Direct Proof – Division into Cases
Lecture 16
Section 4.4
Robb T. Koether
Hampden-Sydney College
Mon, Feb 11, 2013
Robb T. Koether (Hampden-Sydney College)
Direct Proof – Division into Cases
Mon, Feb 11, 2013
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1
The Quotient-Remainder Theorem
2
The Operators / and % in C
3
Proof by Cases
4
Leap Years
5
Assignment
Robb T. Koether (Hampden-Sydney College)
Direct Proof – Division into Cases
Mon, Feb 11, 2013
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Outline
1
The Quotient-Remainder Theorem
2
The Operators / and % in C
3
Proof by Cases
4
Leap Years
5
Assignment
Robb T. Koether (Hampden-Sydney College)
Direct Proof – Division into Cases
Mon, Feb 11, 2013
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The Quotient-Remainder Theorem
Theorem
Let n and d be integers, d 6= 0. Then there exist unique integers q and
r such that
n = qd + r
and
0 ≤ r < |d|.
q is the quotient and r is the remainder.
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Direct Proof – Division into Cases
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Outline
1
The Quotient-Remainder Theorem
2
The Operators / and % in C
3
Proof by Cases
4
Leap Years
5
Assignment
Robb T. Koether (Hampden-Sydney College)
Direct Proof – Division into Cases
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The Operators / and % in C
The operators / and % in C are based on this theorem.
If a and b are positive integers, then
q = a/b;
and
r = a % b;
where 0 ≤ r < b.
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Direct Proof – Division into Cases
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The Operators / and % in C
Thus,
a == (a/b)*b + (a % b)
is true for all positive integers a and b.
Therefore,
a % b == a - (a/b)*b
is true for all positive integers a and b.
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Direct Proof – Division into Cases
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The Operators / and % in C
What are a/b and a % b when a or b is negative?
C requires that the expression
a % b == a - (a/b)*b
be true for all integers (where b != 0)).
Certainly,
a/b < 0 if a < 0 and b > 0 or if a > 0 and b < 0.
a/b > 0 if a < 0 and b < 0.
Thus, the value of a % b is determined by the expression
a % b == a - (a/b)*b
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Direct Proof – Division into Cases
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The Operators / and % in C
Determine the values of the following expressions.
52 % 11
-52 % 11
52 % -11
-52 % -11
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Direct Proof – Division into Cases
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Outline
1
The Quotient-Remainder Theorem
2
The Operators / and % in C
3
Proof by Cases
4
Leap Years
5
Assignment
Robb T. Koether (Hampden-Sydney College)
Direct Proof – Division into Cases
Mon, Feb 11, 2013
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Proof by Cases
Theorem
If n is odd, then 8 | (n2 − 1).
Proof.
Let n be an odd integer.
Then n = 8q + r for some integers q and r with 0 6= r < 8.
First, note that
n2 − 1 = (8q + r )2 − 1
= 64q 2 + 16qr + r 2 − 1
= 8(8q 2 + 2qr ) + (r 2 − 1).
Robb T. Koether (Hampden-Sydney College)
Direct Proof – Division into Cases
Mon, Feb 11, 2013
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Proof by Cases
Proof.
Thus we need only show that 8 | (r 2 − 1).
Because n is odd, we know that r = 1, 3, 5 or 7.
We will consider these four cases.
Case 1:
Case 2:
Case 3:
Case 4:
If r
If r
If r
If r
= 1, then r 2 − 1 = 1 − 1 = 0 = 8 · 0.
= 3, then r 2 − 1 = 9 − 1 = 8 = 8 · 1.
= 5, then r 2 − 1 = 25 − 1 = 24 = 8 · 3.
= 7, then r 2 − 1 = 49 − 1 = 48 = 8 · 6.
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Direct Proof – Division into Cases
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Proof by Cases
Proof.
In every case, 8 | (r 2 − 1).
Therefore, 8 | (n2 − 1) for all odd integers n.
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Direct Proof – Division into Cases
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Proof by Cases
Theorem
For any odd integer n, n3 − n is a multiple of 12.
Proof.
Let n be an odd integer.
Then n = 12q + r for some integers q and r with 0 6= r < 12.
First, note that
n3 − n = (12q + r )3 − (12q + 4)
= (123 q 3 + 3 · 122 q 2 r + 3 · 12qr 2 + r 3 ) − (12q + r )
= 12(122 q 3 + 3 · 12q 2 r + 3qr 2 − q) + (r 3 − r ).
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Direct Proof – Division into Cases
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Proof by Cases
Proof.
Thus we need only show that 12 | (r 3 − r ).
Because n is odd, we know that r = 1, 3, 5, 7, 9 or 11.
We will consider these six cases.
Case 1:
Case 2:
Case 3:
Case 4:
Case 5:
Case 6:
If r
If r
If r
If r
If r
If r
= 1, then r 3 − r = 1 − 1 = 0 = 12 · 0.
= 3, then r 3 − r = 27 − 3 = 24 = 12 · 2.
= 5, then r 3 − r = 125 − 5 = 120 = 12 · 10.
= 7, then r 3 − r = 343 − 7 = 336 = 12 · 28.
= 9, then r 3 − r = 729 − 9 = 720 = 12 · 60.
= 11, then r 3 − r = 1331 − 11 = 1320 = 12 · 120.
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Direct Proof – Division into Cases
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Proof by Cases
Proof.
In every case, 12 | (r 3 − r ).
Therefore, 12 | (n3 − n) for all odd integers n.
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Direct Proof – Division into Cases
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Outline
1
The Quotient-Remainder Theorem
2
The Operators / and % in C
3
Proof by Cases
4
Leap Years
5
Assignment
Robb T. Koether (Hampden-Sydney College)
Direct Proof – Division into Cases
Mon, Feb 11, 2013
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Leap Years
A year y is a leap year if
y is a multiple of 4, but not a multiple of 100 (a century year), or
y is a multiple of 400, but not a multiple of 100.
Write an expression that will be true if and only if y is a leap year.
Today (Feb 11, 2013) is a Monday.
What day of the week will be Feb 11, 2113?
Robb T. Koether (Hampden-Sydney College)
Direct Proof – Division into Cases
Mon, Feb 11, 2013
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Outline
1
The Quotient-Remainder Theorem
2
The Operators / and % in C
3
Proof by Cases
4
Leap Years
5
Assignment
Robb T. Koether (Hampden-Sydney College)
Direct Proof – Division into Cases
Mon, Feb 11, 2013
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Assignment
Assignment
Read Section 4.4, pages 180 - 189.
Exercises 1, 2, 5, 6, 14, 15, 19, 21, 23, 29, 36, 41, 47, 48, page
189.
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Direct Proof – Division into Cases
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