Given a pair of reduced fractions, e.g., `1/30` and `1/42`, or `3/10` and `1/5`, we add them with the LCD Algorithm.

**If the sum reduces, then it will reduce by a factor of the GCD of the denominators.**

#### Example `1/30+1/42`

- Find the LCM

`30=2 * 3 * 5` and `42=2 * 3 * 7`

so the LCM `=2 * 3 * 5 * 7 = 210` and the GCD `=2* 3=6`.
- Modify the first fraction

`210/30=7` so we rewrite `1/30 = (7*1)/(7*30) = 7/210`
- Modify the second fraction

`210/42=5` so `1/42 = (5*1)/(5*42) = 5/210`
- Combine the fractions with the common denominator

`1/30+1/42 = 7/210+5/210 = 12/210`
- Notice if they reduce

`12/210= (12 -: 6)/(210 -: 6) = 2/35`

#### Example `11/30+1/42`

`11/30+1/42 = 77/210+5/210 = 82/210= (82 -: 2)/(210 -: 2) = 41/105`

#### Proof

Applying the LCD algorithm `a/c+b/d=(au+bv)/lcm`

`lcm=LCM(c,d)`

Note that the LCM of two numbers is the product of the highest power of each prime from the prime factorizations of the two numbers.

`gcd=GCD(c,d)`

Note that the GCD is the lowest power of the primes common to the two numbers.

These sets of prime factors are disjoint and complementary to the product of the two numbers, i.e.,

`c* d = lcm * gcd `(1)

Alternatively,

`lcm/c=d/gcd`(2)

To apply the LCD Algorithm we need to find `u` and `v`, the complementary factors of the `lcm` for `c` and `d` respectively.

`u=lcm/c` and `v=lcm/d`(3)

Since `u=lcm/c=d/gcd`, then `u` is made from factors of `d` that are not common to `c`. Similarly, `v` is made from factors of `c` that are not common to `d`. Hence `u` and `v` have no common factors, i.e., `GCD(u,v)=1`.

Notice that `u*v = lcm/c * lcm/d` and `gcd=(c*d)/lcm`

Hence, `lcm = gcd * u * v`

so `a/c+b/d=(au+bv)/(gcd * u * v)`

If this sum is reducible, then there exists an Integer `n` such that

`(au+bv)/n` and `(gcd * u * v)/n`

If `n` divides `u` then n must divide `bv`. However, we know that `u` is relatively prime to `v`. Additionally, since `b` is relatively prime to `d`, `b` is also relatively prime to `u`. Hence `n` can’t divide `bv` and then must not divide `u`,

So if `n` is to be the GCD of the numerator and denominator of the sum, it must divide the `GCD(c,d)`.

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