1. ## Venn

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

Write the following set in their simplest form.
( A $\cap$ B) $\cup$ (A $\cap$ B')

2. What ideas have you had so far?

3. Ok in this part ( A $\cap$ B), A intersects B. So it's the intersection between the two that is shaded. Then (A $\cap$ B') A intersects B', so its B that remains unshaded.

So how do I go from here, to bring these two together?

4. Originally Posted by yorkey
Ok in this part ( A $\cap$ B), A intersects B. So it's the intersection between the two that is shaded. Then (A $\cap$ B') A intersects B', so its B that remains unshaded.

So how do I go from here, to bring these two together?
Well, what is the operation you're asked to perform between your two (correct) sets?

5. 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.

6. 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?

7. Originally Posted by yorkey
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.
$A\cap B$ is everything in A and B, belongs to both.
$A\cap B'$ is everything in A and not in B, belongs only to A.
Put those two together what do you have?

8. Originally Posted by Ackbeet
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?
Then the answer would be the intersection of the two sets. What is that called?

Originally Posted by Plato
$A\cap B$ is everything in A and B, belongs to both.
$A\cap B'$ is everything in A and not in B, belongs only to A.
Put those two together what do you have?
Isn't $A\cap B$ 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.

9. 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).

10. 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.

11. 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").

12. Alright, so then the answer is B? I really don't know what to say or do more now.

13. 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?

14. 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?

15. Originally Posted by yorkey
You're a slave driver!
As all the best teachers are (not that that inherently makes me a great teacher, though).

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?
I would agree that A is the final answer. However, I can only follow your explanation because I know to which Venn diagrams you are referring. Your explanation wouldn't convince me out of the blue that you have the right answer. Maybe referring to Venn diagrams is all you need for your particular problem. I would rather see a logic-based explanation for your answer.

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