ICS 6D ( Discrete Mathematic): Solution to Selected Problems on Homework #3
Section 3.4
2.
1

a
because there exists integer
n
, namely
n
=
a
, s.t.
a
=
n
∗
1.
a

0 because there exists integer
n
, namely
n
= 0, s.t. 0 =
n
∗
a
.
6.
By assumptions that
a

c
and
b

d
, there exist integers
n
1
, n
2
s.t.
c
=
n
1
a
and
d
=
n
2
b
.
Therefore
cd
= (
n
1
n
2
)
ab
, i.e. there exists integer
n
, namely
n
=
n
1
n
2
, s.t.
cd
=
n
(
ab
), and therefore
ab

cd
.
8.
This is not true. For example 8

(4
∗
2) but it’s
not true
that either 8

4 or 8

2.
12.
Note that we showed the opposite direction in class! This direction appears a bit more complicated.
Let’s show this by arguing the counterpositive, i.e. that if (
a
mod
n
)
negationslash
= (
b
mod
n
) then
a
negationslash≡
b
(mod
n
).
By the division algorithm theorem there exist
unique
remainders
r
1
and
r
2
in
Z
N
s.t.
n

(
a
−
r
1
) and
n

(
b
−
r
2
). (The theorem also says that the quotients are unique, but we don’t need it here.) Let
q
1
, q
2
be
integers s.t.
nq
1
=
a
−
r
1
and
nq
2
=
b
−
r
2
. Therefore
n
(
q
1
−
q
2
) = (
a
−
b
) + (
r
2
−
r
1
). Therefore we have
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 Fall '07
 Jarecki
 Division, Prime number, Greatest common divisor, Euclidean algorithm

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