Tag Archives: solution disagrees with answer key

How To Prove It by Velleman

Prove It by Velleman, Section 2.3, Problem 11

December 27, 2014 etoix Leave a comment

Problem

Give an example of sets A and B for which $\mathcal P ( A \cup B ) \neq \mathcal P(A) \cup \mathcal P(B)$ .

Solution

A = { 1 }

B = { 2 }

$\mathcal P(A)$ = { $\emptyset$ , { 1 } }

$\mathcal P(B)$ = { $\emptyset$ , { 2 } }

$\mathcal P(A) \cup \mathcal P(B)$ = { $\emptyset$ , { 1 }, { 2 } }

$A \cup B$ = { 1, 2 }

$\mathcal P(A \cup B)$ = { $\emptyset$ , { 1 }, { 2 } { 1, 2 } }

$\{ \emptyset, \{ 1 \}, \{ 2 \} \} \neq \{ \emptyset, \{ 1 \}, \{ 2 \}, \{ 1, 2 \} \}$

How To Prove It by Velleman

Prove It by Velleman, Section 2.2, Problem 1

December 5, 2014 etoix 1 Comment

Problem

Negate these statements and then reexpress the results as a equivalent positive statement. (See Example 2.2.1)

Everyone who is majoring in math has a friend who needs help with his homework.
Everyone has a roommate who dislikes everyone.
$A \cup B \subseteq C \setminus D$
$\exists x \forall y [y > x \to \exists z (z^2 + 5z = y)]$

Solution

Everyone who is majoring in math has a friend who needs help with his homework.

Let M(x) = x majors in math

Let F(x and y) = x and y are friends

Let H(x) = x needs help

Answer Key says $\exists [M(x) \land \forall y ( F(x,y) \to \neg H(y))]$ . I don’t see why it should be an implication.

$\forall x (M(x) \to \forall y [F(x,y) \land H(y)])$
Negate

$\neg \forall x (M(x) \to \forall y [F(x,y) \land H(y)])$

Quantifier Negate

$\exists x \neg (M(x) \to \forall y [F(x,y) \land H(y)])$

Conditional

$\exists x \neg ( \neg M(x) \lor \forall y [F(x,y) \land H(y)])$

DeMorgan

$\exists x ( M(x) \land \neg \forall y [F(x,y) \land H(y)])$

Quantifier Negation

$\exists x ( M(x) \land \exists y \neg [F(x,y) \land H(y)])$

DeMorgan

$\exists x ( M(x) \land \exists y [\neg F(x,y) \lor \neg H(y)])$

There exists a math major and that math major has no friends at all or has friends that don’t need help.

Everyone has a roommate who dislikes everyone.

Let R(x, y) = x has a roommate y

Let L(x, y) = x likes y

$\forall x \exists y[R(x,y) \land \forall z \neg L(y,z)]$

Negate

$\neg \forall x \exists y[R(x,y) \land \forall z \neg L(y,z)]$

Quantifier Negate

$\exists x \neg \exists y[R(x,y) \land \forall z \neg L(y,z)]$

Quantifier Negate

$\exists x \forall y \neg [R(x,y) \land \forall z \neg L(y,z)]$

DeMorgan

$\exists x \forall y [\neg R(x,y) \lor \neg \forall z \neg L(y,z)]$

Quantifier Negate

$\exists x \forall y [\neg R(x,y) \lor \exists z L(y,z)]$

Someone does not have any roommates or has a roommate that likes at least one person.

