Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**iamaditya****Member**- From: Planet Mars
- Registered: 2016-11-15
- Posts: 766

Practice makes a man perfect.

There is no substitute to hard work

All of us do not have equal talents but everybody has equal oppurtunities to build their talents.-APJ Abdul Kalam

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,371

hi iamaditya

When I've got a proof like this I find it helpful to consider an example first.

Let's say a = 48 = 2 x 2 x 2 x 2 x 3

and b = 56 = 2 x 2 x 2 x 7

The hcf = 2 x 2 x 2

and lcm = 2 x 2 x 2 x 2 x 3 x 7

so hcf x lcm = (2 x 2 x 2) x (2 x 2 x 2 x 2 x 3 x 7) = (2 x 2 x 2 x 2 x 3) x (2 x 2 x 2 x 7) = 48 x 56

The common factors occur in both the hcf and lcm and the not-common factors occur in the lcm. So the re-arrangement allows us to pick out one set of common factors together with one set of not-common factors for the first number and what is left is the factors for the other. Here's an attempt to make that rigorous:

Suppose a = hcf x N where N is the not-common factors, and similarly b = hcf x M

Then the lcm = all the common factors once and the not-common factors from both = hcf x N x M

So hcf x lcm = (hcf) x ( hcf x N x M) = (hcf x N) x ( hcf x M) = a x b

Hope that helps,

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

Pages: **1**