Pure Mathematics for Beginners –
Accelerated and Expanded Edition
Just 19.99 on Amazon
Hi everyone! Pure Mathematics for Beginners – Accelerated and Expanded Edition is now available in paperback from Amazon. Similar to its predecessor, this book was written to provide a rigorous introduction to Logic, Set Theory, Abstract Algebra, Number Theory, Real Analysis, Topology, Complex Analysis, and Linear Algebra. The book consists of 16 lessons. Explanations to all the problems in the book are included as a downloadable PDF file.
So, what do I mean by “accelerated and expanded” edition?
By “accelerated” I mean that the book covers most of the material from the standard edition within the first half of the book. For example, the first lesson on set theory now covers relations, functions and equinumerosity (in addition to all the basics). However, nothing is left out. Everything from the original edition is included. In fact, more exposition has been added to the original content, as well as more examples and additional clarifying remarks.
By “expanded,” I mean that a huge amount of additional content has been added to the book. In fact, most of the content in Lessons 9 through 16 consists of material that is not covered in the original edition (although some of the content can be found in my other books from this series such as Real Analysis for Beginners and Abstract Algebra for Beginners).
The paperback is now on sale on Amazon for only $19.99. Note that once the sale ends (in about 24 hours), the price of this book will go up to $64.99.
The promotion is now over. Thanks to all who participated. The book is available at Amazon here: Pure Mathematics for Beginners – Accelerated and Expanded Edition
You can get the solution guide in paperback here: Pure Mathematics for Beginners – Accelerated and Expanded Edition – Solution Guide (Note that you can download the solution guide as a PDF for free, but many readers prefer to have a physical copy of the solution guide.)
If you have any questions, feel free to contact me at the following email:
Thank you all for your continued support!
A Trick For Free Two Day Shipping
I would like to finish this post with a little trick you can use to get free 2 day shipping on any of the books you decide to purchase without making any additional purchases. If you have never used Amazon Prime you can sign up for a free month using the following link.
Sign Up For Amazon Prime For Free
If you have already had a free trial of Amazon Prime you can simply open up a new Amazon account to get a new free trial. It just takes a few minutes! You will need to use a different email address than the one you usually use.
This next part is very important! After you finish your transaction, go to your Account, select “Manage my prime membership,” and turn off the recurring billing. This way in a month’s time Amazon will not start charging you for the service.
After shutting off the recurring billing you will still continue to receive the benefit of free 2 day shipping for one month. This means that as long as you use this new Amazon account for your purchases you can do all of your shopping on Amazon for the next month without having to worry about placing minimum orders to get free shipping.
Just be aware that certain products from outside sellers do not always qualify for free shipping, so please always check over your bill carefully before you check out.
Well I hope you decide to take advantage of this very special offer, or at the very least I hope you will benefit from my Amazon “free 2 day shipping trick.” Here is the link one more time:
Sign Up For Amazon Prime For Free
If you think your friends might be interested in this special offer, please share it with them on Facebook:
Thank you all for your continued support!
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Draw Your Own Figure
Question 4 with Solution
Last week, I went over a strategy for solving certain math problems on standardized tests by drawing a figure, and I gave you four problems to try on your own. You can see that post here: Math Strategy: Draw Your Own Figure
Today I would like to give a solution to the last of those four problems. You can see solutions for the first three problems here: Draw Your Own Figure: Q1 Q2 Q3.
Example: In rectangle PQRS, point T is the midpoint of side PQ. If the area of quadrilateral QRST is 0.8, what is the area of rectangle PQRS ?
Try to solve the problem yourself before checking the solution below.
Solution: Let’s begin by drawing a picture.
This picture alone really sheds some light on the situation. Lets now chop up our picture into 4 equal parts.
To get the area of one of those pieces simply divide 0.8 by 3. In our calculator we get
.266666666666667 or 4/15 if we change back to a fraction.
This is 1/4 of the area of the rectangle, so we simply multiply this result by 4 to get the answer 16/15.
More Problems with Explanations
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Draw Your Own Figure
Question 3 with Solution
Last week, I went over a strategy for solving certain math problems on standardized tests by drawing a figure, and I gave you four problems to try on your own. You can see that post here: Math Strategy: Draw Your Own Figure
Today I would like to give a solution to the third of those four problems. You can see solutions for the first two problems here: Draw Your Own Figure: Q1 Q2.
Example: Point A is a vertex of a 6-sided polygon. The polygon has 6 sides of equal length and 6 angles of equal measure. When all possible diagonals are drawn from point A in the polygon, how many triangles are formed?
Try to solve the problem yourself before checking the solution below.
Solution: We draw a picture.
Observe that the number of triangles is 4.