$A \cup B \subseteq C \setminus D$

$\forall x [(x \in A \lor x \in B) \to (x \in C \land x \notin D)]$

Negate

$\neg \forall x [(x \in A \lor x \in B) \to (x \in C \land x \notin D)]$

Conditional

$\neg \forall x [\neg (x \in A \lor x \in B) \lor (x \in C \land x \notin D)]$

Quantifier Negation

$\exists x \neg [\neg (x \in A \lor x \in B) \lor (x \in C \land x \notin D)]$

DeMorgan

$\exists x [(x \in A \lor x \in B) \land \neg (x \in C \land x \notin D)]$

DeMorgan

$\exists x [(x \in A \lor x \in B) \land (x \notin C \land x \in D)]$

Associative

$\exists x [(x \in A \lor x \in B) \land x \notin C \land x \in D]$

$\exists x \forall y [y > x \to \exists z (z^2 + 5z = y)]$

Negate

$\neg \exists x \forall y [y > x \to \exists z (z^2 + 5z = y)]$

Quantifier Negate

$\forall x \neg \forall y [y > x \to \exists z (z^2 + 5z = y)]$

Quantifier Negate

$\forall x \exists y \neg [y > x \to \exists z (z^2 + 5z = y)]$

Conditional

$\forall x \exists y \neg [\neg (y > x) \lor \exists z (z^2 + 5z = y)]$

DeMorgan

$\forall x \exists y [(y > x) \land \neg \exists z (z^2 + 5z = y)]$

Quantifier Negate

$\forall x \exists y [(y > x) \land \forall z \neg (z^2 + 5z = y)]$

$\forall x \exists y [(y > x) \land \forall z (z^2 + 5z \neq y)]$

How To Prove It by Velleman

Prove It by Velleman, Section 2.1, Problem 1

November 26, 2014 etoix Leave a comment

Problem

Analyze the logical forms of the following statements.

Anyone who has forgiven at least one person is a saint.
Nobody in the calculus class is smarter than everybody in the discrete math class.
Everyone likes Mary, except Mary herself.
Jane saw a police officer, and Roger saw one too.
Jane saw a police officer, and Roger saw him too.

Solution

Anyone who has forgiven at least one person is a saint.

Let S(x) = x is a saint.

Let F(x, y) = x forgives y.

$\forall x [\exists y F(x, y) \to S(x) ]$

Nobody in the calculus class is smarter than everybody in the discrete math class.

Let C(x) = x is in calculus

Let D(x) = x is in discrete math

Let S(x,y) = x is smarter than y

$\neg \exists x [C(x) \land \forall y [D(y) \land S(x,y)]]$

Answer key is $latex $\neg \exists x [C(x) \land \forall y [D(y) \to S(x,y)]] $. Why implication instead of and?

Everyone likes Mary, except Mary herself.

Let L(x,y) = x likes y

Let M(x) = x is Mary

$\forall x [\neg M(x) \land L(x, Mary)]$

Answer key is $latex $ \forall x [\neg (x = m) \to L(x , m)] $. Why implication instead of and? (I can see this more than the last question.)

Jane saw a police officer, and Roger saw one too.

Let S(x, y) = x saw y

Let P(x) = x is a police officer

$\exists x [P(x) \land S(Jane, x)] \land \exists y [P(y) \land S(Roger, y)]$

Jane saw a police officer, and Roger saw him too.

$\exists x [P(x) \land S(Jane, x) \land S(Roger, y)]$

Calculus by Spivak

Calculus by Spivak, Chapter 1, Problem 4

September 28, 2014 etoix Leave a comment

Problem

Find all numbers $x$ for which

$4 -x < 3-2x$
$5-x^2<8$
$5-x^2<-2$
$(x-1)(x-3) >0$ (When is a product of two numbers positive?)
$x^2-2x+2>0$
$x^2+x+1>2$
$x^2-x+10 >16$
$x^2+x+1>0$
$(x- \pi) (x+5)(x-3) >0$
$(x- \sqrt[3]{2} ) (x- \sqrt{2} ) > 0$
$\sqrt{x}{2} < 8$
$x + 3^x < 4$
$\frac{1}{x} + \frac{1}{1-x} > 0$
$\frac{x-1}{x+1} > 0$

Solution

$4 -x < 3-3x$

Collect the terms.

$x < -1$

$5-x^2<8$

Move terms around and combine.

