Chapter 5
159
7. Let
S
=
{a,
b}
be a set of two elements. The only 2element subset of
S
is
S
itself. Also, for
n
=
2,
~n(
n
1) = 1,
so the result is true. Now assume that
k
2::
2 and the result is true for sets of
k
elements.
Let
S
be a set of
k
+
1 elements and let
a
be an element of
S.
By the induction hypothesis, there are
~k(k

1)
2element subsets of
S"
{a}.
The only other 2element subsets of
S
contain
a,
and there
are
k
of these. Altogether, the number of 2element subsets of
S
is
~k(k
1)
+
k
=
~k(k
+
1),
which
is the desired formula for
n
=
k
+
1. By the Principle of Mathematical Induction, we conclude that
the result is true for all
n
2::
2.
8. (a) Any logarithm (0, 00)
~
R is an example.
(b) For
n
=
1,
f(XI)
=
f(XI),
which is the desired result. Now assume that
k
2::
1 and that
f(XIX2
...
Xk)
=
f(XI)
+
f(X2)
+ .
.. +
f(Xk)
for any real numbers
Xl,
X2,
... ,
Xk
E
A.
Given
Xl,
X2,···, Xk+l
E
A,
we have
f(XIX2
...
Xk+l)
=
f((XIX2'" Xk)Xk+I)
since we are given that
f(ab)
=
f(a)
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.
 Summer '10
 any
 Graph Theory, Sets

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