Division into Cases and the Quotient

```Discrete Structures
Chapter 4: Elementary Number Theory and Methods of
Proof
4.4 Direct Proof and Counter Example IV: Division into
Cases and the Quotient Remainder Theorem
Be especially critical of any statement following the word “obviously”.
– Anna Pell Wheeler, 1883-1966
4.4 Direct Proof and Counter Example IV:
Division into Cases and the Quotient
Remainder Theorem
1
Theorem 4.4.1 – The QuotientRemainder Theorem
Given any integer n and positive integer d, 
unique integers q and r s.t.
n = dq + r
and 0  r < d
4.4 Direct Proof and Counter Example IV:
Division into Cases and the Quotient
Remainder Theorem
2
Definitions
• Given an integer n and a positive integer d,
n div d = the integer quotient obtained when n is
divided by d.
n mod d = the nonnegative integer remainder obtained
when n is divided by d.
Symbolically, if n and d are integers and d > 0, then
n div d = q and n mod d = r  n = dq + r
Where q and r are integers and 0  r < d.
4.4 Direct Proof and Counter Example IV:
Division into Cases and the Quotient
Remainder Theorem
3
Example – pg. 189 # 8 & 9
• Evaluate the expressions.
8. a. 50 div 7
b. 50 mod 7
9. a. 28 div 5
b. 28 mod 5
4.4 Direct Proof and Counter Example IV:
Division into Cases and the Quotient
Remainder Theorem
4
Theorems
• Theorem 4.4.2 – The Parity Property
Any two consecutive integers have opposite
parity.
• Theorem 4.4.3
The square of any odd integer has the form
8m + 1 for some integer m.
4.4 Direct Proof and Counter Example IV:
Division into Cases and the Quotient
Remainder Theorem
5
Definition
• For any real number x, the absolute value of
x, is denoted |x|, is defined as follows:
 x if x  0
x 
  x if x  0
4.4 Direct Proof and Counter Example IV:
Division into Cases and the Quotient
Remainder Theorem
6
Lemmas
• Lemma 4.4.4
For all real numbers r, -|r|  r  |r|
• Lemma 4.4.5
For all real numbers r, |-r| = |r|
• Lemma 4.4.6 – The Triangle Inequality
For all real numbers x and y, |x + y|  |x| + |y|
4.4 Direct Proof and Counter Example IV:
Division into Cases and the Quotient
Remainder Theorem
7
Example – pg. 189 # 24
• Prove that for all integers m and n, if
m mod 5 = 2 and n mod 5 = 1, then
mn mod 5 = 2.
4.4 Direct Proof and Counter Example IV:
Division into Cases and the Quotient
Remainder Theorem
8
Example – pg. 190 #36
• Prove the following statement.
The product of any four consecutive integers is
divisible by 8.
4.4 Direct Proof and Counter Example IV:
Division into Cases and the Quotient
Remainder Theorem
9
Example – pg. 190 #42
• Prove the following statement.
Every prime number except 2 and 3 has the
form 6q + 1 or 6q + 5 for some integer q.
4.4 Direct Proof and Counter Example IV:
Division into Cases and the Quotient
Remainder Theorem
10
Example – pg. 190 #45
• Prove the following statement.
For all real numbers r and c with c  0, if
-c  r  c, then |r|  c.
4.4 Direct Proof and Counter Example IV:
Division into Cases and the Quotient
Remainder Theorem
11
Example – pg. 190 # 49
• If m, n, and d are integers, d > 0, and
m mod d = n mod d, does it necessarily follow
that m = n? That m – n is divisible by d?