Unformatted text preview: , n
into a finite oneone sequence is n! Solution: We use mathematical induction. The exercise is asking us to prove that if, for each positive integer n we write F n for the set of all oneone functions from the set 1, 2, , n onto
1, 2, , n then for each n we have card F n n!.
For each positive integer n we take p n to be the assertion that card F n n! The assertion p 1 is
obvious.
Now suppose that n is any positive integer for which the assertion p n is true. From part a and from
Exercise 5 we see that 57 card F n1 card F n card 1, 2, , n, n 1 n! n 1 n 1 !
12. Given that r and n are integers and that 0 r n, the binomial coefficient
n!
n
.
r
n r !r! n
r is defined by a. Prove that if n and r are integers and 1 r n then
n
n
n1 .
r
r
r1
The identity follows at once when we combine the terms on the left side. b. Suppose that r is a nonnegative integer. Prove that for every integer n
members, the number of subsets of S that have r members is n .
r r and every finite set S that has n Solution: For each nonnegative integer n we define p n to be the assertion that, whenever r is an
integer and 0 r n, the number of subsets with exactly r members of any given set that has exactly
n members is n . The statement p 0 is obviously true. Now suppose that n is a nonnegative integer for
r
which the statement p n is true, suppose that r is an integer, that 0 r n 1 and that S is a set with
exactly n 1 members. Choose a member q of the set S. We observe that the set S
q has exactly n
members.
Now we define A to be the family of all subsets of the set S
q and having exactly r members and B
to be the family of all sets of the form q Þ E where E is a subset of S
q that has exactly r 1
members. (In the event that r 0, this definition makes the set B empty.) We see that
card A n
r
and
n
r1
and therefore, by Exercise 3 we deduce that the number of subsets of S that have exactly r members is
n
card A Þ B card A card B n
n1 .
r
r
r1
The desired assertion therefore follows by mathematical induction.
card B c. Prove that if n is a positive integer then
n n n 2n.
n
1
0
n
An intuitive approach to this question is to conclude from part b that the number j0 n is
j
the number of subsets of 1, 2, , n which we know from Exercise 6 to be 2 n . We could write a
more careful proof using mathematical induction but that proof would be the special case of
Exercise 13 obtained when a b 1.
13. Prove that if n is a nonnegative integer and a and b are any real numbers then
a b n n anb0 n an 1b1 n an rbr n a0bn.
r
n
0
1
This equation, as you may know, is known as the binomial theorem.
The solution of this exercise would be the standard proof by mathematical induction of the binomial
theorem. Of course the calculus part of the text shows much more elegant ways of deducing the
binomial theorem. Exercises on Infinite Sets
1. Prove that a subset E of Z is equivalent to Z if and only if for every integer n there...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.
 Fall '08
 STAFF
 Math, Calculus

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