Show that if a and b are positive integers then ab lcma b

Show that if a and b are positive integers, then ab = lcm(a, b) · gcd(a, b).

Theorem: lcm(a, b) × gcd(a, b) = ab for any positive integers a, b.

Proof: First a

Lemma: If m > 0, lcm (ma, mb) = m × lcm (a, b).

Since lcm(ma, mb) is a multiple of ma, which is a multiple of m, we have m | lcm (ma, mb).

Let mh1 = lcm(ma, mb), and set h2 = lcm(a, b).

Then ma | mh1 a | h1 and mb | mh1 b | h1 .

That says h1 is a common multiple of a and b; but h2 is the least common multiple,

so h1 h2. (1)

Next, a | h2 am | mh2 and b | h2 bm | mh2.

Since mh2 is a common multiple of ma and mb, and mh1 = lcm(ma, mb), we have mh2 mh1,

i.e. h2 h1. (2)

From (1) and (2),

h1 = h2.

Therefore, lcm(ma, mb) = mh1 = mh2 = m × lcm(a, b); proving the Lemma.

Conclusion of Proof of Theorem:

Let g = gcd(a, b). Since g | a, g | b, let a = gc and b = gd.

From a result in the text, gcd(c, d) = gcd(a/g, b/g) = 1.

Now we will prove that lcm(c, d) = cd. (3)

Since c | lcm(c, d), let lcm(c, d) = kc.

Since d | kc and gcd(c, d) = 1, d | k and so dc kc.

However, kc is the least common multiple and dc is a common multiple, so kc dc.

Hence kc = dc, i.e. lcm(c, d) = cd.

Finally, using the Lemma and (3),

we have: lcm(a, b) x gcd(a, b) = lcm(gc, gd) x g = g x lcm(c, d) × g = gcdg = (gc)(gd) = ab.