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### Proof Regarding LCM

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Date: 12/05/2001 at 20:49:28
From: James kingsbery
Subject: Proof regarding LCM

Independently of other sources I came up with the idea that given
integers a and b, a*b = GCF(a,b) * LCM(a,b). I have searched the
Internet and have found a couple of sources, including this site, that
confirm my idea, but I cannot flesh out a proof, nor have I been able
to find one. Is there a proof of this equation?

Thanks!
```

```
Date: 12/06/2001 at 10:11:10
From: Doctor Paul
Subject: Re: Proof regarding LCM

First we need to notice a preliminary fact about lcm(a,b) and
gcd(a,b).

Consider the prime factorization of a and b:

a = p_1^a_1 * p_2^a_2 * ... * p_L^a_L

b = p_1^b_1 * p_2^b_2 * ... * p_L^b_L

Notice that in general two numbers won't have the same number of
primes in their prime factorization. So what we're going to do is
allow some of the exponents to be zero to ensure that each number is
written as the product of L prime powers. An example (below) will make
this more clear.

Then gcd(a,b) = the product (big Pi symbol), i = 1..L, of
p_i^(min{a_i,b_i})

So if a = 24 = 2^3 * 3

and if b = 15 = 3 * 5

Then write:

a = 2^3 * 3^1 * 5^0

b = 2^0 * 3^1 * 5^1

gcd(a,b) = 2^0 * 3*1 * 5^0 = 3 (notice I take the minimum of the two
exponents each time)

Similarly,

lcm(a,b) = the product (big Pi symbol), i = 1..L, of p_i^
(max{a_i,b_i})

In our example,

a = 2^3 * 3^1 * 5^0

b = 2^0 * 3^1 * 5^1

so lcm(a,b) = 2^3 * 3^1 * 5*1 = 15 * 8 = 120

Now we can prove the statement:

Recall we had the prime factorization of a and b:

a = p_1^a_1 * p_2^a_2 * ... * p_L^a_L

b = p_1^b_1 * p_2^b_2 * ... * p_L^b_L

Then:

lcm(a,b) * gcd(a,b) = Product, i = 1..L, of

p_i^(min{a_i,b_i}) * p_i^(max{a_i,b_i})

which is equal to:

Product, i = 1..L, of

p_i^[(min{a_i,b_i}) + (max{a_i,b_i})]

Now notice that min{m,n} + max{m,n} = m+n (can you verify this?) for
all real numbers m and n.

So we can write the above product as:

Product, i = 1..L, of

p_i^(a_i + b_i)

which is equal to:

Product, i = 1..L, of

p_i^a_i * p_i^b_i

which is equal to ab.

So we have established that fact that lcm(a,b) * gcd(a,b) = ab

some more.

- Doctor Paul, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Number Theory

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