$-3 < x^2$

$x^2$ will always be positive so any $\mathbb{R}$ will work.

$5-x^2<-2$

Move terms around and combine.

$7 < x^2$

If this were $x^2 = 7$ , we see that $x = \pm \sqrt{7}$ .

We looking for when $x^2$ is greater than 7, so the solution is $x < -\sqrt{7}$ or $x > \sqrt{7}$ .

$(x-1)(x-3) >0$

Our critical points are $x= 1$ and $x = 3$ .

If $x = 0$ , then we would have $- * - = + > 0$ .

If $x = 2$ , then we would have $+ * - = - < 0$ .

If $x = 4$ , then we would have $+ * + = + > 0$ .

So the inequality works when $x < 1$ or $x > 3$ .

$x^2-2x+2>0$

If this were $x^2 - 2x + 2 = 0$ , the solution would be $\frac{2 \pm \sqrt{-4}}{2}$ . The solution is imaginary, so the parabola is entirely above or below the x-axis.

Since the first term is positive, we know this is a parabola facing up, so it’s entirely above the x-axis.

Therefore any $\mathbb{R}$ will work.

$x^2+x+1>2$

$x^2 + x -1 >0$

If this were an equality and we solve for x, we’d get $x = \frac{-1 \pm \sqrt{5}}{2}$ . This is a real solution so the parabola does cross the x-axis and there are two solutions. The parabola faces up so we want the two extreme branches. Therefore $x < \frac{-1-\sqrt{5}}{2}$ or $x > \frac{-1+\sqrt{5}}{2}$ .

$x^2-x+10 >16$

$x^2 -x-6 > 0$

$(x-3)(x+2) > 0$

The critical points are $x = 3$ and $x = -2$ .

If $x = -3$ , we would have $- * - = + > 0$ .

If $x = 0$ , we would have $- * + = - < 0$ .

If $x = 4$ , we would have $+ * + = + > 0$ .

Therefore, we want $x < -2$ or $x > 3$ .

$x^2+x+1>0$

If this were $x^2 -x + 1 = 0$ , the solution would be $\frac{-2 \pm \sqrt{-3}}{2}$ . The solution is imaginary, so the parabola is entirely above or below the x-axis.

Since the first term is positive, we know this is a parabola facing up, so it’s entirely above the x-axis.

Therefore any $\mathbb{R}$ will work.

$(x- \pi) (x+5)(x-3) >0$

The critical points are $x = \pi$ , $x = -5$ , and $x = 3$ .

If $x = -6$ , we would have $- * - * - = - < 0$ .

If $x = 0$ , we would have $- * + * - = + > 0$ .

If $x = 3.1$ , we would have $- * + * + = - < 0$ .

If $x = 4$ , we would have $+ * + * + = + > 0$ .

Therefore, the solution is $-5 < x < 3$ or $x > \pi$ .

$(x- \sqrt[3]{2} ) (x- \sqrt{2} ) > 0$

The critical points are $x = \sqrt[3]{2}$ and $x = \sqrt{2}$ .

If $x = 0$ , we would have $- * - = + > 0$ .

If $x = 1.3$ , we would have $+ * - = - < 0$ .

If $x = 2$ , we would have $+ * + = + > 0$ .

The solution is $< \sqrt[3]{2}$ or $x > \sqrt{2}$ .

$\sqrt{x}{2} < 8$

2 is positive. $2^3 = 8$ . Any exponent greater than 3 should be greater than 8. So $x > 3$ .

$x + 3^x < 4$

By visual inspection, it’s clear that if $x = 1$ , $1 + 3^1 = 4$.

So any x less than 1 should be less than 4.

$x<3$

$\frac{1}{x} + \frac{1}{1-x} > 0$

The critical points are $x = 0$ and $x = 1$ .

