How To Prove It by Velleman

Prove It by Velleman, Section 3.1, Problem 10

January 10, 2015 etoix Leave a comment

Problem

Suppose x is a real number and $x \neq 0$ . Prove that if $\frac{\sqrt[x]{x} + 5}{x+6} = \frac{1}{x}$ then $x \neq 8$ .

Solution

Given:

x is real
x \neq 0
$\frac{\sqrt[x]{x} + 5}{x+6} = \frac{1}{x}$

Goal:

$x \neq 8$

$\frac{\sqrt[x]{x} + 5}{x+6} = \frac{1}{x}$

$x^{4/3} + 5x = x + 6$

$x^{4/3} + 4x = 6$

If x were 8, then the left side would be 48.

It might be easier to prove the contrapositive.

Form:

Supppose $x \neq 8$ is false.

— Proof that $\frac{\sqrt[x]{x} + 5}{x+6} \neq \frac{1}{x}$ goes here.

Therefore, if $\frac{\sqrt[x]{x} + 5}{x+6} = \frac{1}{x}$ then $x \neq 8$ .

Proof: We will prove the contrapositive. Multiplying both sides by (x + 6) and x, we get $x^{4/3} + 4x \neq 6$ . If x = 8, we conclude that $48 \neq 6$ . Thus, if $\frac{\sqrt[x]{x} + 5}{x+6} = \frac{1}{x}$ then $x \neq 8$ .

How To Prove It by Velleman

Prove It by Velleman, Section 3.1, Problem 9

January 9, 2015 etoix Leave a comment

Problem

Suppose a and b are real numbers. Prove that if a < b then $\frac{a+b}{2} < b$ .

Solution

Given:

a and b are real
a < b

Goal:

$\frac{a+b}{2} < b$

a – b < 0. Add 2b on both sides to get a + b < 2b. Divide by 2.

Form:

Suppose a < b

— Proof of $\frac{a+b}{2} < b$ goes here.

Therefore, if a < b then $\frac{a+b}{2} < b$ .

Proof: Suppose a < b. Adding b to both sides we get a + b < 2b. Dividing by 2 we get $\frac{a+b}{2} < b$ as required. Thus, if a < b, then $\frac{a+b}{2} < b$ .

How To Prove It by Velleman

Prove It by Velleman, Section 3.1, Problem 8

January 8, 2015 etoix Leave a comment

Problem

Suppose $A \setminus B \subseteq C \cap D$ and $x \in A$ . Prove that if $x \notin D$ then $x \in B$ .

Solution

Given:

$A \setminus B \subseteq C \cap D$
$x \in A$
$x \notin D$

Goal:

$x \in B$

So the logic version is $\forall x (x \in A \land x \notin B \to x \in C \land x \in D)$ . Since $x \notin D$ , the right side is false. This is a conditional statement, so the only way the statement can be true is if the left side is also false. Since $x \in A$ the only way to make the left side false is if $x \in B$ .

Form:

Suppose $x \notin D$

— Proof of $x \in B$ goes here.

Therefore, if $x \notin D$ , then $x \in B$ .

Proof: Suppose $A \setminus B \subseteq C \cap D$ , $x \in A$ , and $x \notin D$ . We conclude that $x \notin C \cap D$ . Similarly $x \notin A \setminus B$ . However, we know that $x \in A$ , so we conclude that $x \in B$ , as required. Therefore if $x \notin D$ then $x \in B$ .

How To Prove It by Velleman

Prove It by Velleman, Section 3.1, Problem 7

January 7, 2015 etoix Leave a comment

Problem

Suppose that a is a real number. Prove that if $a^3 > a$ then $a^5 > a$ . (Hint: One approach is to start by completing the following equation: $a^5 - a = (a^3 - a) \cdot ?$ .)

Solution

Given:

a is real
$a^3 > a$

Goal:

$a^5 > a$

The tactic is to figure out the ratio between $a^3$ and $a^5$ .