More Problems with Explanations
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Draw Your Own Figure
Question 2 with Solution
Last week, I went over a strategy for solving certain math problems on standardized tests by drawing a figure, and I gave you four problems to try on your own. You can see that post here: Math Strategy: Draw Your Own Figure
Today I would like to give a solution to the second of those four problems. You can see a solution for the first problem here: Draw Your Own Figure: Question 1.
Example: If line m is perpendicular to segment PQ at point R, and PR = RQ, how many points on line m are equidistant from point P and point Q ?
A) One
B) Two
C) Three
D) More than three
Try to solve the problem yourself before checking the solution below.
Solution: Recall first that equidistant means at the same distance. So we’re looking for points on line m that are at the same distance from P as they are from Q. Let’s begin by drawing a picture:
Notice that R is equidistant from P and Q, so there is at least one. Let’s draw some more.
Since there are 5 shown in the above picture, the answer is choice D.
Note: Every point on line m is actually equidistant from P and Q. m is the perpendicular bisector of line segment PQ.
The perpendicular bisector of a line segment is a line perpendicular to the segment that passes through the midpoint of the segment.
More Problems with Explanations
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Draw Your Own Figure
Question 1 with Solution
Last week, I went over a strategy for solving certain math problems on standardized tests by drawing a figure, and I gave you four problems to try on your own. You can see that post here: Math Strategy: Draw Your Own Figure
Today I would like to give a solution to the first of those four problems. I will give solutions to the other three problems throughout this week.
Example: What is the area of a right triangle whose sides have lengths 14, 48, and 50?
Try to solve the problem yourself before checking the solution below.
Solution: Let’s begin by drawing a picture
Remember: The hypotenuse of a right triangle (the side opposite the right angle) is always longer than both legs.
In a right triangle we can always take the two legs to be the base and the height (in either order). So b = 14, h = 48, and
A = 1/2 bh = 1/2(14)(48) = 336.
More Problems with Explanations
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Draw Your Own Figure
Today I would like to remind you of a very simple but effective strategy for solving math problems on standardized tests such as the ACT, SAT, and GRE.
If a math problem does not have a figure above it, then do not hesitate to draw your own. Sometimes drawing a quick picture of a situation makes a problem very easy, or at least easier. This is especially helpful with geometry problems.
Example: Segment PQ has midpoint M. If the length of PM is t, what is the length of PQ in terms of t ?
Try to solve the problem yourself before checking the solution below.
Solution: Let’s begin by drawing a picture
From the picture, we see that PQ has twice the length of PM. Thus, the length of PQ is 2t.
Here are a few more problems for you to try. Try to draw a picture. I will provide solutions to these over the next few days.
1. What is the area of a right triangle whose sides have lengths 14, 48, and 50?
2. If line m is perpendicular to segment PQ at point R, and PR = RQ, how many points on line m are equidistant from point P and point Q ?
A) One
B) Two
C) Three
D) More than three
3. Point A is a vertex of a 6-sided polygon. The polygon has 6 sides of equal length and 6 angles of equal measure. When all possible diagonals are drawn from point A in the polygon, how many triangles are formed?
4. In rectangle PQRS, point T is the midpoint of side PQ. If the area of quadrilateral QRST is 0.8, what is the area of rectangle PQRS ?
More Problems with Explanations
If you are preparing for the SAT, ACT, or an SAT math subject test, you may want to take a look at the Get 800 collection of test prep books.
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Geometry Question with Solution
Yesterday I went over a method for solving certain geometry problems by moving the sides of a figure around. You can see that post here: Moving the Sides of a Figure Around
Today I would like to solve yesterday’s problem.
Problem: In the figure below, AB = 2, BC = 8, and AD = 10. What is the length of line segment CD ?
Solution: The problem becomes much simpler if we “move” BC to the left and AB to the bottom as shown below.
We now have a single right triangle and we can either use the Pythagorean Theorem, or better yet notice that 10 = 5 ⋅ 2 and 8 = 4 ⋅ 2. Thus, the other leg of the triangle is 3 ⋅ 2 = 6. So we see that CD must have length 6 – 2 = 4.
Remark: If we didn’t notice that this was a multiple of a 3-4-5 triangle, then we would use the Pythagorean Theorem as follows.
(x + 2)2 + 82 = 102
(x + 2)2 + 64 = 100
(x + 2)2 = 36
x + 2 = 6
x = 4
More Problems with Explanations
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Geometry Strategy – Moving the Sides of a Figure Around
Today I would like to teach you a more advanced geometry strategy that can sometimes be used on standardized tests such as the ACT, SAT, and GRE.
A seemingly difficult geometry problem can sometimes be made much easier by moving the sides of the figure around.
This procedure is best understood with an example:
Example: What is the perimeter, in meters, of the figure below?
Try to solve the problem yourself before checking the solution below.
