Proving two sets equal

• Jan 24th 2008, 11:18 PM
sfitz
Proving two sets equal
Here's the problem:

Let L={(x,y) | 2x+3y=6} and L'={a(3,0) + b(0,2) | a+b=1}
Prove that L=L'

What I've done:
I've picked a few examples, so I believe that this is true. I know that I have to prove that L<L' and L'<L to conclude that L=L'.
I just can't seem to get how to go about doing it.

I think i did the second part:
if (m,n) is in L', then (m,n)=a(3,0)+b(0,2) where a+b=1
so, m=3a and n=2b
but a+b=1, so b=1-a
so (m,n)=(3a,2b)=(3a,2(1-a))
so then I tried to prove that this point is in L:
2x+3y=2(3a)+3(2(1-a))=6a+6-6a=6
I think this proves the second half, L'=L. Does it?

For the first part, I have:
(p,q) is in L, so 2p+3q=6
and I didn't know what to do so I tried to work backwards from the conclusion to get an idea
(p,q)=(p,0)+(0+q)=(p/3)(3,0)+(q/2)(0,2)
does p/3 + q/2 =1? but I can't figure out how to get any further.

Any help or hints or ideas or anything would be greatly appreciated!
• Jan 25th 2008, 04:27 AM
sfitz
Well, the first part is to pick a point in L and show that it is in L', and the second part is to pick a point in L' and show that it is in L. That's how I've always learned to prove that one set is equal to another. Does that make sense?
• Jan 25th 2008, 04:41 AM
Plato
Quote:

Originally Posted by sfitz
Well, the first part is to pick a point in L and show that it is in L', and the second part is to pick a point in L' and show that it is in L. That's how I've always learned to prove that one set is equal to another.

That is the correct method.
$\begin{array}{l}
2p + 3q = 6 \\
\frac{p}{3} + \frac{q}{2} = 1 \\
\frac{p}{3}\left( {3,0} \right) + \frac{q}{2}\left( {0,2} \right) = \left( {p,q} \right) \\
\end{array}$
• Jan 25th 2008, 04:44 AM
sfitz
Quote:

Originally Posted by Plato
That is the correct method.
$\begin{array}{l}
2p + 3q = 6 \\
\frac{p}{3} + \frac{q}{2} = 1 \\
\frac{p}{3}\left( {3,0} \right) + \frac{q}{2}\left( {0,2} \right) = \left( {p,q} \right) \\
\end{array}$

I'm not sure how to go from the first line to the second two. What am I missing?
• Jan 25th 2008, 07:26 AM
Plato
Quote:

Originally Posted by sfitz
I'm not sure how to go from the first line to the second two. What am I missing?

You are missing division by 6.