If x is negative, then $\frac{1}{x} < 0$ and $\frac{1}{1-x} > 0$ , but the absolute value of $\frac{1}{x}$ will ways be bigger.

x	$\frac{1}{x}$	$\frac{1}{1-x}$
-1	-1	$\frac{1}{2}$
-2	$-\frac{1}{2}$	$\frac{1}{3}$
-3	$-\frac{1}{3}$	$\frac{1}{4}$
-4	$-\frac{1}{4}$	$\frac{1}{5}$

Therefore, when $x < 0$ , $\frac{1}{x} + \frac{1}{1-x} < 0$ .

If $x > 1$ , then $\frac{1}{x} > 0$ and $\frac{1}{1-x} < 0$ , but the absolute value of $\frac{1}{1-x}$ will ways be bigger.

x	$\frac{1}{x}$	$\frac{1}{1-x}$
2	$\frac{1}{2}$	-1
3	$-\frac{1}{3}$	$-\frac{1}{2}$
4	$-\frac{1}{4}$	$-\frac{1}{3}$
5	$-\frac{1}{5}$	$-\frac{1}{4}$

Therefore, when $x > 1$ , $\frac{1}{x} + \frac{1}{1-x} < 0$ .

So let’s consider $0 < x < 1$ . Consider $x = \frac{1}{2}$ .

$\frac{1}{\frac{1}{2}} + \frac{1}{(1 - \frac{1}{2})}$

Notice that for $0 < x < 1$ both terms must be positive. Therefore, the only solution is $0 < x < 1$ .

Here I have a disagreement which the answer key. It claims that the solution is $x > 1$ or $0 < x < 1$ . However, I don’t see how $x> 1$ could be a solution. Below is a plot of $\frac{1}{x} + \frac{1}{1-x}$ . It is clearly negative where $x > 1$ .

$\frac{x-1}{x+1} > 0$

The critical points here are $x = 1$ and $x = -1$ .

If $x = -2$ , we would have $\frac{-}{-} = + > 0$ .

If $x = 0$ , we would have $\frac{-}{+} = - < 0$ .

If $x = 2$ , we would have $\frac{+}{+} = + > 0$ .

So the solution is $x < -1$ or $x > 1$ .

Ordinary Differential Equations by Tenenbaum & Pollard

ODE by Tenenbaum & Pollard, Exercise 3, Problem 4

September 24, 2014 etoix Leave a comment

Problem

Determine whether the equations on the right define implicit functions of x. For those which do, determine whether they are implicit solutions of the differential equations on the left.

$y^2 - 1 - (2y + xy) y\prime = 0$	$y^2 - 1 = (x+2)^2$
$e^{x-y} + e^{y-x} \frac{dy}{dx} = 0$	$e^{2y}+e^{2x} = 1$
$\frac{dy}{dx} = -\frac{y}{x}$	$x^2 + y^2 + 1 = 0$

Solution

We start with $y^2 - 1 = (x+2)^2$ .

Solve for y and get $y = \pm \sqrt{(x+2)^2 + 1}$ .

So we know that this function is defined for all $\mathbb{R}$ .

Let implicitly differentiate and get:

$2yy\prime = 2(x+2)$

Simplfy:

$yy\prime = x+2$

The derivative is $y\prime = \frac{x+2}{y}$ .

Let’s substitute into $y^2 - 1 - (2y + xy) y\prime = 0$ to see if we get identity.

$y^2 - 1 - (2y + xy ) \frac{x+2}{y} = 0$

$y^2 - 1 - (2[x+2] + x[x+2]) = 0$

$y^2 - 1 - (x^2 + 4x + 4) = 0$

Now let’s substitute with $y^2 - 1 = (x+2)^2$ .

$latex (x+2)^2 – (x^2 + 4x + 4) = 0$

We have identity, so $y^2 - 1 = (x+2)^2$ does implicitly solve $y^2 - 1 - (2y + xy) y\prime = 0$ .

Keep in mind that y has two solutions: $y = \pm \sqrt{(x+2)^2 + 1}$ . So you need to pick one to complete the answer.

We start with $e^{2y}+e^{2x} = 1$ .

