Download Quiz 12 (1) Factor the following positive integers into primes. • 42

Department of Mathematical Sciences
Instructor: Daiva Pucinskaite
Discrete Mathematics
November 18, 2015
Quiz 12
(1) Factor the following positive integers into primes.
• 42 000 = 42·1000 = |{z}
6 · 7 · |10 · 10
{z · 10} = 2·3·7·2·5·2·5·2·5
1000
42
4
3
=2 ·3·5 ·7
2
· 53 · 7}
• 42 0002 = 24 · 3 · 53 · 7 = |24 · 3{z
· 53 · 7} · 2| 4 · 3{z
42 000
42 000
= 28 · 32 · 56 · 72
• 3 · 42 0002 = 3 · 2| 8 · 32{z
· 56 · 7}2 = 28 · 33 · 56 · 72
42 000
(2) Let n ∈ N. Show that
(a) 3 appears an even number (or zero) of times in the prime
factorization of n2 .
Let n = p1 · p1 · · · p1 · p2 · p2 · · · p2 · · · pi · pi · · · pi · · · pm · pm · · · pm
|
{z
} |
{z
} | {z }
|
{z
}
n1 -times
=
pn1 1
·
pn2 2
n2 -times
· · · pni i
ni -times
nm -times
· · · pnmm
be the prime factorization of n, where p1 , p2 , . . . , pi , . . . , pm
are pairwise different (i.e. pi 6= pj if i 6= j).
Example: In the prime factorization of n = 42 000
42 000 = 2
· 2 · 2} · |{z}
3 · |5 · {z
5 · 5} · |{z}
7 = 2 4 · 3 · 53 · 7
| · 2{z
4-times
1-time
3-times
1-time
we have m = 4, because the primes p1 = 2, p2 = 3, p3 = 5, p4 = 7 in the prime
factorization of 42 000 are pairwise different, and n1 = 4, n2 = 1, n3 = 3, n4 = 1,
because
• p1 = 2 appears n1 = 4 times in the prime factorization of n,
• p2 = 1 appears n2 = 1 time in the prime factorization of n,
• p3 = 5 appears n3 = 3 times in the prime factorization of n,
• p4 = 7 appears n4 = 1 times in the prime factorization of n.
We have that
n2 = pn1 1 · pn2 2 · · · pni i · · · pnmm
2
m
= p12n1 · p22n2 · · · pi2ni · · · p2n
m
is the prime factorization of n2 , because all numbers are
prime in this product. (When a composite number is written as a
product of prime numbers, we have the prime factorization of the number,
by the ”Fundamental Theorem of Arithmetic”)
• If for some 0 ≤ i ≤ m we have pi = 3, then 3 appears
2·ni times in the prime factorization of n2 . In other words:
If pi = 3, then 3 appears an even number of times in the
prime factorization of n2 .
• If 3 does not appear in the prime factorization of n, then
3 does not appear in the prime factorization of n2 . In this
case 3 appears 0 times in the prime factorization of n2 .
Since 0 is an even number too, we have that in any case 3
appears an even number of times in the prime factorization
of n2 (for any positive number n).
(b) 3 appears an odd number of times in the prime factorization of 3 · n2 .
Let n = p1 · p1 · · · p1 · p2 · p2 · · · p2 · · · pi · pi · · · pi · · · pm · pm · · · pm
|
{z
} |
{z
} | {z }
|
{z
}
n1 -times
=
pn1 1
·
pn2 2
· · · pni i
n2 -times
ni -times
nm -times
· · · pnmm
be the prime factorization of n, where p1 , p2 , . . . , pm are
pairwise different (i.e. pi 6= pj if i 6= j). Then n2 =
2ni
2nm
1
2
is the prime factorization of n2 .
· · · pm
p2n
· p2n
1
2 · · · pi
Therefore
2nm
3 · n2 = 3 · p12n1 · p22n2 · · · pi2ni · · · pm
is the prime factorization of 3 · n2 , because all numbers are
prime in this product (by the ”Fundamental Theorem of Arithmetic”
this is the prime factorization of 3 · n2 ).
