# Rational Numbers - LCM and HCF

I was reading a text book and came across the following approach to find the LCM and HCF of rational numbers/fractions:

• LCM of fractions = LCM of numerators/HCF of denominators
• HCF of fractions = HCF of numerators/LCM of denominators

Can someone please help me understand why the above formula holds true or how the above is logically deduced?

• I think you start with the observation that for integers $a, b, k$ we have $\gcd(ka, kb) = k \gcd(a, b)$ and $\text{lcm}(ka, kb) = k \text{lcm}(a, b)$ and then just extend this to rational $k$. You should get the same result as the above. – Qiaochu Yuan Jun 11 '11 at 21:24
• Which textbook is this? – Hans Giebenrath Jul 3 '16 at 5:54

Notation: $\text{HCF}$ is denoted below as $\text{gcd}$.

Assume you have two fractions $\frac{a}{b},\frac{c}{d}$ reduced to lowest terms. Let

$$\begin{eqnarray*} a &=&\underset{i}{\prod } p_{i}^{e_{i}(a)},\qquad b=\underset{i}{\prod } p_{i}^{e_{i}(b)}, \\ c &=&\underset{i}{\prod } p_{i}^{e_{i}(c)},\qquad d=\underset{i}{\prod } p_{i}^{e_{i}(d)}. \end{eqnarray*}$$

be the prime factorizations of the integers $a,b,c$ and $d$. Then

$$\frac{\underset{i}{\prod }\ p_{i}^{\max \left( e_{i}(a),e_{i}(c)\right) }}{% \prod_{i}\ p_{i}^{\min \left( e_{i}(b),e_{i}(d)\right) }}$$

is a fraction which is a common multiple of $\frac{a}{b},\frac{c}{d}$. It is the least one because by the properties of the $\text{lcm}$ and $\gcd$ of two integers, $\prod_{i}\ p_{i}^{\max \left( e_{i}(a),e_{i}(c)\right) }$ is the least common multiple of the numerators and $\prod_{i}\ p_{i}^{\min \left( e_{i}(b),e_{i}(d)\right) }$ is the greatest common divisor of the denominators. Hence

$$\begin{eqnarray*} \text{lcm}\left( \frac{a}{b},\frac{c}{d}\right) &=&\text{lcm}\left( \frac{{\prod_{i}\ p_{i}^{e_{i}(a)}}}{\prod_{i}\ p_{i}^{e_{i}(b)}},\frac{% \prod_{i}\ p_{i}^{e_{i}(c)}}{\prod_{i}\ p_{i}^{e_{i}(d)}}\right)=\frac{\prod_{i}\ p_{i}^{\max \left( e_{i}(a),e_{i}(c)\right) }}{% \prod_{i}\ p_{i}^{\min \left( e_{i}(b),e_{i}(d)\right) }}=\frac{\text{lcm}(a,c)}{\gcd (b,d)}.\quad(1) \end{eqnarray*}$$

Similarly

$$\begin{eqnarray*} \gcd \left( \frac{a}{b},\frac{c}{d}\right) =\gcd \left( \frac{{\prod_{i}\ p_{i}^{e_{i}(a)}}}{\prod_{i}p_{i}^{e_{i}(b)}},\frac{% \prod_{i\ }p_{i}^{e_{i}(c)}}{\prod_{i}p_{i}^{e_{i}(d)}}\right) =\frac{\prod_{i}\ p_{i}^{\min \left( e_{i}(a),e_{i}(c)\right) }}{% \prod_{i}\ p_{i}^{\max \left( e_{i}(b),e_{i}(d)\right) }} =\frac{\gcd (a,c)}{\text{lcm}(b,d)}.\quad(2) \end{eqnarray*}$$

The repeated application of these relations generalizes the result to any finite number of fractions.

Below is a proof that works in any GCD domain, using the universal definitions of GCD, LCM. These ideas go back to Euclid, who defined the greatest common measure of line segments. Nowadays this can also be viewed in terms of fractional ideals or Krull's $v$-ideals.

Theorem $\rm\quad\ \ (a/b,A/B)\: =\: (a,A)/[b,B]\ \$ if $\rm\ \ (a,b) = 1 = (A,B)$
Proof
$\rm\begin{eqnarray} &\rm c &|&\rm a/b,A/B \\ \quad\iff&\rm Bbc &|&\rm aB,Ab \\ \iff&\rm Bbc &|&\rm (aB,Ab) \\ \iff&\rm Bbc &|&\rm (aB, (A,aB)(b,aB))\ \ &\rm by\quad (x,yz) = (x,y(z,x)),\ \ see\ [2] \\ \iff&\rm Bbc &|&\rm (aB, (A,a) (b,B))\ \ &\rm by\quad (a,b) = 1 = (A,B) \\ \iff&\rm Bbc &|&\rm (a,A) (b,B)\ \ &\rm by\quad (A,a)\ |\ a,\ (b,B)\ |\ B \\ \iff&\rm c &|&\rm (a,A)/[b,B]\ \ &\rm by\quad (b,B)\:[b,B] = bB, \ \ see\ [3] \end{eqnarray}$

Here are links to proofs of the gcd laws used: law [2] and law [3].

This is how I understand it,

Let $\dfrac 23$, $\dfrac 56$ and $\dfrac 79$ be three fractions. Let $\dfrac AB$ be their LCM. Now all the fractions must divide $\dfrac AB$, that is, A/B ÷ 2/3, A/B ÷ 5/6 and A/B ÷ 7/9 all must be natural numbers. Or A3/B2, A6/B5 and A9/B7 all must be natural numbers. This requires that A must be divisible by 2,5 and 7 each and B must be a factor of 3,6 and 9 each. In other words A must be a multiple of 2,5 and 7. Now A/B must be lowest possible value which requires us to choose highest possible value of B(HCF of numerators) and lowest possible value of A(LCM of denominators).

Similarly we can understand how to find HCF of a given number of fractions.

Let's try to find LCM of (a/b) and (c/d). If b and d were same it was easy to find LCM since if denominators are same, we just need to find LCM of numerators, hence LCM of (a/b) and (c/b) would be LCM(a,c)/b. So we have to first make denominators of both the fractions same.

So here are the steps to find LCM of a/b and c/d

1. Find the LCM of b and d = LCM(b,d)
2. Multiply numerator and denominator of first fraction by LCM(b,d)/b. Multiply numerator and denominator of first fraction by LCM(b,d)/d. After this multiplication, denominator of both fractions are same.
3. Find LCM of new numerators.
4. The answer is LCM(numerators)/LCM(b,d)

Source : LCM of Fractions