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# Counting proofs, help? watch

1. Anyway, I'm struggling like hell to understand counting proofs.

Corollary For a positive integer n, suppose that is a collection of n pairwise disjoin finite sets
(i.e. )

Then

All its says is this is an easy proof by induction. I know you need to use addition principle, but then thats it.

Also, the proof of multiplication principle is like impossible to understand.

This chapter is gay. I wonder if I can get away with not learning it
as the exam gives you option of choosing three out of five questions.

P.S. Yeah, this is so gay. I hate counting. Why do we even need to count or proof stuff about counting.
P.P.S. Its stupid as two chapters depends on me not sucking on this chapter. I think I can memorize the proofs and stuff. Or maybe I should try and draw pictures of what is happening.
2. (Original post by Simplicity)
Anyway, I'm struggling like hell to understand counting proofs.

Corollary For a positive integer n, suppose that is a collection of n pairwise disjoin finite sets
(i.e. )

Then

All its says is this is an easy proof by induction. I know you need to use addition principle, but then thats it.
Can you prove that |X union Y| = |X| + |Y| if X and Y are disjoint? Then write , which is a union of two disjoint sets...

(Original post by Simplicity)
Also, the proof of multiplication principle is like impossible to understand.
Care to state it?

(Original post by Simplicity)
P.S. Yeah, this is so gay. I hate counting. Why do we even need to count or proof stuff about counting.
This is maths. You thought you'd get away without counting?

(Original post by Simplicity)
P.P.S. Its stupid as two chapters depends on me not sucking on this chapter.
By which you mean "if I don't learn the basics I can't learn the more advanced stuff"? Yes, that is stupid.
3. (Original post by Simplicity)
disjoin finite sets
(i.e. )
Also, this is wrong.
4. (Original post by generalebriety)
Can you prove that |X union Y| = |X| + |Y| if X and Y are disjoint?
Yes.

Yeah, I get what you mean now.

(Original post by generalebriety)
Care to state it?
Yeah, I think I understand now thanks for the help as I would have be starring into the book for half an hour trying to understand it.

(Original post by generalebriety)
By which you mean "if I don't learn the basics I can't learn the more advanced stuff"? Yes, that is stupid.
Yeah. But still if it involved just learning technique then I wouldn't have said it. I don't like conceptual understanding as it takes more time to understand it. Its like something just clicks and then its easy. Pretty annoying.

(Original post by generalebriety)
This is maths. You thought you'd get away without counting?
Actually, I read somewhere that there is a program on maths that wants to get rid off all counting arguments and replace it with something else. Something called categorification.

P.S. I have noticed that you have been helping alot during the summer holiday. Are you doing some studying during the summer holiday? as it seems most people have just given up on maths for summer holiday. I really should do the same.
5. (Original post by Simplicity)

But now, if is a bijection, then gives a bijection so that . It follows from the previous corollary(the one in this thread) that .

Anyway, I understand everything intill the last paragraph. I don't undestand or how the corollary that needed to be proven by induction will show the last bit.
You've written X*Y as a union of n sets of the form {x_i}*Y for i = 1, 2, ..., n. Each of these sets is obviously disjoint, because if i =/= j, every element in {x_i}*Y differs from every element in {x_j}*Y in the first component.

Now suppose you order all the elements in Y, say y_1, y_2, ...., y_m. (This is all your function g is doing: g(1) = y_1, ..., g(m) = y_m.) Then every element in the set {x_k}*Y is of the form (x_k, y_j), simply by definition of the Cartesian product. This gives an obvious bijection between {x_k}*Y and Y, and so |{x_k}*Y| = |Y| = m.

Now put these things together. We've taken the union of n disjoint sets each of which has size m, so the size of the union is m+m+...(n times)...+m = mn.

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