$e^{anything} > 0$ . In order for two positive real numbers to add up to 1, they both need to be less than 1.

Below is a plot:

Notice that the plot approaches both axes asymptotically.

The domain is $x < 0$ . Here is where I disagree with the answer key, which assert $x \neq 0$ .

Let’s implicitly differentiate and get:

$2 e^{2y} y\prime + 2 e^{2x} = 0$

Simplify:

$e^{2y} y\prime + e^{2x} = 0$

Divide by $e^y$ .

$e^{y} y\prime + e^{2x-y} = 0$

Divide by $e^x$ .

$e^{y-x} y\prime + e^{x-y} = 0$

This is exactly the differential equation we’re looking for, so $e^{2y}+e^{2x} = 1$ is a solution for $e^{x-y} + e^{y-x} \frac{dy}{dx} = 0$ .

Let’s start with $x^2 + y^2 + 1 = 0$ .

This is $x^2 + y^2 = -1$ .

$x^2$ and $y^2$ both must be positive and cannot be -1.

Therefore, this implicit function is not defined in $\mathbb{R}$ .

Ordinary Differential Equations by Tenenbaum & Pollard

ODE by Tenenbaum & Pollard, Exercise 3, Problem 2

August 28, 2014 etoix Leave a comment

Problem

Prove that the function sin the right-hand column below are solutions of the differential equation sin the left-hand columns [sic]. (Be sure to state the common interval for which solution and differential equation make sense.)

$y^{\prime} + y = 0$	$y = e^{-x}$
$y^{\prime} = e^x$	$y = e^x$
$\frac{d^2y}{dx^2} = \frac{1}{\sqrt{1-x^2}}$	$y = x \arcsin{x} + \sqrt{1 - x^2}$
$f^{\prime}(x) = f^{\prime\prime}(x)$	$y = e^x + 2$
$xy^{\prime} = 2y$	$y = x^2$
$(1 + x^2) y^{\prime} = xy$	$y = \sqrt{1 + x^2}$
$\cos(\theta) \frac{dr}{d\theta} - 2r \sin{\theta} = 0$	$r = a \sec^2(\theta)$
$y^{\prime\prime} - y = 0$	$y = a e^x + b e^{-x}$
$f^{\prime}(x) = \frac{1}{3} f(x)$	$f(x) = 2 e^{\frac{x}{3}}$
$xy^{\prime} + y = y^2$	$y = \frac{2}{x+2}$
$x + y y^{\prime} = 0$	$y = \sqrt{16 - x^2}$

Solution

$y = e^{-x}$

$y^{\prime} = -e^{-x}$ .

$-e^{-x} + e^{-x} = 0$

The differential equation and solution are valid for $-\infty < x < \infty$ .

$y = e^x$

$y^{\prime} = e^x$

The differential equation and solution are valid for $-\infty < x < \infty$ .

$y = x \arcsin{x} + \sqrt{1 - x^2}$

$\frac{dy}{dx} = \arcsin(x) + \frac{x}{\sqrt{1-x^2}} - \frac{2x}{2 \sqrt{1-x^2}}$

$\frac{dy}{dx} = \arcsin(x)$

$\frac{d^2y}{dx^2} = \frac{1}{\sqrt{1-x^2}}$

The differential equation and solution are valid for $-1 \leq x \leq 1$ .

$y = e^x + 2$

$y^{\prime} = e^x$

$y^{\prime\prime} = e^x$

$f^{\prime}(x) = f^{\prime\prime}(x)$

The differential equation and solution are valid for $-\infty < x < \infty$ .

$y = x^2$

$y^{\prime} = 2x$

$x (2x) = 2 ( x^2)$

The differential equation and solution are valid for $-\infty < x < \infty$ .

The answer key says that the domain is $x \neq 0$ . It is unclear why this is so.