$a^5 - a = (a^3 - a) \cdot x$

$x = \frac{a^5 -a}{a^3 - a}$

Factor

$x = \frac{(a^2 + 1) (a^2 - 1)}{a^2 - 1}$

$x = a^2 + 1$

We know that $a^2$ is a positive number, so $a^2 + 1$ is a positive number greater than 1.

So we know the multiplier to from from $a^3 - a$ to $a^5 - a$ is a positive number greater than 1.

Form:

Suppose $a^3 > a$ .

— Proof of $a^5 > a$ goes here.

Therefore, if $a^3 > a$ , then $a^5 > a$ .

Proof: Suppose $a^3 > a$ . Subtracting a from both sides yields $a^3 - a = 0$ . Multiply both sides by $(a^2 + 1)$ we get $(a^3 - a) \cdot (a^2 + 1) = a^5 - a > 0$ . Adding a to both sides, we get $a^5 > a$ as required. Thus if $a^3 > a$ then $a^5 > a$ .

How To Prove It by Velleman

Prove It by Velleman, Section 3.1, Problem 6

January 6, 2015 etoix Leave a comment

Problem

Suppose a and b are real numbers. Prove that if 0 < a < b then $\frac{1}{b} < \frac{1}{a}$ .

Solution

Given:

a and b are real
0 < a < b

Goal:

$\frac{1}{b} < \frac{1}{a}$

We know that a and b are positive, no sign change. We can divide by $\frac{1}{ab}$ .

Form:

Suppose 0 < a < b

— Proof of $\frac{1}{b} < \frac{1}{a}$ goes here.

Therefore, if 0 < a < b, then $\frac{1}{b} < \frac{1}{a}$ .

Proof: Suppose 0 < a < b, Multiplying the inequality a < b by the positive number $\frac{1}{ab}$ we get $\frac{1}{b} < \frac{1}{a}$ as required. Thus if 0 < a < b, then $\frac{1}{b} < \frac{1}{a}$ .

How To Prove It by Velleman

Prove It by Velleman, Section 3.1, Problem 5

January 5, 2015 etoix Leave a comment

Problem

Suppose a and b are real numbers. Prove that if a < b < 0 then $a^2 > b^2$ .

Solution

Scratch:

Given:

a and b are real
– a < b < 0

Goal:

$a^2 > b^2$

We know that a and b are negative, so multiplying would flip the sign.

Form:

Suppose a < b < 0

— Proof of $a^2 > b^2$ goes here

Therefore, if a < b < 0, then $a^2 > b^2$ .

Proof: Suppose a < b < 0. Multiplying the inequality a < b by the negative number a we can conclude $a^2 > ab$ and similarly multiplying by the negative number b we get $ab > b^2$ . Therefore $a^2 > ab > b^2$ , so $a^2 > b^2$ , as required. Thus if a < b < 0, then $a^2 > b^2$ .

How To Prove It by Velleman

Prove It by Velleman, Section 3.1, Problem 4

January 4, 2015 etoix Leave a comment

Problem

Complete the following alternative proof of the theorem in Example 3.1.2.

Proof. Suppose 0 < a < b. Then b – a > 0.

[Fill in a proof of $b^2 - a^2 > 0$ here.]

Since $b^2 - a^2 > 0$ , it follows that $a^2 < b^2$ . Therefore if 0 < a < b then $a^2 < b^2$ .

Solution

Since a and b are positive, a + b > 0. Multiply both sides:

$(b - a) (b + a) > 0 (b + a)$

$b^2 - a^2 > 0$ :

How To Prove It by Velleman

Prove It by Velleman, Section 3.1, Problem 3

January 3, 2015 etoix Leave a comment

Problem

Consider the following incorrect theorem:

Incorrect Theorem. Suppose n is a natural number larger than 2, and n is not a prime number. Then 2n + 13 is not a prime number.

What are the hypotheses and conclusion of this theorem? Show that the theorem is incorrect by finding a counterexample.

