=?

I say by definition is or .

Since , then

I don't know whether this is correct.

How do you draw a Venn Diagram for ?

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- Jan 28th 2010, 04:36 AMnoviceUnion of two sets
=?

I say by definition is or .

Since , then

I don't know whether this is correct.

How do you draw a Venn Diagram for ? - Jan 28th 2010, 05:06 AMDinkydoe
Is the union of A with an extra element " ".

Or, you can also turn the picture around: It is the container united with all elements of A.

After the union we have: and .

Quote:

I say http://www.mathhelpforum.com/math-he...e8028d22-1.gif by definition is http://www.mathhelpforum.com/math-he...6fa34dea-1.gif or http://www.mathhelpforum.com/math-he...6fc21079-1.gif.

Since http://www.mathhelpforum.com/math-he...0d7358d9-1.gif, then

http://www.mathhelpforum.com/math-he...7aa06b42-1.gif

In that case your example is still incorrect. Provided that we have

If then - Jan 28th 2010, 07:17 AMnovice
You painted a good picture using and . I understood this part, but I am still not sure I have a full grasp of the concept.

Let's see: Assuming that is a nonempty container and is some candy, by definition

or .

Since , I can conclude .

I am thinking since there isn't any candy in the empty container, , and that is a nonempty container, I may conclude that the candy is in .

For this, suppose I make the question like

I say since , thus - Jan 28th 2010, 07:44 AMDinkydoeQuote:

Let's see: Assuming that http://www.mathhelpforum.com/math-he...2eacbe29-1.gif is a nonempty container and http://www.mathhelpforum.com/math-he...155c67a6-1.gif is some candy, by definition

http://www.mathhelpforum.com/math-he...a195745e-1.gif or http://www.mathhelpforum.com/math-he...d0d2f845-1.gif.

Quote:

. The empty set is an element of this set. You can write the emptyset as: . It's a container without elements. But a container without elements, can still be an element of another container! Namely:

And how do you conclude ? The empty-set may, or may not be in . It depends on how we define to be.

Quote:

For this, suppose I make the question like http://www.mathhelpforum.com/math-he...b11670ed-1.gif

I say since http://www.mathhelpforum.com/math-he...7941f460-1.gif, thus http://www.mathhelpforum.com/math-he...20d23d34-1.gif

- Jan 28th 2010, 07:54 AMnovice
- Jan 28th 2010, 02:30 PMDrexel28
- Jan 28th 2010, 08:30 PMnovice
- Jan 28th 2010, 08:35 PMDrexel28
I think you are making this much harder on yourself than it is supposed to be. You are doing what is called Naive Set Theory. In other words, it's naive because the concepts are supposed to be taken axiomatically and are intuitive. You are not rigorously defining what means, you are just supposed to realize that means "take everything in and everything in and put it in this new set while making sure to take out duplicates."

- Jan 28th 2010, 09:02 PMVonNemo19
- Jan 28th 2010, 09:03 PMDrexel28
- Jan 29th 2010, 02:02 AMcanyiahset identities
[IMG]file:///C:/Users/VCOX/AppData/Local/Temp/moz-screenshot-3.png[/IMG]i attached some set identities. a book by rosen discrete math and its applications is a really good book. there is a wikiibook on discrete math to although i havent really looked at it. did your problem come with what elements are in A I would think that you would need that information to solve for A U "set of nothing".

also an set of "nothing" is NOT equal to a set of a set of "nothing" which is what {0} represents. 0 means the empty set and {0} means a set of an empty set (NOTE: zero represents the empty set i didnt want to open word and have to print screen add to paint then insert the picture)

"nothing" is an element of any set but a set of "nothing" isnt an element of every set. sets and elements are different a set has elements in them(group of things ie the entire alphabet is a set "a to z"). elements goes into sets(things ie individual alphabets "a" are considered elements). - Jan 29th 2010, 03:06 AMHallsofIvy
- Jan 29th 2010, 06:41 AMnovice
I sounds very funny.(Rofl)

That is a clever definition. It would be nice if that was in all the math books.

I told a friend that I was bothered by the fact that the set wasn't being defined. He told me the point was to show is no longer two sets but one---just as you explained it.

I think I understand the concept thoroughly this time. - Jan 29th 2010, 01:23 PMcanyiahRosen Discrete Math link
here is a link to rosens book site has alot of interactive stuff on it you can practice problems and other things

Discrete Mathematics and Its Applications