# Is the logic correct?

I had the following on an exam today.

Let the universe = $Z^+$, $Z = (1,2,3,....100)$

$|A| = 10$

$|B| = 10$

$|A\cap B| = 0$

(a) What is $|A^c \cup B^c|$

I thought about it like this.

Let set A be dogs

Let set B be cats.

Or let A be even numbers and B be odd numbers. Sticking with dogs and cats though.

A-compliment is cats

B-compliment is dogs

their union is the whole Z so $|A^c \cup B^c|= 100$ is this correct?

(b) What is $|A^c \cap B^c|$ I said this was 0 using the same logic above

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Maybe it is better to ask "Is this logic correct?" or "Is this line of reasoning correct?", removing the self from the topic of study will remove the unhelpful element of "your logic" from the answer. Also it will decouple you from the method being studied. Logic should stand on it's own, even if it wrong, it is not "you" or "your logic" that is wrong but the "logical method being used" Regards – Arjang Mar 10 '11 at 23:09
You mean $|A \cap B| = 0$? – Calle Mar 10 '11 at 23:09
yes, I just made the correction – lampShade Mar 10 '11 at 23:11
It is customary to write your $A$-compliment as $A^c$ (although some books have their own notation). From what I surmise, you are asking for the quantity $|A^c \cup B^c|$. This follows easily from De Morgan's laws and the fact that $|A \cup B| = |A| + |B| - |A \cap B|$, as we are double-counting the elements in $A \cap B$. I'll leave it to you to fill in the details. – JavaMan Mar 10 '11 at 23:12
@Ross Millikan yes it is as you assumed. – lampShade Mar 10 '11 at 23:15

Let $A^c$ and $B^c$ be the complements of $A$ and $B$ respectively. We know that $|A \cap B| = 0$, that is, $A$ and $B$ have no elements in common, which implies that $A \subset B^c$ and $B \subset A^c$.

So $A^c$ contains all elements, except $A$. $B^c$ contains all elements except $B$. But $A^c$ contains $B$ and $B^c$ contains $A$, so $A^c \cup B^c = Z$, thus $|A^c \cup B^c| = 100$.

Your reasoning is not correct, however. If you mean that $A$ is all dogs in $Z$ and $B$ is all cats in $Z$, then there are 80 other things in $Z$ which are neither cats or dogs, say elephants. Thus $A^c$ is the set of all cats and elephants and $B^c$ is the set of all dogs and elephants.

If you mean that $A$ consist of only dogs, but not necessarily all dogs in $Z$, $B$ consists only of cats, but not necessarily all cats in $Z$, and $Z$ contains only dogs and cats, the situation is a bit different. Then three things are possible: $A^c$ contains dogs, $B^c$ contains cats, or both.

As for the b-part. We know that $|A^c| = 90$ and $|B^c| = 90$. As per above, we know that $A \subset B^c$ and $B \subset A^c$, but $A$ is not part of $A^c$ and $B$ is not part of $B^c$. However, "the rest" of $A^c$ and $B^c$ is the same, so $|A^c \cap B^c| = 80$.

Using the cats, dogs and elephants: $A^c$ is all cats and elephants, $B^c$ is all dogs and elephants. The elements in common are the elephants.

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Your answer is the correct way to think about it. Thanks – lampShade Mar 10 '11 at 23:42

The reasoning is more or less correct.

Your analogies to other $A$ and $B$ (cats/dogs/even/odd) are misleading, because they have $A$ and $B$ as complementary sets themselves. Because $|A|=10$ and $|B|=10$ and there are 100 elements in the universe, there are at least 80 other elements in the universe that are not in $A$ and $B$.

As to part (a), I tend to think about it this way: Since $A$ and $B$ have no common elements, all of $B$ is contained in the complement of $A$, so the union of a set that contains all of $B$ with the complement of $B$ has to be the whole universe.

Beyond that, Calle's answer does a good job of answering the question.

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-1: For the misleading first sentence. The reasoning by OP is incorrect, as we can see from Calle's answer. – Aryabhata Mar 10 '11 at 23:30
@Moron: Ahh—in light of the added (b) in the question, it's clear that you're correct on that (I took the reasoning to be more generally about the mutual-exclusivity). – Isaac Mar 10 '11 at 23:35
so part b is wrong then? – lampShade Mar 10 '11 at 23:37
@lampShade: Yes, (b) is wrong. Since you know the universe is the integers 1 to 100, it might help to consider specific sets $A$ and $B$ consisting of integers in the range 1 to 100 that satisfy the given information. – Isaac Mar 10 '11 at 23:43