COT 3100 Homework # 4 Solutions
Spring 2000
1)
Define a relation T
⊆
N x N such that T = {(a,b)| a
∈
A
∧
b
∈
A
∧
a – b = 2c+1 for some
integer c}.
(N is the set of non-negative integers.)
a)
Prove that this relation is not reflexive.
We have that a – a = 0 for all integers a. But 0
≠
2c + 1 for any integer c. Thus,
for all integers a we have (a,a)
∉
R. So, not only is the relation not reflexive, it is
in fact irreflexive.
b)
Prove that this relation is symmetric.
If we have that (a,b)
∈
R, that means that a – b = 2c+1, for some integer c. Now
consider the quantity b – a.
b – a = -(a – b)
= -(2c+1)
= -2c – 1
= -2c –2 + 1
=2(-c-1) + 1
Thus, we have that (b,a)
∈
R as well.
c) Define the term anti-transitive as the following:
Given a set A and a relation R,
if for all a,b,c
∈
A, (aRb
∧
bRc
∧
cRa)
⇒
(a = b
∨
b = c)
Prove that the relation T is anti-transitive.
The relation is vacuously anti-transitive. That is because it is impossible for
(a,b)
∈
R, (b,c)
∈
R, (c,a)
∈
R all at the same time. To see this consider the
following :
a –b
= 2d +1 for some integer d
b – c = 2e +1 for some integer e, so
a – c = 2d + 2e +2 = 2(d+e+1). But this makes a – c even, which means (a,c)
∉
R.
Proving that (a,b)
∈
R, (b,c)
∈
R, (c,a)
∈
R can NOT all be true. This also proves that
the relation is anti-transitive.