Addition of Fractions

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Theorem

Let $a, b, c, d \in \Z$ such that $b d \ne 0$.

Then:

$\dfrac a b + \dfrac c d = \dfrac {a D + B c} {\lcm \set {b, d} }$

where:

$B = \dfrac b {\gcd \set {b, d} }$
$D = \dfrac d {\gcd \set {b, d} }$
$\lcm$ denotes lowest common multiple
$\gcd$ denotes greatest common divisor.


Proof

\(\ds \dfrac a b + \dfrac c d\) \(=\) \(\ds \dfrac {a d} {b d} + \dfrac {b c} {b d}\)
\(\ds \) \(=\) \(\ds \dfrac {a d + b c} {b d}\)
\(\ds \) \(=\) \(\ds \dfrac {a d + b c} {\gcd \set {b, d} \lcm \set {b, d} }\) Product of GCD and LCM
\(\ds \) \(=\) \(\ds \dfrac {a D \gcd \set {b, d} + B \gcd \set {b, d} c} {\gcd \set {b, d} \lcm \set {b, d} }\) substituting for $b$ and $d$ as defined above
\(\ds \) \(=\) \(\ds \dfrac {a D + B c} {\lcm \set {b, d} }\) dividing top and bottom by $\gcd \set {b, d}$

$\blacksquare$


Examples

Example: $\frac 1 2 + \frac 1 3$

$\dfrac 1 2 + \dfrac 1 3 = \dfrac 5 6$


Sources

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