Download MTH280 Sections 3.1 - 3.2 Revised: Fall 2007 First, we need to

MTH280
Sections 3.1 - 3.2
Revised: Fall 2007
First, we need to establish what background knowledge we may use in doing proofs
throughout the quarter. We will assume the following. Retain the list for reference. I will
add to the list as needed. If in doubt about what may be assumed, please ask me. When
using one of these to justify a step in a proof, use its name, if it has one. Otherwise, refer to
it by number and a brief description (e.g. ABK 7–mult. by a neg. number)
Assumed Background Knowledge (ABK)
1. Arithmetic (Examples: 1 + 1 = 2; “minus times minus is plus” and other rules of
signs; dividing is the same as multiplying by the reciprocal; subtracting is adding the
opposite; common denominators; . . . )
2. Commutative and associative properties of addition and multiplication of real numbers;
distributive property of real numbers
3. Adding, subtracting, or multiplying both sides of an equality by the same real number
preserves equality.
4. Dividing both sides of an equality by the same nonzero real number preserves equality.
5. Adding or subtracting both sides of an inequality by the same real number preserves
the inequality.
6. Multiplying or dividing both sides of an inequality by the same positive real number
preserves the inequality. (Example: If x < 2, then 3x < 6.)
7. Multiplying or dividing both sides of an inequality by the same negative real number
reverses the inequality. (Example: If x < 2, then −3x > −6.)
8. Transitivity of inequality (Example: If a ≥ b and b ≥ c, then a ≥ c.)
9. Law of Trichotomy. If a and b are real numbers, then exactly one of the following is
true: a = b, a < b, or a > b.
10. Closure. C, R, Q, Z are closed under addition, subtraction, multiplication, and division
by nonzero numbers. N is closed under addition and multiplication. (Example: If a
and b are real numbers, then a + b is a real number.)
11. If a is a real number and n is an odd natural number, then a has a unique nth root in
the real numbers.
Homework
1. Let’s do some proofs from vector algebra.
First, the definitions.
Definition 1 R3 := {< x, y, z >: x, y, z ∈ R}.
Definition 2 For every < a, b, c >, < u, v, w > in
R3 and s ∈ R, we define
i. < a, b, c > + < u, v, w > to be < a + u, b + v, c + w >;
ii. < a, b, c > · < u, v, w > to be au + bv + cw; and
iii. s < a, b, c > to be < sa, sb, sc >.
Prove (outline form is OK) each of the following theorems.1 In doing the proofs, you
may assume the familiar properties of real numbers (e.g., commutative property of
addition, distributive property). Each time you use one such property, you must cite
it by name.
(a) Suppose < p, q, r >∈ R3 and < f, g, h >∈ R3 . Then
< p, q, r > · < f, g, h >=< f, g, h > · < p, q, r >.
(b) Suppose < p, q, r >∈ R3 , < f, g, h >∈ R3 and t ∈ R. Then
t (< p, q, r > + < f, g, h >) = t < p, q, r > +t < f, g, h >.
(c) Suppose < p, q, r >∈ R3 , < f, g, h >∈ R3 and t ∈ R. Then
(t < p, q, r >) · < f, g, h >=< p, q, r > · (t < f, g, h >).
2. Basic proofs: Do Section 3.1, problems 5, 6, 8, 9, 10
3. More proofs and other matters: Do Section 3.1, problems 1, 3, 15, 16
4. Below are some conjectures. Find counterexamples for as many of them as you can.
(Some of them may be true statements. The true ones, of course, won’t have counterexamples.)
(a) If n ∈ N, then 3n − 1 is not prime.
(b) If n ∈ N with n ≥ 2, then 3n − 1 is not prime.
(c) If n ∈ N where n is prime, then 3n − 1 is prime.
(d) If n ∈ N where n is prime, then 3n − 2n is prime.
(e) If n ∈ N where n > 0 and n is not prime, then 3n − 1 is not prime.
(f) If n ∈ N and n is not a multiple of 5, then 3n − 2n is prime.
5. Do Section 3.2, problems 3, 4, 5, 7, 9, 10, 11
1
These theorems, by the way, are stated in any treatment of vector algebra. For example, they are in
multivariable calculus textbooks. The proofs in calculus books, however, are sometimes not very good.