Solution

Hypotheses: $n \in \mathcal{N}$ , n > 2, and n is not prime.

Conclusion: 2n + 13 is not prime.

$n = 3$

$2 \cdot 3 + 13 = 19$

19 is prime, so the theorem is false.

How To Prove It by Velleman

Prove It by Velleman, Section 3.1, Problem 2

January 2, 2015 etoix Leave a comment

Problem

Consider the following theorem. (The theorem is correct, but we will not ask you to prove it here.)

Theorem. Suppose that $b^2 > 4ac$ . Then the quadratic equation $ax^2 + bx + c = 0$ has two real solutions.

1. Identify the hypotheses and conclusion of the theorem.
2. To give an instance of the theorem, you must specify values for a, b, and c, but not x. Why?
3. What can you conclude from the theorem in the case a = 2, b = -5, c = 3? Check directly that this conclusion is correct.
4. What can you conclude from the theorem in the case a = 2, b = 4, c = 3?

Solution

1. Identify the hypotheses and conclusion of the theorem.

Hypothesis: $b^2 > 4ac$

Conclusion: $ax^2 + bx + c = 0$ has two real solutions

2. To give an instance of the theorem, you must specify values for a, b, and c, but not x. Why?

The general solution for a quadratic formula is $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ . The question concerns us with real versus imaginary solutions for x. That is determined solely by the term \sqrt{b^2-4ac}. If there were two real solutions, \sqrt{b^2-4ac} would be positive, or $b^2 > 4ac$ so it is sufficient to consider a, b, and c, but not x.

3. What can you conclude from the theorem in the case a = 2, b = -5, c = 3? Check directly that this conclusion is correct.

$(-5)^2 - 4 \cdot 2 \cdot 3$

$25 - 24$

This is positive so we can conlude that there are two real solutions.

4. What can you conclude from the theorem in the case a = 2, b = 4, c = 3?

$4^2 - 4 \cdot 2 \cdot 3$

$16 - 24$

This is negative so we can conclude there wouldn’t be two real solutions.

How To Prove It by Velleman

Prove It by Velleman, Section 3.1, Problem 1

January 1, 2015 etoix Leave a comment

Problem

Consider the following theorem. (This theorem was proven in the introduction.)

1. Identify the hypotheses and conclusion of the theorem. Are the hypotheses true when n = 6? What does the theorem tell you in this instance? Is it right?

2. What can you conclude from the theorem in the case n = 15? Check directly that this conclusion is correct.

3. What can you conclude from the theorem in the case n = 11?

Solution

1. Identify the hypotheses and conclusion of the theorem. Are the hypotheses true when n = 6? What does the theorem tell you in this instance? Is it right?

The hypotheses are: $n \in \mathcal{Z}$ , n > 1, and n is not prime.

The conclusion is $2^n - 1$ is not prime.

6 is an integer. 6 > 1. 6 is not prime. The hypotheses are true when n = 6.

$2^6 - 1 = 63$ , which is not prime, so this instance of the theorem is correct.

2. What can you conclude from the theorem in the case n = 15? Check directly that this conclusion is correct.

The hypotheses are true when n = 15.

$2^{15} - 1 = 32767$ which is not prime (32767 is divisible by 7). So this instance of the theorem is correct.

3. What can you conclude from the theorem in the case n = 11?

11 is not prime so the hypothesis, n is not prime, is false. Therefore, the theorem does not apply.

e^(ix)

Prove It by Velleman, Section 3.1, Problem 10

Prove It by Velleman, Section 3.1, Problem 9

Prove It by Velleman, Section 3.1, Problem 8

Prove It by Velleman, Section 3.1, Problem 7

Prove It by Velleman, Section 3.1, Problem 6

Prove It by Velleman, Section 3.1, Problem 5

Prove It by Velleman, Section 3.1, Problem 4

Prove It by Velleman, Section 3.1, Problem 3

Prove It by Velleman, Section 3.1, Problem 2

Prove It by Velleman, Section 3.1, Problem 1

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