There is no escape...believe it or not. Mathematics is everywhere.
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I think so, but I haven't tried it out for anything more than a set of 3 elements. But the procedure should go something like this:
Write each element of the set down.
Draw a circle around each element.
Draw a circle around every possible combination of elements (this would be the tricky part).
Draw one big circle around all the elements.
Draw one circle off to the side for the empty set.
Shouldn't that Venn Diagram every element of the powerset, overlapping the sets that share elements in common? I'm not sure if this is what you meant.
I just tried it for 4 elements. It gets messy and hard to follow. Maybe if you use different colors...
Hope this helps!
thanks for your valuable reply sir and i didn't get about why you have written null set separately?????
Dylan Faullin said:I think so, but I haven't tried it out for anything more than a set of 3 elements. But the procedure should go something like this:
Write each element of the set down.
Draw a circle around each element.
Draw a circle around every possible combination of elements (this would be the tricky part).
Draw one big circle around all the elements.
Draw one circle off to the side for the empty set.
Shouldn't that Venn Diagram every element of the powerset, overlapping the sets that share elements in common? I'm not sure if this is what you meant.
I just tried it for 4 elements. It gets messy and hard to follow. Maybe if you use different colors...
Hope this helps!
Should be self explanatory :-)
Should be self explanatory :-)
The empty set is here but can't be seen :-)
(intersection of none, part of outside)
I think that nested sets are not convenient because it became quickly messy.
My representation does not give the feeling that "{A,B}'" is contained in "{A,B,C}" but it has the advantage to be clear.(Even if I had to put coloureds dots to help reader figure it out)
But I couldn't get it work for 3 items or more.
And it's quite normal.
If you had 1 set, each set would contain 1 group of dot.(1 possibility)
If you had 2 sets, each set would contain 2 groups of dots.(3 possibilities)
If you had 3 sets, each set would contain 4 groups of dots.(7 possibilities)
If you had 4 sets, each set would contain 8 groups of dots.(15 possibilities)
As you can see in my diagram, each set contain 7 groups of dots. My diagram is wrong :-) (missing a triplet)
The diagram you did is, at my sense, better matching reality because you've the feeling that these are included sets. But it's messy.
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