiples of any one value. For (2,pi), we cannot get LCM as their ratio (2/pi) is irrational. – Bhargav Roy Jun 17 '22 at 01:41