Why does the least common denominator work?

Question

Take for instance the following problem. You have two beakers of the same height. One has tick marks that break it into thirds. The other has tick marks that separate it into fourths. The water levels are 1/3 and 1/4 respectively. If I did not know about the concept of LCDs, how would I figure out how much water there is all together? Please walk me through your reasoning.

Note: I understand the need to find a common scale between the two beakers. I don't know how I would find that 12 is the smallest possible common scale, if I had never been introduced to the concept of LCDs/LCMs.

Answer 1

You are looking for numbers say $x$ and $y$ such that: $$x \times \left(\frac{1}{3} + \frac{1}{4}\right)$$ i.e. $$\frac{x}{3} + \frac{x}{4} = y,$$ where $y$ is an integer. Assuming you do not know about LCM, you will try numbers $x = 1, 2, 3, \ldots$ and $x = 12$ will be the first number for which you will get an integer (7 in this case) as an answer. So you have $$12 \times \left(\frac{1}{3} + \frac{1}{4}\right) = 7.$$ Hence $$\frac{1}{3} + \frac{1}{4} = \frac{7}{12}.$$

Continuing this way, we find for $x = 24$, we have $y = 14$, for $x = 36$, we have $y = 21$, etc. And clearly 12 is the least value of $x$ for which $y$ is an integer.

Answer 2

If you are trying to figure out how much water there is total, pour the contents of one beaker into the other. You can then use any sort of interpolation method, such as a permanent marker and a shoelace, to make the scale of the beaker precise enough to measure how much water there is total.

Answer 3

The fractions $\ x = \dfrac{a}b,\ y = \dfrac{c}d\ $ are, by definition, the (unique!) solutions of

the equations $\ bx = a,\ dy = c.\ $ To find their sum $\ x+ y \ $ it suffices to find

some equation $\ k (x + y) = j,\ $ giving the fraction $\, x+y = \dfrac{j}k\ $ for their sum.

By scaling $\ mbx = ma,\ ndy = nc.\ $ They have common coefficient $\,k\,$ of $\,x\,$ and $\,y\,$

whenever $\ \, mb\, =\, k\, = nd,\ $ i.e. whenever $\,k\,$ is a common multiple of $\,b\,$ and $\,d.$

Then $\ k(x+y) = mbx + ndy = ma+nc\ $ so $\ x+y = \dfrac{ma+nc}k = \dfrac{ma}{mb}+\dfrac{nc}{nd}$

Remark $\ $ Conversely, we show that only common multiples are common denominators of $\rm\color{#c00}{reduced}$ fractions. If $\,k\,$ is a common denominator for fractions $\,x = a/b,\ y = c/d, \,$ i.e. $ $ if $\,kx = ka/b\in\Bbb Z,\,$ and $\, ky = kc/d\in\ \Bbb Z\,$ then $\,b\mid ak\,\Rightarrow\,b\mid k,\,$ by $\,\color{#c00}{a,b\ \ \rm coprime}$ and Euclid's Lemma. Similarly $\,d\mid k,\,$ so $\,k\,$ is a common multiple of the (unique) least denominators $\,b,d.\,$ Beware that this converse can fail over more general domains that lack unique factorization.

Answer 4

Because it readily bridges the gap between different measures. If you didn't know about least common denominators, you'd probably wind up rediscovering the concept and wondering why no one else had thought of it before.

Let's say you swap the contents of the two beakers. You find that $$\frac{2}{4} > \frac{1}{3} > \frac{1}{4} > 0.$$ But this doesn't tell you precisely how much more $\frac{1}{3}$ is than $\frac{1}{4}$. It's clear that you need finer tick marks. If you're smart, you won't bother trying fifths. But you might try sixths, in which case you find that $$\frac{3}{6} > \frac{2}{6} > \frac{1}{4} > \frac{1}{6} > 0.$$ So the total content is more than $\frac{3}{6}$ but less than $\frac{4}{6}$. So you need finer tick marks still: $$\frac{6}{12} > \frac{4}{12} > \frac{3}{12} > 0.$$ The denominators are all the same now. Since $\frac{1}{3} = \frac{4}{12}$ and $\frac{1}{4} = \frac{3}{12}$ you can now add up these fractions with ease: $$\frac{4}{12} + \frac{3}{12} = \frac{4 + 3}{12} = \frac{7}{12}.$$

You might figure out that 6 did not work because $6 = 2 \times 3$ yet $4 = 2^2$, and that 12 did work because $12 = 2^2 \times 3$. We could say that 12 brings together 3 and 4.

You might name this concept "Joe's finest tick mark denominator method," write up a pamphlet about it and give it to a lot of people. Then someone comes up to you and tells you that the ancient Egyptians or the ancient Chinese had already figured this out a long time ago.