I haven't done arithmetics during the past few years, so I'm filling the gaps before I'm starting out in math in a month, so I have little understanding of the numerics. I've come across such a gap when operating on fractions. From what I can tell, I could pinpoint the issue of the question as to why use the lowest common denominator for applying the addition/subtraction operations to the fractional terms of an expression - at the relationship between the GCD and the LCM, that's fine as I perceive it, I can research further. But the stumbling stone remains the why are we compelled to use the LCM as the LCD? Why no other common multiple?

Here's an example, assume an expression $$\frac{a}{bc}+\frac{d}{be}$$

The LCD is then defined as $lcd(b,c,e):=b\cdot c\cdot e=:lcm(b,c,e)$ as we cannot factorize these variables. Here's a mistake, thanks @J. W. Perry for pointing this out, this is the correct definition I refer to: $$lcd(b,c,e):=\frac{b\cdot c\cdot e}{gcd(c,d,e)}=:lcm(b,c,e)$$

Which is commonly stated as being equivalent to the LCM. Albeit without any reference to why this is the case, as far as I'm currently aware of.

So it appears my question is actually branching into two:

  1. Why $LCD\equiv LCM$, as expanded above, and
  2. Is the multiple in the notion of the LCM a strict concept, i.e., one that requires us to not factorize out the operands? - As I'm trying to concretize in the next paragraph:

Why couldn't we rightfully assume a "correct" (speaking from my perspective: what is acceptable by the faculty, I can't reason yet what is mathematically correct) common divisor to be, say, $\bf{b}$ as in this expression:


I mean, from the trivial examples: $lcm(4,6)=12$ (1) and $lcm(2,3)=6$ (2) where the coefficient of $2$ remains stable, it is obvious that the resulting number in (2) is smaller by the factor by which both arguments were divided. That on one hand, on the other, the result in (2) is wrong compared with (1) if left unmultiplied by that factor and the task is to find the LCM of 4 and 6.

I've got a feeling that there's some well-understood theorem in number theory that would state something like that the denominator in a division of any two operands must be the LCM. If so, I would love to see the argumentation behind it to understand why this is the case, I think the name and source would do it.

So if someone could please shed the light onto that identity, and if possible, also, just tell if there's indeed a link between an arithmetical fraction with an LCD and the numerical notion of the LCM and GCD or that I'm following the wrong or futile trace, I'd appreciate it much. A simple reference to some monograph would also be helpful.

Edit1: An attempt to concretize

I'm afraid I can't clearly explain what I mean, maybe the reason for this actually is my lack of knowledge in this field and maybe the link is more obvious than I think. So let me try to show you how I came up with this question from the specific example that I first solved wrongly:

$$\frac{x-2}{x(x-1)^2 (x+1)}+\frac{2}{x(x-1)(x+1)^2 }$$

The task is to find the lcd here. My first take was divide and multiply--as does J. W. Perry referring to abiessu's comment--the first subterm by $x(x-1)^2$ and the second by $x(x^2-1)$ we could posit that then $x\ne\{-1,0,1\}$ but it's beyond the question. This would bring us to a quasi simplified representation of that fraction:


In the end we've simplified the denominator by introducing a fraction in the numerator. We've found a common denominator. But now:

  1. Is this common denominator the least if considered isolated from the numerator modification?
  2. Is this numerator fractioning acceptable from the point of the total fraction? Or put differently: From what principle do we derive the need to simplify a fraction? Or is it just a convention?

    a) If so, then--as in this case--should we want to simplify straightforwardly only the common denominator as I did that, or

    b) is it a number of operations left undone that is decisive?

    c) Or is it rather a question of an appealing visual elegance?

  3. If we're interested in the lcd, is it always the lcm, and if so why? I tend to believe that this has to do with the theory around the primes as atomic integers.

I do understand that there is a numeral identity between $\frac{1}{4}\text{and}\frac{256}{1024}$ if "converted" to the decimal representation or, well, divided and multiplied by $256$ which doesn't change the relationship in the quotient. Over time I've become addicted to consistent chains of derivation in argumentation, so I'm sort of trying to find it here.

