Chapter 2, Section 2.4

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Chapter 2, Section 2.4

• The Integers and Division

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Recall that the integers are not closed under
division. That is, when we divide one integer by
another we are not guaranteed to get another
integer. Sometimes we do get another integer, ex.
39 divided by 3 is 13. Other times we end up
with a rational number that is not an integer,
ex. 25 divided by 8 is 3.125. Other times we
dont even get a real number, ex. 5 divided by 0.
Def Let a and b be integers with a ? 0. We say
that a divides b if there is an integer c such
that b ac. When a divides b we say that a is a
factor of b and that b is a multiple of a. To
denote that a divides b we use a b. We slash
through the symbol to denote that a does not
divide b. Dont get confused, ab and a/b
Ex 4 divides 12 since 12 43. So 4 is a
factor of 12 and 12 is a multiple of 4. 5 does
not divide 12 since there is no integer c such
that 12 5c.
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Ex Let n and d be positive integers. How many
positive integers not exceeding n are divisible
by d? That is, how many members of the set 1,
2, , n are divisible by d?
The positive integers divisible by d are d
itself, 2d, 3d, So what were really asking
is, what is the largest positive integer k, such
that kd ? n. Or equivalently, such that k ?
n/d. Recall the floor function there are ?n/d?
positive integers not exceeding n that are
divisible by d.
Ex Based on the previous example, there are
?100/7? ?14.285714? 14 positive integers not
exceeding 100 that are divisible by 7. They are
7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84,
91, and 98.
Ex Does 11 divide 0?
Yes. 0 11c (where c 0) so 11 does divide 0.
Remark k divides 0 for any k ? Z since 0 k0
always.
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• Theorem Let a, b, and c be integers. Then
• If a b and a c, then a (b c). Division
  of Sum
• If a b, then a bx for all integers
  x. Division of Multiples
• If a b and b c, then a c. Transitivity

Proof (of 1) Let a, b, c ? Z such that a b
and a c.
Since a b, then a ? 0 and b ar for some r ?
Z. Since a c, then a ? 0 and c as for some s
? Z.
Now b c ar as a(r s).
So a b c since a ? 0 and b c a(r s),
where r s ? Z.?
Corollary If a, b, and c are integers such that
a b and a c, then a (mb nc) for all m, n
? Z.
Justification Let a, b, c ? Z such that a b
and a c. Then by (2) it follows that a mb
and a nc for all m, n? Z. Then apply (1).
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Consider a positive integer k. Since k 1k and
k k1 we can see that it is always the case
that 1 k and k k. That is, every positive
integer has at least 2 divisors, 1 and itself.
except 1 Some positive integers, such as 7,
have only these two positive integer divisors.
No other positive integer divides 7. Other
positive integers, such as 8, have additional
positive integer divisors. In addition to 1 and
8, both 2 and 4 also divide 8.
Def A positive integer p greater than 1 is
called prime if the only positive factors of p
are 1 and p. A positive integer greater than 1
that is not prime is called composite.
Remark A positive integer n gt 1 is composite if
and only if there exists an integer a such that a
n where 1 lt a lt n. Such an a is called a
proper factor of n. 1 and n are called trivial
factors of n.
Remark What we have done is to partition the
set of positive integers greater than 2 into two
disjoint sets, the primes (2, 3, 5, 7, 11, 13,
) and the composites (4, 6, 8, 9, 10, 12, 14,
15, ).
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Ex Is 57 prime?
No. 57 319 so 3 57. So 57 is not prime.
Also 19 57
Ex Is 59 prime?
Yes. We have to determine that no positive
integer n, 1 lt n lt 59, divides 59. If you try
this you will start to find shortcuts.
Ex The number 1 is neither prime nor composite.
Only positive integers greater than 1 are prime
or composite.
Ex The set of primes less than 100 is 2, 3, 5,
7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47,
53, 59, 61, 67, 71, 73, 79, 83, 89, 97. If you
take any of these numbers you will find that the
only two positive divisors of the number are 1
and itself.
Observation If a b then a ? b. When we
are looking for possible divisors of a number, we
only need to check integers of lesser magnitude.
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Theorem (Fundamental Theorem of Arithmetic)
Every positive integer greater than 1 can be
written uniquely as a prime or as the product of
two or more primes where the prime factors are
written in order of nondecreasing size.
Ex 100 2 2 5 5 2252 999 3
3 3 37 3337 59 59 1024 2
2 2 2 2 2 2 2 2 2 210
The unique factorization into a product of primes
for a number is called its prime factorization.
The unique factorization of a prime number is
simply the number itself. The unique
factorization of a composite number is a product
of two or more primes.
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Theorem If n is a composite integer, then n has
a prime divisor less than or equal to ?n.
Proof Let n be a composite integer.
Then n has a (proper) factor r with 1 lt r lt n.
That is, n rq for some integer r where 1 lt r lt
n.
It must be the case that 1 lt q lt n as well since
n rq.
In fact, either r or q must be ? ?n because if
both were greater than ?n then their product
would exceed n, which it doesnt.
So we have established that n has a proper factor
less than or equal to ?n (call it z).
Now apply the Fundamental Theorem of Arithmetic
to z to get its prime factorization. Each prime
factor in the prime factorization of z must be
less than or equal to z. Any one of these primes
is hence ? ?n. So n has a prime factor ? ?n.?
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Ex 1000 10 100. So 10 and 100 are factors
of 1000. The square root of 1000 is about 31.6.
And we see that 10 ? 31.6. 10 is not prime, but
its prime factorization is 25. So 5 is a prime
factor of 1000 not exceeding ?1000.
Ex 101 is prime. The square root of 101 is just
a bit over 10. So we can check only the prime
numbers not exceeding 10. We can see that none
of 2, 3, 5, or 7 divides 101. So 101 is prime
since it doesnt have a prime factor ? ?101.
Ex Find the prime factorization of 7007. 2 does
not divide 7007. 3 does not divide 7007. 5 does
not divide 7007. But 7007 71001. Now we
continue to factor 1001. Start with 7. Why?.
1001 7143. Now continue to factor 143. Start
with 7. 7 does not divide 143. But 143 1113.
So the prime factorization of 7007 771113.
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Preliminaries