Solution: Recall that to compute the perimeter of the figure we need to add up the lengths of all 8 line segments in the figure. We “move” the two smaller horizontal segments up and the two smaller vertical segments to the right as shown below.
Note that the “bold” length is equal to the “dashed” length. Thus, the perimeter is
(2)(7) + (2)(5) = 14 + 10 = 24.
Warning: Although lengths remain unchanged by moving line segments around, areas will be changed. This method should not be used in problems involving areas.
Here is one more problem for you to try. See if you can find the best way to move the sides of the figure around to make the problem easy to solve. I will provide a solution to this problem tomorrow.
Example: In the figure below, AB = 2, BC = 8, and AD = 10. What is the length of line segment CD ?
More Problems with Explanations
If you are preparing for the SAT, ACT, or an SAT math subject test, you may want to take a look at the Get 800 collection of test prep books.
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Ratio Question 3 with Solutions
Earlier this week I went over my method for setting up a ratio, and I gave you three ratio problems to try on your own. You can see that post here: Setting Up a Ratio
Today I would like to give a solution to the third of those three ratio problems. You can see solutions for the first and second problems here: Ratio Q1 Ratio Q2
Problem 3: The height of a solid cone is 22 centimeters and the radius of the base is 15 centimeters. A cut parallel to the circular base is made completely through the cone so that one of the two resulting solids is a smaller cone. If the radius of the base of the small cone is 5 centimeters, what is the height of the small cone, in centimeters?
For those students that are getting the hang of this, here’s the quick computation right away:
Quick solution: (22/15) ⋅ 5 = 22/3.
And here are the details for the rest of us:
Detailed solution. A picture of the problem looks like this:
In the above picture we have the original cone together with a cut forming a smaller cone. We have also drawn two triangles that represent the 2 dimensional cross sections of the 2 cones. Let’s isolate the triangles:
The two triangles formed are similar, and so the ratios of their sides are equal. We identify the 2 key words “height” and “radius.”
height 22 h
radius 15 5
We now find h by cross multiplying and dividing.
22/15 = h/5
110 = 15h
h = 110/15 = 22/3
Definition: Two triangles are similar if they have the same angles.
Notes: (1) To show that two triangles are similar we need only show that two angles of one of the triangles are equal to two angles of the other triangle (the third is free because all triangles have 180 degrees).
(2) Similar triangles do not have to be the same size.
(3) Similar triangles have sides that are all in the same proportion.
More Problems with Explanations
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Ratio Question 2 with Solutions
Earlier in the week I went over my method for setting up a ratio, and I gave you three ratio problems to try on your own. You can see that post here: Setting Up a Ratio
Today I would like to give a solution to the second of those three ratio problems. You can see solutions for the first problem here: Ratio Question 1 with Solutions.
Problem 2: A copy machine makes 4800 copies per hour. At this rate, in how many minutes can the copy machine produce 920 copies?
Solution: We identify 2 key words. Let’s choose “copies” and “minutes.”
copies 4800 920
minutes 60 x
At first glance it might seem to make more sense to choose “hours” as our second key word, but choosing the word “minutes” is more efficient because
- The answer that we’re looking for must be in minutes.
- It’s extremely simple to convert 1 hour into 60 minutes.
We now find x by cross multiplying and dividing.
4800/60 = 920/x
4800x = 55200
x = 11.5
So, the answer is 11.5. ⋅
* Quick computation: (920/4800) ⋅ 60 = 11.5.
More Problems with Explanations
If you are preparing for the SAT, ACT, or an SAT math subject test, you may want to take a look at the Get 800 collection of test prep books.
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Ratio Question 1 with Solutions
Yesterday I went over my method for setting up a ratio, and I gave you three ratio problems to try on your own. You can see that post here: Setting Up a Ratio
Today I would like to give a solution to the first of those three ratio problems. I will give solutions to the other two problems throughout this week.
Problem 1: Running at a constant speed, a cheetah traveled 200 miles in 5 hours. At this rate, how many miles did the cheetah travel in 4 hours?
Solution: As in yesterday’s problem we identify 2 key words. This time let’s choose “miles” and “hours.”
miles 200 x
hours 5 4
We now find x by cross multiplying and dividing.
200/5 = x/4
800 = 5x
x = 800/5 = 160.
Alternate solution: Using d = r ⋅ t (distance = rate ⋅ time), we have
200 = r ⋅ 5
r = 200/5 = 40 mph
Using d = r ⋅ t again, we have d = 40 ⋅ 4 = 160.
* Mental math: 200 miles in 5 hours is 40 miles per hour (divide 200 by 5). Thus, the cheetah traveled 4 ⋅ 40 = 160 miles in 4 hours.
More Problems with Explanations
If you are preparing for the SAT, ACT, or an SAT math subject test, you may want to take a look at the Get 800 collection of test prep books.
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