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Unformatted text preview: et of real numbers that has exactly n 1 members. Choose a member 89 a of the set S. Since the set S
a has exactly n members, it follows from the assertion P n that
S
a has a largest member that we shall call b. The larger of the two numbers a and b is clearly
the largest member of the set S.
Since P 1 is true and since the condition P n P n1 holds for every positive integer n it follows from
the principle of mathematical induction that P n is true for every positive integer n.
7. Given that S Z , prove that the following three conditions are equivalent: a. The set S is infinite.
b. The set S has no largest member.
c. The set S is unbounded above.
It it probably easier to explain why the denials of these three statements are equivalent to one
another. We shall show that the following three satements are equivalent:
a. The set S is finite.
b. The set S has a largest member.
c. The set S is bounded above.
We already know that every finite set of numbers must have a largest member. Therefore condition
a implies condition b. If a set of numbers has a largest member then that largest member is clearly
an upper bound of the set. Therefore condition b implies condition c. Finally, we know that every
nonempty set of integers that is bounded above must have a largest member. If the largest
member of a set of postive integers is n then that set cannot have more than n members and must
therefore be finite. Therefore condition c implies condition a.
8. Prove that for every positive integer n, if n horses run in a race and no two horses tie then there are exactly n!
possible outcomes.
For each positive integer n we define P n to be the assertion that if n norses run in a race and no
two horses tie then there are exactly n! possible outcomes.
The assertion P 1 is obviously true. Now suppose that n is any positive integer for which P n is true.
To show that the assertion P n1 is also true, suppose that n 1 horses run in a race tha that no two
horses tie. Choose one of the horses whom we shall call Dobbin and, for the moment let’s focus
our attention on the other n horses. Those other horses present us with n! possible outcomes. For
each of those outcomes, if we add in Dobbin then there are n 1 positions in which Dobbin can
place among the other horses. Therefore the number of outcomes, including Dobbin is
n 1 n! n 1 !.
Since P 1 is true and since the condition P n P n1 holds for every positive integer n it follows from
the principle of mathematical induction that P n is true for every positive integer n.
9. Given nonnegative integers n and r satisfying r n, the binomial coefficient
n!
n
.
r
n r !r! n
r is defined by the equation a. Prove that if n and r are nonnegative integers and r n then
n
n
n1 .
r
r
r1
Hint: No special techniques are needed for this proof. Just combine the expressions on the left side.
b. Prove that if n and r are nonnegative integers and r n then every set which has n members must have
exactly n subsets that have r...
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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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