MTH280
Sections 3.1 - 3.2: Partial Solutions
Composed: Spring 2009
1. Section 3.1 Problem 5
Suppose a and b are real numbers. Prove that if a < b < 0, then a2 > b2 .
Proof outline.
a and b are real numbers
Assume a < b < 0
a2 > ab
premise/hypothesis/given
begin the basic if-then strategy
a is negative by hypothesis. Multiply
both sides of a < b by a. The inequality
is reversed, see ABK 7.
ab > b2
b is negative by hypothesis. Multiply
both sides of a < b by b. The inequality
is reversed, see ABK 7.
a2 > b2
by transitivity of inequality, ABK 8,
applied to the previous two inequalities
Therefore, if a < b < 0, then a2 > end of basic if-then strategy
b2
2. Section 3.1 Problem 6
Suppose a and b are real numbers. Prove that if 0 < a < b, then 1/b > 1/a.
Proof outline.
a and b are real numbers
Assume 0 < a < b
a/(ab) < b/(ab)
premise/hypothesis/given
begin the basic if-then strategy
a and b are both positive by hypothesis, so ab > 0 by arithmetic (ABK 1).
Divide both sides of a < b by ab. The
inequality is preserved, see ABK 6.
1/b < 1/a
reduce the fractions in the previous assertion (ABK 1)
Therefore, if 0 < a < b, then 1/b > end of basic if-then strategy
1/a.
3. Section 3.1, Problem 8
Suppose A\B ⊆ C ∩ D and x ∈ A. Prove that if x ∈
/ D, then x ∈ B.
Proof outline.
A\B ⊆ C ∩ D
premise/hypothesis/given
x∈A
premise/hypothesis/given
We’ll prove the contrapositive: if contrapositive strategy
x∈
/ B, then x ∈ D.
Assume x ∈
/B
begin if-then strategy (applied to contrapositive)
x ∈ A\B
by the def. of A\B since we already
have x ∈ A (hyp.) and x ∈
/ B (previous
line)
x∈C ∩D
by the def. of ⊆ (since A\B ⊆
C ∩ D by hyp. and x ∈ A\B from
previous line)
x ∈ C and x ∈ D
def. of ∩
In particular, x ∈ D
(just emphasizing the part of the previous line that we need)
Therefore, if x ∈
/ B, then x ∈ D.
end of basic if-then strategy (applied to
contrapositive)
Therefore, if x ∈
/ D, then x ∈ B.
(Contrapositive)
4. Section 3.1 Problem 3
Solution.
The hypotheses (premises) are: n is a natural number greater than 2 and n is not
prime.
The conclusion is: 2n + 13 is not a prime number.
A counterexample is n = 8. The hypotheses are true in this instance because 8 is a
natural number greater than 2 and 8 is not prime. The conclusion is false because
2 · 8 + 13 = 29 which is a prime number. Therefore, 8 is a counterexample.
5. Section 3.1 Problem 15(b) Theorem. Suppose x is a real number and x 6= 4. If
2x−5
= 3, then x = 7.
x−4
Correct proof.
x is a real number and x 6= 4
Assume 2x−5
=3
x−4
2x − 5 = 3(x − 4)
2x − 5 = 3x − 12
−x − 5 = −12
−x = −7
x=7
Therefore, if 2x−5
= 3, then x = 7.
x−4
premise/hypothesis/given
begin if-then strategy
multiply both sides by x − 4, ABK 3
distributive property ABK 2
subtract 3x from each side, ABK 3
add 5 to both side ABK 3
multiply both sides by -1 ABK 3
end if-then strategy
Note: It would be acceptable to combine some of the steps in the proof that just involve
the simple algebra from ABK 3.
6. Below are some conjectures. Find counterexamples for as many of them as you can.
(Some of them may be true statements. The true ones, of course, won’t have counterexamples.)
First note than none of them have premises and each is an if-then statement. Therefore,
a counterexample is an instance that makes the antecedent true and the consequent
false.
(a) If n ∈ N, then 3n − 1 is not prime.
Counterexample. n = 1 is a counterexample because 2 ∈ N (so the antecedent is
true), but 31 − 1 = 2 is prime (so the consequent is false).
(b) If n ∈ N with n ≥ 2, then 3n − 1 is not prime.