• Reminder HW5 is due Wednesday
• Pass back HW4
• average 90, median 91
• Emmanuel graded 1.5 27, 30, and 64 (green)
• Pascal graded the rest -- S1.6, S1.7 (red)
• http//www.cs.virginia.edu/cmt5n/cs202/hw4/

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Theorem For any positive integer n, there is a
prime number greater than n.
Proof Let n be a positive integer.
Consider the number n! 1.
Recall n! 1 123n 1.
By the FTOA, this number has a unique prime
factorization.
It turns out that none of the numbers 2, , n can
divide n! 1. This is because we know any such
number divides n!. So if that number also
divided n! 1 then it would have to also divide
the sum (-1)(n!) 1(n! 1) -(n!) n! 1
1. (Recall if a b and a c, then a (mb
nc) for all m, n ? Z). But none of these
numbers divides 1. Hence no number 2, , n can
divide n! 1.
So all of the factors in the prime factorization
of n! 1 must be greater than n. Hence there
must be a prime greater than n.?
Corollary There are infinitely many primes.
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Theorem (The Division Algorithm) Let a, d ? Z
with d gt 0. Then there are unique integers q and
r, with 0 ? r lt d, such that a dq r. We call
d the divisor, a the dividend, q the quotient,
and r the remainder. This is what we do in long
division
Ex Let a 101 and d 11. The unique q and r
which satisfy the division algorithm are q 9
and r 2 since 101 119 2.
Ex Let a -11 and d 3. The unique q and r
which satisfy the division algorithm are q -4
and r 1 since -11 3(-4) 1.
It is tempting to say q -3 and r -2 since -11
3(-3) (-2) however this r does not satisfy 0
? r lt d.
Remark The integer a is divisible by the
integer d if and only if the remainder r is zero
when a is divided by d.
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Def Let a and b be integers, not both 0. The
largest integer d such that d a and d b is
called the greatest common divisor of a and b.
We denote this as gcd(a, b).
Remark The greatest common divisor of two
integers which are not both zero exists because
this set of common divisors is always finite and
non-empty. This is due to the fact that the set
of divisors of a non-zero integer is always
finite and non-empty.
Ex Let a 24 and b 36. The set of positive
divisors of 24 is 1, 2, 3, 4, 6, 8, 12, 24 and
for 36 is 1, 2, 3, 4, 6, 9, 12, 18, 36. So the
set of common divisors of 24 and 36 is the
intersection of these two sets 1, 2, 3, 4, 6,
12. So gcd is 12.
Ex Let a 17 and b 22. Then the set of
positive divisors of 17 is 1, 17 and for 22 is
1, 2, 11, 22. So the set of common divisors of
17 and 22 is 1. So gcd is 1.
There is an efficient algorithm (called the
Euclidean algorithm) for finding the gcd of two
numbers. It is discussed in Section 2.5.
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Def The integers a and b are relatively prime
if gcd(a, b) 1.
Ex We saw that 17 and 22 are relatively prime.
Ex 35 and 49 are not relatively prime since
gcd(35, 49) 7.
Def The integers a1, a2, , an are pairwise
relatively prime if gcd(aj, ak) 1 whenever 1 ?
i lt j ? n. That is, the integers are pairwise
relatively prime if every pair of them is
relatively prime.
Ex The integers 10, 19, and 24 are not pairwise
relatively prime since gcd(10, 24) 2.
