76
Solutions
to
Exercises
(c) By the result
of
part (a),
319
566
has
1+
LloglO
319
566
J
digits. Since
10glO
319
566
=
566
log
10
319
~
1417.15,
the integer
319
566
has
1418
digits.
Exercises
4.2
1.
(a) [BB] Not totally ordered; for example, 6 and
21
are not comparable.
(b) [BB]
1
is a minimum element since
1
1
n
for all
n
E
N,
but there is no maximum element since
given any
n
E
N,
n
1
2n,
so
n
can't
be
maximum.
2.
(a) The greatest lower bound
of
natural numbers is their greatest common divisor. Since the gcd
of
any two elements
of
this poset is, in each case, still an element
of
the poset, every pair
of
elements
has a glb.
(b) The least upper bound
of
natural numbers is their least common multiple. The lcm
of
4 and 6 is
12.
Since
12
is not in the poset,
4
and
6
have no lub.
(c) This is not a lattice (because not all pairs
of
elements have a lub.)
3.
(a)
[BB]
4
6
V1
0
0
(6)
8
9
2
3
5
7
7
1
4. (a)
2,
3,
5
and
7
are minimal;
4,5, 6
and
7
are maximal. There are no minimum nor maximum
elements.
(b)
1
is minimal and minimum;
6,
7,
8,
9
and
10
are maximal. There is no maximum.
5. [BB] As in Problem 6, write
a
=
nq
+
r,
where
0
::;
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 Summer '10
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 Graph Theory, Order theory, Prime number, Greatest common divisor

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