Addition of Fractions

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 $\gcd \set {b, d}$

$\blacksquare$

Sources

• 1971: George E. Andrews: Number Theory ... (previous) ... (next): $\text {2-2}$ Divisibility: Exercise $7$