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?
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!