Homework #10 – 02/10/03
Chapter 3, Section 4
8.)
(a)
37 div 10 = 3
(b)
37 mod 10 = 7
13.)
Use: DayN = (DayT + N) mod 7
Let DayT = 1 and N = 30
DayT = (1 + 30) mod 7
= 31 mod 7
=3
∴ , Your friend will meet you on a Wednesday
Chapter 3, Section 5
2.)
10 / 3 = 3.3 3  = 3
10 / 3 = 3.3 3  = 4
5.)
Use the floor notation to express 589 div 12 and 589 mod 12.
589 div 12 = 589 / 12
= 49.08
= 49
589 mod 12 = 589 – 12 * 589 / 12
= 589 – 12 * 49
= 589 – 588
=1
12.)
Prove that if n is any even integer, then n / 2 = n / 2
Proof: Let n be any even integer. Then by the definition of even n = 2k.
Then,
n / 2 = 2k / 2
= k 
=k
by substitution
by simplification
since k is an integer and k ≤ k < k + 1 by
definition of floor.
We know that since n = 2k ⇒ k = n / 2. n / 2 = k ⇒ n / 2 = n / 2.
∴ , If n is any even integer, then n / 2 = n / 2.
Homework #10 – 02/10/03
16.)
Prove for all real numbers x, x 2  = x  2.
By counterexample: suppose x = 1.5:
x 2  = 1.5 2  = 2.25
2
2
2
x  = 1.5 = 1 = 1
2.25 is not equal to 1, so the statement is not true for all x.
Chapter 3, Section 6
1.)
Proof:
Suppose not. Suppose that there is a real number x such that x is positive
and x ≤ y for all positive real numbers y. Consider the number x / 2. Then x / 2
is a positive real number because x is positive and x / 2 < x because if we
multiplying both sides of the inequality 1 / 2 < 1 by x, which is positive, gives
x / 2 < x. Hence, x / 2 is a positive real number that is less than x, which is a
contradiction. [Thus the supposition is false, and so there is no least positive real
number.]
3.)
Prove there is no greatest negative real number.
Proof (by contradiction):
Negation of statement: There is a greatest negative real number
Suppose not. Suppose there is a greatest negative real number, N. Then N
is a negative real number and N ≥ n for every negative real number n. [Deduce a
contradiction]. Let M = N/ 2, then M is a negative real number because it is a sum
of real numbers, and a negative divided by a positive is negative. Also, M > N
since M = N/2 and N is negative. This contradicts the supposition that N ≥ n for
every negative real number n. Hence, the supposition is false and the original
statement is true.
6.)
Prove by contradiction that the difference of any rational number and any
irrational number is irrational.
Proof (by contradiction):
Suppose not. Suppose that there exists a rational number x and any
irrational number y such that x – y is rational. By definition of rational, x = a / b
and x – y = c / d for some integers a, b, c, and d where b and d both ≠ 0. Then by
substitution,
c/d =
(a / b) – y
(c / d) – (a / b) =
–y
(by subtracting a / b from both sides)
(a / b) – (c / d) =
y
(by multiplying each side by –1)
Homework #10 – 02/10/03
[(a * d – c * b)] / (b * d) =
y
(by getting a common denominator)
Since a, b, c, and d are integers, the product and difference of integers is
an integer, and b * d ≠ 0. Therefore, y is rational, by definition of rational.
Hence, the supposition is false and the original statement it true.
7.)
Prove if a product of two positive real numbers is greater than 100, then at least
one of the numbers is greater than 10.
Proof (by contraposition):
[It must be shown that for every positive real number r and s, that if r ≤
10 and s ≤ 10, then r * s ≤ 100.] Let use suppose that r and s are positive real
numbers and r ≤ 10 and s ≤ 10.
r ≤ 10, multiply both sides by s to get r * s ≤ 10 * s
s ≤ 10, multiply both sides by 10 to get 10 * s ≤ 10 * 10 = 100
By the transitivity of ≤ , r * s = 100. This is what was to be shown.
20.)
Prove or disprove that the difference of any two irrational numbers is irrational.
Counterexample:
Suppose a = π and b = π, then,
a–b=π–π
=0
which is a rational number (i.e., 0 = 0 / 1)
∴ , The difference of any two irrational numbers will not always be
irrational.