Ex The integers 10, 19, and 21 are pairwise
relatively prime since gcd(10, 19) 1, gcd(10,
21) 1, and gcd(19, 21) 1.
Note that a group of integers is pairwise
relatively prime if and only if there is no
positive integer greater than 1 which divides
more than one of the integers in the group.
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We can take advantage of the prime factorizations
of two integers to find their gcd. Of course
this method will only apply if both integers are
nonzero since 0 has no prime factorization.
Ex 120 23 3 5 and 500 22 53. What is
gcd(120, 500)?
gcd(120, 500) 22 5 20. We simply take the
minimum of the exponents that each prime is
raised to in the prime factorization.
If we had taken the maximum of the exponents that
each prime is raised two then we would get the
least common multiple. Def The least common
multiple of two positive integers a and b is the
smallest positive integer that is divisible by
both a and b.
Ex lcm(120, 500) 23 3 53 3000.
Theorem Let a and b be positive integers.
Then ab gcd(a, b) lcm(a, b).
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Def Let a, b, m ? Z with m gt 0. Then a is
congruent to b modulo m if m divides a b. We
use the notation a ? b (mod m) to indicate that a
is congruent to b modulo m.
Remark Dont confuse this notation with the
notation a mod m which denotes the remainder when
a is divided by m. The following theorem
indicates the connection between the two.
Theorem Let a, b, m ? Z with m gt 0. Then a ? b
(mod m) if and only if a mod m b mod m. That
is, two integers are congruent modulo m if and
only if they have the same remainder when divided
by m.
Ex 17 ? 5 (mod 6) by the definition since 6
(17 5). We also see that 17 and 5 both leave
the same remainder when divided by 6. So by the
previous theorem we also see 17 ? 5 (mod 6).
Ex 24 is not congruent to 14 modulo 6. We see
that 6 does not divide (24 14) 10. Further,
24 leaves a remainder of 0 when divided by 6
while 14 leave a remainder of 2 when divided by 6.
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Theorem Let a, b, m ? Z with m gt 0. Then a is
congruent to b modulo m if and only if there is
an integer k such that a b km.
Proof Let a, b, m ? Z with m gt 0. (?) Assume
that m (a b). Then a b mk for some k ?
Z. So a b mk for some integer k. (?) Assume
that there exists an integer k such that a b
km. Then a b km, so m (a b).? We could
have shown 1 ? 2 ? 3 ? 1 to show the 3 equivalent
Def Let a, m ? Z with m gt 0. The set of all
integers congruent to a modulo m is called the
congruence class of a modulo m.
Ex The congruence class of 2 modulo 6 is
, -10, -4, 2, 8, 14,
6k 2 k ? Z .
Ex The congruence class of 0 modulo 8 is
, -16, -8, 0, 8, 16,
8k 0 k ? Z . all multiples of 8
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Theorem Let a, b, c, d, m ? Z with m gt 0. If a
? b (mod m) and c ? d (mod m) then (1) a c ? b
d (mod m) (2) ac ? bd (mod m)
Ex If it is 1700 hours now, what time will it
be in 60 hours?
(17 60) mod 24 77 mod 24 5. So it will be
0500 hours.
We could have applied the above theorem instead
to find that since 17 ? 17 (mod 24) and 60 ? 12
(mod 24) then (17 60) ? (17 12) (mod 24). So
instead of computing (17 60) mod 24, we can use
(17 12) mod 24 29 mod 24 5.