82
Solutions to Exercises
(d) (2,2) and (2,3) are minimal; (4,3) and (4,4) are maximal; there are no minimum nor maximum
elements.
(e)
a
b
glb(a,
b)
lub(a, b)
i.
(2,2)
(3,3)
*
*
ii.
(4,2)
(3,4)
(3,2)
(4,4)
iii.
(3,2)
(2,4)
(2,2)
(3,4)
iv.
(3,2)
(3,4)
(3,2)
(3,4)
(f) Let
Al
=
A2
=
Z and let
~I
and
~2
be
the usual (total) order relation on the integers. Then
Z x Z is not totally ordered since, for example, (1,2) and (0,3) are not comparable.
35. (a) First we note that
(A,
I )
is a poset, in the same way that (N,
I )
is a poset. Now we show that any
two elements
a, b
E
A
have a glb and a lub. Certainly lcm(
a, b)
has the properties required of a
least upper bound and gcd(
a,
b)
of a greatest lower bound. The key point is to show that these
elements lie in
A,
and this follows from the definitions of gcd and lcm.
If
a,
b
E
A,
then
a
I
n
and
bin,
hence,
lcm(a, b)
I
n
by Exercise 29, so lcm(a,
b)
E
A.
Furthermore, since
gcd(a, b)
I
a
and
a
I
n,
we have
gcd(a, b)
I
n
and so
gcd(a, b)
E
A
too.
(b) [BB]
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 Summer '10
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 Graph Theory, Order theory, greatest lower bound, unique integers, maximum elements

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