So this isn't really a homework question and I'm sorry if I posted in the wrong section. Please feel free to move it if needed.
I'm doing a project on groups. I will be working on groups of order 12 through 16. I have no questions to follow or really have no idea on what to do with them...
But I don't have the LHS = RHS anywhere. For a proof like this I need to get the original LHS to look like the original RHS, then I need to get the original RHS to look like the original LHS. I have done neither.
On the LHS I'm stuck at A^c n B^c n C, I'm trying to get it to equal the original RHS. As far as everything I did on the RHS, I was trying to get it to equal the LHS
(A U B)^c - C^c = A^c - (B U C^c)
A^c n B^c n C = (A^c n B^c) n (A^c n C)
A^c n B^c n C = (A U B)^c n (A^c n C)
A^c n B^c n C = (A U B)^c n (A^c - C^c)
I'm getting close, is this correct so far. Not sure where to go now?
OK, here is what I got so far, still trying to prove the RHS
(A U B)^c - C^c
A^c n B^c - C^c
Since A^c n B^c = A^c - B
I get
A^c - B - C^c
Now I'm stuck again
Wow, I just wasted a lot of time, I wrote it down wrong. It should have been this
(A U B)^c - C^c = A^c - (B U C^c)
Ok, now that that is settled, I'm trying to prove the RHS first given the LHS.
Like I stated before, I know that
(A U B)^c = A^c and B^c
So I get
A^c and B^c -...
What part is generally false? Maybe this is more clear.
The complement of ( A U B) - The complement of C = The complement of A - The complement of (B U C)
I know that the complement of ( A U B ) = the complement of A and the complement of B
Homework Statement
I need to prove that: (A U B)^c - C^c = A^c - (B U C)^c
Homework Equations
The Attempt at a Solution
I know that (A U B)^c = A^c and B^c
My problem is I'm not sure how to rearrange or distribute the minus sign to make it equal to the other side...
Well I've been staring at this thing for the past hour and I'm not coming up with anything. Am I to be looking at the (x+y)^n side of the binomial theorem or the side with the summation
Homework Statement
(n¦0)-(n¦1)+(n¦2)-. . . ± (n¦n)=0
that reads n choose zero and so on
Homework Equations
Prove this using the binomial theorem
The Attempt at a Solution
I really have no idea where to start. Any help would be greatly appreciated
thanks