$y = \sqrt{1 + x^2}$

$y^{\prime} = \frac{2x}{2 \sqrt{1+x^2}}$

$(1 + x^2) y^{\prime} = xy$

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

The differential equation and solution are valid for $-\infty < x < \infty$ .

$r = a \sec^2(\theta)$

$\frac{dr}{d\theta} = 2 a \sec^2(\theta) \tan(\theta)$

$\cos(\theta) \frac{dr}{d\theta} - 2r \sin{\theta} = 0$

$\cos(\theta) ( 2 a \sec^2(\theta) \tan(\theta) ) - 2 ( a \sec^2(\theta)) \sin(\theta) = 0$

$2 a \sec^2(\theta) \sin(\theta) - 2 ( a \sec^2(\theta)) \sin(\theta) = 0$

The differential equation is valid for $-\infty < \theta < \infty$ but the solution is valid for $\theta \neq \frac{\pi}{2}, -\frac{\pi}{2}, \frac{3 \pi}{2}, - \frac{3 \pi}{2}, \frac{5 \pi}{2}, - \frac{5 \pi}{2}, \dots$ .

$y = a e^x + b e^{-x}$

$y^{\prime} = a e^x - b e^{-x}$

$y^{\prime\prime} = a e^x + b^{-x}$

$a e^x + b^{-x} - ( a e^x + b^{-x} ) = 0$

The differential equation and solution are valid for $-\infty < x < \infty$ .

$f(x) = 2 e^{\frac{x}{3}}$

$f^{\prime}(x) = \frac{2 e^{\frac{x}{3}}}{3}$

$f^{\prime}(x) = \frac{1}{3} f(x)$

$\frac{2 e^{\frac{x}{3}}}{3} = \frac{1}{3} (2 e^{\frac{x}{3}} )$

The differential equation and solution are valid for $-\infty < x < \infty$ .

$y = \frac{2}{x+2}$

$y^{\prime} = - \frac{2}{(x+2)^2}$

$xy^{\prime} + y = y^2$

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

$- \frac{2x}{ (x+2)^2 } + \frac{2x+4}{ (x+2)^2} = \frac{4}{ ( x+ 2)^2}$

The differential equation is valid for $-\infty < x < \infty$ but the solution is valid for $x \neq -2$ .

The answer keys says that the domain is $x \neq -2, 0$ . It is unclear why 0 is a problem. The solution is defined for $x = 0$ .

$y = \sqrt{16 - x^2}$

$y^{prime} = - \frac{2x}{2 \sqrt{16 - x^2}}$

$x + y y^{\prime} = 0$

$x - \sqrt{16 - x^2} ( \frac{2x}{2 \sqrt{16 - x^2}} ) = 0$

$x - x = 0$

The differential equation is valid for $-\infty < x < \infty$ but the solution is valid for $-4 < x < 4$ .

Ordinary Differential Equations by Tenenbaum & Pollard

ODE by Tenenbaum & Pollard, Exercise 2, Problem 16

August 25, 2014 etoix 1 Comment

Problem

Can you apply the method of implicit differentiation as taught in the calculus to the function of problem 14, of problem 15?

Solution

The answer key says no, but this is clearly possible.

$\frac{d}{dx} ( \sqrt{x^2- y^2} + \arccos{\frac{x}{y}} ) = \frac{2x - 2y \frac{dy}{dx}}{2 \sqrt{x^2 - y^2}} - \frac{- \frac{x \frac{dy}{dx}}{y^2} + \frac{1}{y}}{ \sqrt{1 - \frac{x^2}{y^2}}}$

$\frac{d}{dx} ( \sqrt{x^2- y^2} + \arccos{\frac{x}{y}} ) = \frac{2x - 2y \frac{dy}{dx}}{2 \sqrt{x^2 - y^2}} + \frac{- \frac{x \frac{dy}{dx}}{y^2} + \frac{1}{y}}{ \sqrt{1 - \frac{x^2}{y^2}}}$

Ordinary Differential Equations by Tenenbaum & Pollard

ODE by Tenenbaum & Pollard, Exercise 2, Problem 13

August 21, 2014 etoix Leave a comment

Problem

Explain why the procedure followed in problem 12, applied to the relation $x^2 + y^2 + 1 = 0$ and yielding the result

$\frac{dy}{dx} = - \frac{x}{y}$

is meaningless.

