# Set Theory Basic Proof, showing two sets are equal

1. Aug 16, 2012

### LawdyLawdy

Hello,
I am trying to teach myself set theory...main problem is, as an engineer, mathematical proofs were never exactly stressed in my curriculum. (Scary, right?)

The problem is stated as follows:

"Prove the following,
{x$\in$Z|for an integer y, x=6y}={x$\in$Z|for integers u and v, x=2u and x=3v}"

Z is the set of all integers.
(let's call the left side of the equation set A and the right side set B)
I can't seem to get my head around the "prose" of proofs.

I understand in order to prove a set A is equivalent to set B I need to show they have the exact same members ( A$\subseteq$B and B$\subseteq$A )
I also realize that if x=6y is in A then x=2(3y) and x=3(2y) is in B if you consider 3y and 2y the integers u and v, respectively.

I guess my question would be how to write it in an "acceptable" way and if the way I am going is the correct direction.(that is important too I suppose :p )

Apologies if this is a bit cluttered, this is my first time posting a math question on a forum and am working on getting the hang of it. Thanks in advance though for any help.

2. Aug 16, 2012

### SteveL27

As you point out, you need to show that

$A\subseteq B$ and $B\subseteq A$

So you need to show two separate things.

To show that $A\subseteq B$, you go to directly to the set theory definition of $A\subseteq B$; which is:

For all x, if $x \in A$ then $x \in B$

So assume x is in A, and show that it must be in B; and then assume x is in B, and show it must be in A.

In general, it's always helpful to go directly to the texbook definitions, rather than trying to overthink the problem. In other words we have an intuitive idea of what $A\subseteq B$ means, but it's more effective to go directly to the definition. This is a common pattern in doing proofs.

Re-reading your post, it looks like you're going in the right direction.

Last edited: Aug 16, 2012
3. Aug 17, 2012

### HallsofIvy

Yes, you are correct. To prove $A= B$, for sets, prove both $A\subseteq B$ and $B\subseteq A$. And to prove each of those, start "if $p\in A$" and use the properties of both A and B to show "therefore $p\in B$"

Let A= {x$\in$Z|for an integer y, x=6y} and B={x$\in$Z|for integers u and v, x=2u and x=3v}. If $p\in A$ then x= 6y for some y. Let u= 3y and v= 2y. Then x= 6y= 2(3y)= 2u and x= 6y= 3(2y)= 3v. Therefore $p\in B$ so $A\subset B$. Now do the other way. That's slightly harder. You will need to use the fact that 2 and 3 are mutually prime.

Last edited by a moderator: Aug 17, 2012