As you see I'm just trying to figure out a principle or a rule that tells us to apply exactly this algorithm of simplification and some rule that states we should simplify (the latter seems more a convention).

That boldface expression seems to me as an intuitive definition of the gcd. If so, it seems indeed to be revolving around these concepts. I'm sorry I can't explain what I'm trying to understand in proper terms. I'll grab a numerics book and see what I can digest at this point. Maybe it will turn out that the link is obvious or has been answered here, I then immediately will mark the answer as accepted. Thanks for all the answers and comment so far!

Edit2: why numerator modification

It appeals to me from what abiessu writes in his comment to J.W.Perry's answer to assume that $$lcm(a,b)=c⋅lcm(a/c,b/c)$$ may be correct since if $lcm(bc,be)=b\cdot lcm(c,e)$ we could as well assume that, the common factor here, $b$, is substituted for the common factor there,$\frac{c}{c}$. This in turn would explain the the numerator expansion. Does it at least sound sensible?

Man, this feels like wading in the dark and hitting your head against concepts you can't grasp. :)

  • 1
    $\begingroup$ One of the simple observations I have is that the most common way to add two fractions is like so: ${a\over b}+{c\over d}={a\over b}{d\over d}+{c\over d}{b\over b}={ad+ bc\over bd}$. This is the generation of all sums of fractions. Then we must consider GCD$(ad+bc, bd)$ in order to reduce the fraction. $\endgroup$
    – abiessu
    Sep 9, 2013 at 21:06
  • 2
    $\begingroup$ The denominator is not always the LCM. Consider sums like $1/2 + 3/2$, or $1/10 + 1/2$. $\endgroup$ Sep 9, 2013 at 21:07
  • $\begingroup$ Completing this thought, LCM$(b,d)$ is the smallest number that is required as a common denominator for ${a\over b}$ and $c\over d$ and thus be able to add the fractions directly. $\endgroup$
    – abiessu
    Sep 9, 2013 at 21:13
  • $\begingroup$ @abiessu: I can follow this far, but why is this so? $\endgroup$
    – user93986
    Sep 9, 2013 at 21:17
  • 2
    $\begingroup$ To add $\frac{a}{b}+\frac{c}{d}$ one needs to put both fractions in the form $A/B+C/B$ for some number $B$. In practice $B$ can be made to be anything that is a multiple of both $b$ and $d$, which means it can be any multiple of their LCM. There is nobody forcing us to use the LCM, it is just better to use on paper because we get smaller numbers that are easier to work with. $\endgroup$
    – anon
    Sep 9, 2013 at 23:02

2 Answers 2


The elegant theorem you are looking for may not exist, and I have read and written alot of number theory.

Correct me if I am wrong. One of the things you are asking is, what necessitates the common denominator when adding fractions so that we can say $\frac{a}{c}+\frac{b}{c}=\frac{a+b}{c}$. The answer is simpler than you think. It has to do with consistency of fractional quantities. In other words $\frac{1}{2}$ and $\frac{1}{3}$ are not consistent fractional quantities, but $\frac{3}{6}$ and $\frac{2}{6}$ are consistent in that they are both portions of sixths.

I see another sort of question, and that is why must one use the least common multiple in the denominator. You do not have to use the least common multiple. Any common multiple will suffice. It is only necessary that we make the two quantities fractionally consistent with each other. After that we can add the two consistent quantities. Finding the least common multiple just ensures that the fraction is reduced. For example, in your comments Abiessu derives the formula $$\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}.$$

His derivation is proper, and that formula will always work. Also note how he derived this by judiciously multiplying each fraction by the number 1. In the first case $\frac{d}{d}=1$, in the latter, $\frac{b}{b}=1$, and so he did not change the entire quantity, he just made the fractions consistent. However, if the greatest common factor (gcf or gcd) of the two fractions is a number other than 1 then the result could be further reducible. When the gcd is not 1, we may sometimes choose to find the lcm so as to prevent having to reduce the result.

Also I see a third question. "lcm=lcd". Think of the lcd as being "the least common multiple in the denominator". This is a mathematical term issue. The world would not be at a loss if we just used the terms lcm and gcf (with respect to two or more numbers), and got rid of the terms lcd and gcd. The "d" part in lcd probably arose in trying to explain to people how to add fractions.