There are no counterexamples for this one!
(c) If n ∈ N where n is prime, then 3n − 1 is prime.
Counterexample. n = 2 is a counterexample because 2 ∈ N and 2 is prime (so the
antecedent is true), but 32 − 1 = 8 is not prime (so the consequent is false).
(d) If n ∈ N where n is prime, then 3n − 2n is prime.
Counterexample. n = 7 is a counterexample because 7 ∈ N and 7 is prime (so the
antecedent is true), but 37 − 27 = 2059 = 29 · 71 is not prime (so the consequent
is false).
(e) If n ∈ N where n > 0 and n is not prime, then 3n − 1 is not prime.
Counterexample. n = 1 is a counterexample because 1 ∈ N, 1 > 0, and 1 is not
prime (so the antecedent is true), but 31 − 1 = 2 is prime (so the consequent is
false).
(f) If n ∈ N and n is not a multiple of 5, then 3n − 2n is prime.
Counterexample. n = 7 is a counterexample because 7 ∈ N and 7 is not a multiple
of 5 (so the antecedent is true), but 37 − 27 = 2059 = 29 · 71 is not prime (so the
consequent is false).
7. Section 3.2, Problem 3
Suppose A ⊆ C and B and C are disjoint. Prove that if x ∈ A, then x ∈
/ B.
Proof outline.
A ⊆ C.
B and C are disjoint.
Assume x ∈ A
By way of contradiction, assume x ∈ B.
x ∈ C.
Thus, x ∈ C and x ∈ B.
Contradiction.
Therefore, x ∈
/ B.
Therefore, if x ∈ A, then x ∈
/ B.
premise/hypothesis/given
premise/hypothesis/given
begin if-then strategy
begin contradiction strategy
by the def. of ⊆ (since A ⊆ C by hyp.
and x ∈ A from earlier)
reiterating the two previous lines
B and C are disjoint by hyp. but the
previous lines shows they are not disjoint
end of contradiction strategy
end of if-then strategy
8. Section 3.2 Problem 7
Suppose y + x = 2y − x, and x and y are not both zero. Prove that y 6= 0.
Beware the difference between “not both zero” and ”both not zero.”
Proof outline.
y + x = 2y − x
x and y are not both zero
Assume y = 0
0+x=2·0−x
x = −x
2x = 0
x=0
Contradiction
Therefore, y 6= 0.
premise/hypothesis/given
premise/hypothesis/given
begin contradiction strategy
substitute y = 0 into the premise
simplify ABK 1
add x to each side, ABK 3
divide both sides by 2 which is OK b/c
2 6= 0, ABK 4
We have that y = 0 and x = 0, but the
premise is that they are not both 0
contradiction strategy
9. Section 3.2 Problem 9
Suppose x and y are real numbers. Prove that if x2 y = 2x + y, then if y 6= 0, then
x 6= 0.
Two options are given for this one.
Proof outline using contradiction.
Suppose x and y are real numbers.
Assume x2 y = 2x + y.
Assume y 6= 0
By way of contradiction, assume
x = 0.
02 · y = 2 · 0 + y
0=y
Contradiction
Thus, x 6= 0.
Therefore, if y 6= 0, then x 6= 0.
Therefore, if x2 y = 2x + y, then if
y 6= 0, then x 6= 0.
premise/hypothesis/given
begin the first if-then strategy
begin the second if-then strategy
begin contraction strategy
substitute x = 0 into the assumption
simplify ABK 1
We have that y = 0 and y 6= 0.
contradiction strategy
end of second if-then strategy
end of first if-then strategy
Proof outline using contrapositive.
Suppose x and y are real numbers.
Assume x2 y = 2x + y.
We’ll prove the contrapositive: if
x = 0, then y = 0.
Assume x = 0
02 · y = 2 · 0 + y
0=y
Thus, if x = 0, then y = 0.
Therefore, if y 6= 0, then x 6= 0.
Therefore, if x2 y = 2x + y, then if
y 6= 0, then x 6= 0.
premise/hypothesis/given
begin the if-then strategy
contrapositive strategy
begin if-then strategy (applied to contrapositive)
substitute x = 0 into the assumption
simplify ABK 1
end if-then strategy (applied to contrapositive)
contrapositive strategy
end of if-then strategy