1. Direct Proof and Counterexample IV
Lecture 17
Section 3.4
Wed, Feb 14, 2007

2. The Quotient-Remainder Theorem
Theorem: Let n and d be integers, d  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.

3. The C Divide and Mod Operators
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.

4. The C Divide and Mod Operators
Thus,
a = (a/b)*b + (a % b)
is true for all positive integers a and b.
Therefore,
a % b = a – (a/b)*b
for all positive integers a and b.

5. The C Divide and Mod Operators
What are a/b and a % b when a or b is negative? Is
a = (a/b)*b + (a % b)
still true when a or b is negative?

6. Example: QuotientRemainder.cpp

7. Example: Proof by Cases
Theorem: For any integer n, n3 – n is a multiple of 6.
Proof:
Divide n by 6 to get q and r: n = 6q + r, where 0  r < 6.
That is, r = 0, 1, 2, 3, 4, or 5.
Substitute: n3 – n = (6q + r)3 – (6q + r).

8. Proof continued Expand and rearrange: n3 – n =
6(36q3 + 18q2r + 3qr2 – q) + (r3 – r).
Therefore, 6 | (n3 – n) if and only if | (r3 – r).

9. Proof continued Consider the 6 possible cases:
Case 1: r = 0. r3 – r = 03 – 0 = 0 = 60.

10. Proof continued Consider the 6 possible cases:
Case 1: r = 0. r3 – r = 03 – 0 = 0 = 60.
Case 2: r = 1. r3 – r = 13 – 1 = 0 = 60.

11. Proof continued Consider the 6 possible cases:
Case 1: r = 0. r3 – r = 03 – 0 = 0 = 60.
Case 2: r = 1. r3 – r = 13 – 1 = 0 = 60.
Case 3: r = 2. r3 – r = 23 – 2 = 6 = 61.

12. Proof continued Consider the 6 possible cases:
Case 1: r = 0. r3 – r = 03 – 0 = 0 = 60.
Case 2: r = 1. r3 – r = 13 – 1 = 0 = 60.
Case 3: r = 2. r3 – r = 23 – 2 = 6 = 61.
Case 4: r = 3. r3 – r = 33 – 3 = 24 = 64.

13. Proof continued Consider the 6 possible cases:
Case 1: r = 0. r3 – r = 03 – 0 = 0 = 60.
Case 2: r = 1. r3 – r = 13 – 1 = 0 = 60.
Case 3: r = 2. r3 – r = 23 – 2 = 6 = 61.
Case 4: r = 3. r3 – r = 33 – 3 = 24 = 64.
Case 5: r = 4. r3 – r = 43 – 4 = 60 = 610.

14. Proof continued Consider the 6 possible cases:
Case 1: r = 0. r3 – r = 03 – 0 = 0 = 60.
Case 2: r = 1. r3 – r = 13 – 1 = 0 = 60.
Case 3: r = 2. r3 – r = 23 – 2 = 6 = 61.
Case 4: r = 3. r3 – r = 33 – 3 = 24 = 64.
Case 5: r = 4. r3 – r = 43 – 4 = 60 = 610.
Case 6: r = 5. r3 – r = 53 – 5 = 120 = 620.

15. Proof continued Therefore, 6 | (r3 – r) in general.
In every case, 6 | (r3 – r).
Therefore, 6 | (r3 – r) in general.
Therefore, 6 | (n3 – n) for all integers n.

16. max(x, y) = ((x + y) + |x – y|)/2.
Example: Max(x, y)
Theorem: For all real numbers x and y,
max(x, y) = ((x + y) + |x – y|)/2.
Proof:

17. max(x, y) = ((x + y) + |x – y|)/2.
Example: Max(x, y)
Theorem: For all real numbers x and y,
max(x, y) = ((x + y) + |x – y|)/2.
Proof:
Consider two cases…

18. Other Formulas What is a similar formula for min(x, y)?
What is a formula for max(x, y, z)?