Also note that $$\dfrac{\dfrac{a}{c}+\dfrac{d}{e}}{b}\neq\dfrac{a}{c}+\dfrac{d}{e},$$ except when $b=1$, or $a=d=0$, or when $d=-a$ and c=e ....

Also note that it is not necessarily true that $\operatorname{lcm}(b,c,e)=b \cdot c\cdot e$. The lcm is the smallest multiple of the three numbers $b,c,e$. For example $$\operatorname{lcm}(5,10,15)=30 \neq 5 \cdot 10 \cdot 15.$$

When the greatest common divisor (aka. greatest common factor) of the three numbers is 1, then the least common multiple will be their product.

Also note that you are not following "the wrong or futile trace." When I see someone think with the energy that must be going through you mind to write this question, I think "future mathematician". Just keep exploring, and don't ever let your spirit evaporate.

Enjoy your maths.

  • $\begingroup$ I think it's worth mentioning as well that his example, $\frac{a}{bc}+\frac{d}{be}$, actually has the common denominator $b$LCM$(c,e)$ rather than LCM$(b,c,e)$. $\endgroup$
    – abiessu
    Sep 10, 2013 at 2:10
  • $\begingroup$ @abiessu: thanks, this is a missing block in my reasoning and supports my intuitive assumption of what I call maybe improperly "factoring out." $\endgroup$
    – user93986
    Sep 10, 2013 at 9:34
  • $\begingroup$ Regarding the first part: the need for consistency in fraction quantities is clear. With respect to the second implicit question of mine: do you mean there is nothing compelling us to simplify, so it's rather a convention which as I'd purport arose from either elegance or school exercises? Considering whether $lcm\equiv lcd$: that's a nice insight. Thanks for pointing out the mistake in the formula for $lcm$, fixed it. Thanks for the kudos, I appreciate that. I believe it's math that lends me the energy by evoking my curiosity, if it hadn't I might've never decided to study it.Thank you! $\endgroup$
    – user93986
    Sep 10, 2013 at 9:51
  • $\begingroup$ A partial list of things that might compel one to simplify a fraction are elegance, convention, school exercise, utilitarian purposes (in that future operations are easier), courtesy (for the benefit of a reader), and the list continues. That being said, as a great professor used to quip (he was actually referring to isomorphisms), you calling me Perry does not mean that I am not Joel. Certainly $\frac{2}{4}=\frac{1}{2}$, but if $\frac{2}{4}$ has more meaning, or aids the understanding, you need not feel compelled to simplify. $\endgroup$ Sep 14, 2013 at 1:46

You can use any other multiple for adding fractions. For example, to add $\frac 16+\frac 18$, the LCM is $24$, but you can certainly say $\frac 16+\frac 18=\frac {1\cdot 8}{6 \cdot 8}+\frac {1\cdot 6}{6 \cdot 8}=\frac 8{48}+\frac 6{48}=\frac {14}{48}=\frac 7{24}$. This works perfectly well. The upside is that you have a truly mechanical approach to find a common denominator. The downside is that the numerators are larger than necessary. You are guaranteed to be able to reduce the fraction at the end, but whether you do may be a matter of taste.

  • $\begingroup$ So no principle governing this? Is this liberty also applicable to a general case where you would factorize the lcm in the denominator so that you'd need to divide the numerator by the same amount to prevent a loss of consistency, i.e., change in the ratio? Which could likely be derived from what abiessu suggests in his comment to J.W.Perry's answer as: $lcm(a,b)=c\cdot lcm(a/c, b/c)$. $\endgroup$
    – user93986
    Sep 10, 2013 at 9:59
  • 1
    $\begingroup$ Usually you try to show the values in the most useful way. People often like fractions in lowest terms, as it is easier to comprehend $\frac 12$ than $\frac {4837}{9674}$, but the second is correct. Sometimes it is clearer to leave a fraction unreduced. On this site I have done things like the above, stopping at $\frac {14}{48}$ because I think it makes it clear where the numbers come from. $\endgroup$ Sep 10, 2013 at 13:08

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