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?

Thanks in advance!

  • 2
    $\begingroup$ 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. $\endgroup$ Jun 11, 2011 at 21:24
  • 1
    $\begingroup$ Which textbook is this? $\endgroup$ Jul 3, 2016 at 5:54

4 Answers 4


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

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


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

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


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,$ \dfrac{A}{B} ÷ \dfrac{2}{3},\dfrac{A}{B} ÷ \dfrac{5}{6}$ and $\dfrac{A}{B} ÷ \dfrac{7}{9} $ all must be natural numbers. Or $\dfrac{A\cdot3}{B\cdot2}, \dfrac{A\cdot6}{B\cdot5}$, and $\dfrac{A\cdot9}{B\cdot7}$ 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 $\dfrac{A}{B}$ must be the lowest possible value, which requires us to choose the highest possible value of $B$ (HCF of denominators) and the lowest possible value of A (LCM of numerators).

Similarly, we can understand how to find the HCF of a given 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 $\rm\ \ (a,b) = 1 = (A,B)$
$\rm\begin{eqnarray} &\rm c &|&\rm a/b,A/B \\ \quad\iff&\rm Bbc &|&\rm aB,Ab \\ \iff&\rm Bbc &|&\rm (aB,Ab) \\ \iff&\rm Bbc &|&\rm (aB, (A,aB)(b,aB))\ \ &\rm by\quad (x,yz) = (x,y(z,x)),\ \ see\ [2] \\ \iff&\rm Bbc &|&\rm (aB, (A,a) (b,B))\ \ &\rm by\quad (a,b) = 1 = (A,B) \\ \iff&\rm Bbc &|&\rm (a,A) (b,B)\ \ &\rm by\quad (A,a)\ |\ a,\ (b,B)\ |\ B \\ \iff&\rm c &|&\rm (a,A)/[b,B]\ \ &\rm by\quad (b,B)\:[b,B] = bB, \ \ see\ [3] \end{eqnarray}$

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


Note that this result holds true under the assumption that the fractions are in their simplest forms... Otherwise it might not hold. eg. 8/12, 4/9 LCM is in fact 4/3. But using your method we get 8/3.


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