Solution

The answer key simply says that “ $x^2 + y^2 + 1 = 0$ does not define a function.” While this is true, I feel that their solution is incomplete. After all, we’ve seen in problem 12 that a non-function equation can be turn into a function simply by restricting the domain or sectioning off parts of the graph.

With $x^2 + y^2 + 1 = 0$ however, we have something different. This particular equation’s domain has no elements in $\mathbb{R}$ at all. There is no domain to restriction. There is no graph to section off. It is simply impossible in $\mathbb{R}$ .

Ordinary Differential Equations by Tenenbaum & Pollard

ODE by Tenenbaum & Pollard, Exercise 2, Problem 9

August 17, 2014 etoix Leave a comment

Problem

Which of the domains in problem 8 are regions.

Solution

The text defines a region is defined as follows:

Each point of the set is the center of a circle whose entire interior consists of points of the set.
Every two points of the set can be joined by a curve which consists entirely of points of the set.

$z = \frac{\sqrt{1-x^2}}{\sqrt{1-y^2}}$

The domain is $-1 \leq x \leq 1$ and $-1 < y < 1$ .

This is not a region, since it does not satisfy requirement 1.

$z = x - y + 2$

The domain is $\mathbb{R}$ for both x and y.

This is a region.

$z = \sqrt{x^2 + (y-1)^2}$

The domain is $\mathbb{R}$ for both x and y.

This is a region.

$z = \sqrt{x^2 + y^2 -9}$

The domain is $x^2 + y^2 \geq 9$ . This is the area outside a circle of radius 3.

This is not a region, since it does not satisfy requirement 1.

$z = \sqrt{- ( x+y)^2}$

The domain is $x+y=0$ . This is a line.

This is not a region, since it does not satisfy requirement 1 or 2.

$z = \sqrt{- (x^2+y^2+3)}$

There is no real domain for this equation. Any z would be imaginary.

This is not a region since there isn’t even a domain.

$z = \frac{1-x}{1-y}$

The domain is $x \in \mathbb{R}$ and $y \neq 1$ .

This is not a region since it does not satisfy requirement 1 or 2. However, the answer key says it is.

On this particular one, I disagree with the answer key. The set of $x \in \mathbb{R}$ and $y \neq 1$ is the xy plane minus the line $= 1$ .

How is it possible that this formula satisfies condition 2? The points {2, 0} and {0, 0} are both in the domain but they cannot be joined.

How is it possible that this formula satisfies condition 1? If the center of the circle were {0, 0} any radius > 1 wouldn’t work.

How To Prove It by Velleman

Prove It by Velleman, Introduction, Problem 3

August 5, 2014 etoix 1 Comment

Problem

The proof of Theorem 3 gives a method for finding a prime number different from any in a given list of prime numbers.

Problem a

Use this method to find a prime different from 2, 3, 5, 7.

Solution

The method in Theorem 3 is as follows: Let $p_{1}$ , $p_{2}$ , $p_{3}$ , … , $p_{n}$ be a list of all prime numbers. Let $m = p_{1} p_{2} ... p_{n} + 1$ . m will be a prime or a product of primes.

$2 \cdot 3 \cdot 5 \cdot 7 + 1 = 211$

211 is prime.

Problem b

Use this method to find a prime different from 2, 5, 11.

Solution

The section on Theorem 3 also discusses Marsenne Primes – if n is prime, then $2^{n} - 1$ might be prime.

$2^{2} -1 = 3$

That is one possible solution.

The answer key lists 3 and 37 and solutions, but it’s unclear how to calculate the 37.

e^(ix)

Tag Archives: solution disagrees with answer key

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