We have two cases: either 3 = pi for some 0 ≤ i ≤ m, or
3 6= pi for all 0 ≤ i ≤ m.
• If 3 = pi , then 3 appears in the prime factorization
2·pi +1
1
2
m
3 · n2 = p2n
· p2n
· · · p2n
1
2 ··· 3
m
of 3 · n2 precisely 2 · pi + 1-times. Since 2 · pi + 1 is an odd
number, we have that 3 appears an odd number of times
2
in the prime factorization of 3 · n2 .
• If 3 6= pi , i.e. 3 does not appear in the prime factorization
of n2 , then 3 appears 1-time in the prime factorization of
3 · n2 . Since 1 is odd, we proved the statement ” 3 appears
an odd number of times in the prime factorization of 3 · n2
for any positive number n”.
a √
(3) There are no numbers a, b such that = 3.
b
√
a
(In other words: If a, b ∈ N, then
b
6=
3).
This statement is logically equivalent to the statement
”If
a √
= 3, then a 6∈ N or b 6∈ N”.
b
√
a
= 3, then a2 has two prime factorizations. In the first prime
b
factorization of a2 the prime 3 appears an even number of times, and in the second
prime factorization of a2 the prime 3 appears an odd number of times. By the
”Fundamental Theorem of Arithmetic” any number has (up to order) unique prime
factorization, thus each prime appears either an even or odd number of times in the
prime factorisation. Therefore a2 is not a number, and consequently a is not a number
(because if a would be a number, then a2 would be a number too).
We will show: If
Proof. Assume
a √
= 3, we have
b
3
a √
a2
= 3 =⇒ 2 = 3
b
b
=⇒ a2 = 3 · b2
Statement (2)(a)) =⇒ 3 appears an even number of times
in the prime factorization of a2
(Statement (2)(b)) =⇒ 3 appears an odd number of times
in the prime factorization of 3 · b2
a2 = 3 · b2 =⇒ a2 has two prime factorizations;
in one prime factorization
3 appears an even number of times,
and in other prime factorization of a2
3 appears an odd number of times
=⇒ a2 can’t be a number, because by the
”Fundamental Theorem of Arithmetic”
each prime appears
either an even or an odd number of times
in the prime factorisation of any number
=⇒ a can’t be a number
4
(1) Let m ∈ N, m > 1, and a, c ∈ Z. Show
(a + c · m) mod m = a mod m.
By definition of a mod m we have
a mod m = r
if there exists an integer q such that
• a = q · m + r,
• 0 ≤ r < m.
We have to show (a + c · m) mod m = r. In other words we
have to show that we can find an integer q̃ such that
a + c · m = q̃ · m + r
(the property 0 ≤ r < m is already satisfied).
a = q · n + r =⇒ a + c · m = q · n + r +c · m
| {z }
a
=⇒ a + c · m = q · m + c · m + r
=⇒ a + c · m = (q + c) m + r
| {z }
q̃
=⇒ a + c · m = q̃ · m + r
=⇒ (a + c · m) mod m = a mod m = r.
(2) Using Mathematical Induction, prove that
n3 mod 3 = n mod 3
for all n ∈ N.
Proof. Basis step: The case n = 0 is true because both sides
of the equation
03 mod 3 = 0 and
0 mod 3 = 0
evaluate to 0.
Induction hypothesis: Suppose the result is true for k ∈ N;
that is, we assume we have
k 3 mod 3 = k mod 3
We must prove that the result is true for k + 1; that is
using the equation
k 3 mod 3 = k mod 3
we must
3
prove (k + 1) mod 3 = (k + 1) mod 3.
(k + 1)3 mod 3 =
=
=
k 3 + 3k 2 + 3k + 1 mod 3
k 1 + 1 + 3k 2 + 3k mod 3
5
2
k
+
1
+
(k
+
k)
·
3
3
| {z } | {z } |{z} mod |{z}
a
(1)
=
m
integer
m
3
3
k
+ 1} mod |{z}
| {z
m
a
The assumption k 3 mod 3 = k mod 3, implies (k 3 + 1) mod 3 = (k + 1) mod 3
This is what we had to show. Therefore n3 mod 3 = n mod 3
for all n ∈ N.
6