De Morgan's Laws

[cindib80]

I'm not sure if this is actually algebra, but wasn't sure where to put it on the MB. The problem I am trying to solve is: Make up 2 sets A & B so that n(A union B) does not equal n(A) + n(B).

I then have to make up 2 more sets so that n(A union B) DOES equal n(A) + n(B).

Then to finish up I have to show how to modify the n(A union B) so that it always equals n(A) + n(B).
I am completely lost!

 

[stapel]

What is the meaning of "n(A)"?

These exercises ask you to think for a while and invent some sets that will work in the prescribed ways. It's kinda hard for other people to do that original thinking and individualized inventing for you. What is it, precisely, that you are requesting?

Thank you.

Eliz.

 

[cindib80]

The problem I posted is taken directly from my assignment. I am not asking for someone to give me the answer. I do not understand how the first one can be done. While doing this chapter I thought that AUB would always equal n(A) +(B). I would just like an example of how this might be solved.

 

[mark07]

I assume that n(A) denotes the number of elements of a set A.

I'll give you an example and you can find more for your homework (life is full of them afterall!).

Let's say our universal set is USA. Let A be the set of all cars in USA that are red and let B be the set of all cars in USA that are hybrid. Then AUB is the set of all cars that are either red or hybrid.

n(AUB) is not equal to n(A) + n(B), because in the latter you count the red hybrids twice! Does that make sense?

 

[cindib80]

That helps so much. I was thinking in terms of numbers and not looking at the whole picture. I appreciate your help so much! (I will pass this class if it kills me!)

 

[soroban]

Hello, cindib80!

Don't make it complicated . . .

(1) Make up 2 sets A and B so that: \(\displaystyle \: n(A\,\cup\,B)\:\neq\:n(A)\,+\,n(B)\)

(2) Make up 2 more sets so that: \(\displaystyle \:n(A\,\cup\,B)\:=\:n(A)\,+\,n(B)\)

(3) Show how to modify \(\displaystyle n(A\,\cup\,B)\) so that it always equals \(\displaystyle n(A)\,+\,n(B)\)

The union of two sets means: "dump all the elements into one set".

\(\displaystyle (1)\;A\,=\,\{4,5,6\},\;B\,=\,\{6,7,8,9\}\)

So \(\displaystyle n(A)\,=\,3\) and \(\displaystyle n(B)\,=\,4\)

But \(\displaystyle A\,\cup\,B\:=\:\{4,5,6,7,8,9}\)
. . so: \(\displaystyle \,n(A\,\cup\,B) \:=\:6\) . . . and not \(\displaystyle 3\,+\,4\:=\:7\)

Why? .Because the two sets have an "overlap".


So just make up two sets with an overlap.

In a deck of cards, there are 4 Kings and 13 Diamonds.
. . That is: \(\displaystyle \,n(\text{Kings})\,=\,4,\;n(\text{Diamonds}) \,=\,13\)

But \(\displaystyle n(\text{Kings }\cup\text{ Diamonds})\:\neq\:4\,+\,13\)

Why? .There is an overlap: \(\displaystyle K\L\diamond\)



(2) Make up two set with no overlap.

That's easy . . . \(\displaystyle A\,=\,\{\text{windmill},\text{ kangaroo}\}\) and \(\displaystyle B\,=\,\{\text{snap},\text{ crackle},\text{ pop}\)\)



(3) If there is an overlap, simply remove it.