1 (a) Since
by Theorem 2.4.3. (h), by (k) 
. (b)
, { }, {\1} {\ } { , {1}},
{ , { }}, {\1}, { } and the given set itself. 2 (a)
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(b)
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3 (a) Let T denote the set to be defined.
Basis:
Induction: If
, then 
and 
. Extremal: Nothing is in T unless it is obtained by Basis and Induction.
(b) Procedure TEST(x)
Input: A number x
Output: "Yes" if
, else "No". if ( -3 < x < 0 or 0 < x < 3)
return No;
else
if x = 0
return Yes.
else
if x > 0
return TEST(x -3).
else
return TEST(x +3).
4. Basis: If i = 0, LHS =
= 0. RHS = 0*1*1/6 = 0. Hence LHS = RHS.
Induction: Assume that
= n(n+1)(2n+1)/6. LHS for n = n+1 is
= 
=
by the assumption.= (n+1)[n(2n+1) + 6(n+1)]/6 by factoring out (n+1) and taking the common denominator.
=
= (n+1)(n+2)(2n+3)/6
= (n+1)[(n+1)+1][2(n+1)+1]/6, which is RHS for n = n+1.