Hi, I'm coming back to you geniuses for another problem.

Write the following set in their simplest form.

( A $\displaystyle \cap$ B) $\displaystyle \cup$ (A $\displaystyle \cap$ B')

Please help figure this one out.

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- May 6th 2011, 10:16 AMyorkeyVenn
Hi, I'm coming back to you geniuses for another problem.

Write the following set in their simplest form.

( A $\displaystyle \cap$ B) $\displaystyle \cup$ (A $\displaystyle \cap$ B')

Please help figure this one out. - May 6th 2011, 10:25 AMAckbeet
What ideas have you had so far?

- May 6th 2011, 10:40 AMyorkey
Ok in this part ( A $\displaystyle \cap$ B), A intersects B. So it's the intersection between the two that is shaded. Then (A $\displaystyle \cap$ B') A intersects B', so its B that remains unshaded.

So how do I go from here, to bring these two together? - May 6th 2011, 10:42 AMAckbeet
- May 6th 2011, 10:46 AMyorkey
The operation between the two is the Union of the two. So... I think the answer is then B. Because, when I sketched it out, B was the only set that wasn't shaded.

- May 6th 2011, 10:51 AMAckbeet
No, I don't think that's correct. The union looks at everything that's in either set (or both). What does that look like?

- May 6th 2011, 10:56 AMPlato
- May 6th 2011, 11:03 AMyorkey
Then the answer would be the intersection of the two sets. What is that called?

Isn't $\displaystyle A\cap B$ just the intersection of the sets? But if what you say is true, then I believe A is the answer, because A is shaded in both sets. - May 6th 2011, 11:08 AMAckbeet
You're guessing. You've correctly identified the set on the left, and the set on the right. What you want to do is super-impose one of those Venn diagrams on the other: whatever is shaded in either of those Venn diagrams is shaded in your answer (because you're doing a union).

- May 6th 2011, 11:13 AMyorkey
So then I'm not really sure.

The set on the left is drawn with only the intersection between the two shaded.

And the set on the right is drawn with everything shaded except B.

So combining the two, the only part that remains unshaded is B. Please tell me how is this wrong? I dont know where I make the mistake. - May 6th 2011, 11:22 AMAckbeet
A intersect B,

*which is part of B,*is shaded for the left set. So that had better show up in your final result. The set on the right isn't exactly "everything shaded except B." It's everything in A that is not also in B (otherwise known as "set difference"). - May 6th 2011, 11:29 AMyorkey
Alright, so then the answer is B? I really don't know what to say or do more now.

- May 6th 2011, 11:52 AMAckbeet
You can do better than that! Think! Put forth some effort! Don't expect to be handed out the solution. You're actually almost there, just one step away. I am most certainly not going to hand you the solution (see my signature for the reason why). I think you know enough to get the answer. You need to be able to convince me that you know the answer. Here's A intersect B, and here's A intersect B'. Put them together! What do you get?

- May 6th 2011, 12:00 PMyorkey
You're a slave driver! Ok, by combining the two sets, the only part that remains unshaded is B. So therefore... the shaded region is the answer, and that is A. Is that it? The shaded part is the final answer?

- May 6th 2011, 12:03 PMAckbeet
As all the best teachers are (not that that inherently makes me a great teacher, though).

Quote:

Ok, by combining the two sets, the only part that remains unshaded is B. So therefore... the shaded region is the answer, and that is A. Is that it? The shaded part is the final answer?