Easy Math Proof, help!

[Quad]

How do you prove this:

If GCD(a,b) = 1 then GCD(a+b, ab) = 1 ?

thanks!

 

[TuxDave]

if it's so easy, why are you asking us?

 

[GoingUp]

looks like you need to prove that

(a,b) + (a+b,ab)

 

[her209]

GCD(a,b) = 1 means that a and b are relatively prime.

Given that, we know:

a+b != xa where x is a positive integer
-and-
a+b != yb where y is a positive integer

(rephrase, a+b is not a multiple of a or b)

We can see that GCD(a+b,ab) = 1

 

I suppose this is abstract algebra with the whole cyclic group and generator thing. A little review or introduction before we present the proof:

We know that gcd(a, b) = 1 implies that a and b are relatively prime. By definition, gcd(a, b) = 1 means that 1 = ma + nb, where 1 is the greatest common divisor; m and n are integers (i.e., m, n E Z); and a and b are positive integers (i.e., a, b E Z+). (Sorry I cannot get the character map for "element" onto the text here, so I am using E to denote "an element of" and Z to denote integers. So Z+ means positive integers.)

Now, let's move on to the proof:

Suppose gcd(a, b) = 1. (Then we must show that gcd(a+b, ab) = 1. In other words show that r(a+b) + s(ab) can be expressed as pa+qb, where p,q E Z. Then we can conclude that 1 = r(a+b) + s(ab), where r, s E Z.)

Well, r(a+b) + s(ab) = ra + rb + sab = ra + (r+sa)b (Since b divides both rb and sab.)

(The following are obvious, but in case you don't see it: sa E Z because the product of a positive integer and an integer is an integer. Therefore (r+sa) E Z, since the sum of two integers is an integer.)

So can rewrite (if you wish) ra + (r+sa)b = ra + kb, where k E Z and k = r+sa.

Since it is given that a and b are relatively prime, we can say 1 = ra + kb = r(a+b) + s(ab)

Hence 1 = r(a+b) + s(ab)

QED

You can try testing it and you'll see that it works. You can try a few examples with particular numbers to have a feel of it. You can skip many steps in the proof. I just wanted you to see how I got from one step to another. I hope that helps if it isn?t coming too late.

 

[Quad]

thanks luvly! that proof was very clear

perhaps you might be able to solve this calculus problem:
find the sum from n=2 to infinite, n^2/2^n

i can't get it for the life of me thanks again!