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:

steve@SATPrepGet800.com 

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!

I have just uploaded a new video on YouTube. I have embedded it below so that you do not have to go to YouTube itself. But be aware that there are many other useful videos I have uploaded on my channel.

For a while, I have wanted to show how easy it is to complete a section on SAT math. By learning SAT specific strategies that decrease the amount of time you spend on each problem, and help you to avoid careless errors, you too can train yourself to correctly complete each SAT math section very quickly.

As you watch the video, have your Blue Book open to section 3 of SAT 1 so that you can follow along. I cannot reproduce the College Board’s questions due to copywrite laws.

Note that when I say I complete the test very quickly, I do not mean that I rush. Quite the opposite: I often take enough time on each question to not only get the right answer but also to make sure that I haven’t been tricked.

The time that I save can be used to redo each question to ensure that I have not made any errors. Ideally, I would redo each question using a different strategy than I had used the first time around.

In the video above I do not talk about any specific product that you can use to gain the skill set to complete each SAT math section this quickly; however, The 32 Most Effective SAT Math Strategies is an excellent resource to develop this skill set.

This video is just Section 3 of Test 1. I have also completed Sections 7 and 8, so keep an eye out. I will be posting those videos soon.

Last week I posted the laws of exponents that can be useful to know for the SAT. These should be memorized by students that are trying to break a 700 in SAT math. I also mentioned that exponent problems can usually be solved by methods other than using the algebraic rules, but by using algebra you will solve them very, very quickly, giving you time to move on to other questions and then check your answers.

Last week, I had this level 5 exponent problem for you to solve:

If y = 7^x, which of the following expressions is equivalent to 49^x  − 7^x+2  for all positive integer values of x?

Let’s solve this algebraically using the exponent laws.

49^x  − 7^x+2 = (7²)^x  − 7^x7² =  (7^x)² − 49(7^x) =  y² – 49y

In the first equality we rewrote 49 as 7² and used the third law in our list to rewrite 7^x+2 as 7^x7². In the second equality we simply rewrote 7² as 49. Finally, in the third equality we replaced 7^x by y twice.

This is choice (E).

And here is an alternative solution using the strategy of “picking numbers”.

Let’s choose a value for x, say x = 2..

Then we have:

y = 7² = 49, and 49^x  − 7^x+2 = 49² − 7⁴ = 0.

Put a nice, big, dark circle around 0. Make it obvious that this is the value you are expecting to be your solution. Now substitute y = 49 into each answer choice:

Since (A), (B), (C) and (D) are incorrect, we can eliminate them. Therefore the answer choice is (E).

As you can see the method of picking numbers is more time consuming than the algebraic method. But you will get the right answer as long as you are careful.

Whichever method you choose, you should try to check your answer with a different method if possible. If you can only solve the problem by picking numbers, then at least pick different numbers when you are “checking.”

I have had responses to the competition and many of you did get the right answer. I will be contacting you personally to send you your choice of 28 SAT Math Lessons to Improve Your Score in One Month: Intermediate or Advanced.

Students seem to have a lot of trouble remembering the basic rules of exponents. On the SAT, you can often do problems involving exponents by using some basic strategies that avoid using these rules altogether. Using the exponent laws, however, usually leads to the quickest solutions. Students that are trying to get an SAT math score higher than 700 will want to become more proficient in using these laws of exponents. Here is a basic review.

Now let’s try a level 5 question:

If y = 7^x, which of the following expressions is equivalent to 49^{x  −} 7^x+2  for all positive integer values of x?

The first 5 people to comment on this post with a correct answer will get an electronic copy of 28 SAT Math Lessons to Improve Your Score in One Month.

I would also like to see how you got to the answer, so please show your calculations!

Answers have to be received on this blog post by Sunday, November 18, 2012. (Answers will remain hidden until the 18th).

Winners can have a choice of the intermediate course or, the not yet released (!!), advanced course. The intermediate course is for students that are currently scoring between 500 and 600 in SAT math and want to get to the next  score level. The advanced course is for students scoring above 600 and ideally would like to get a perfect score of 800.

Good luck!

Plugging In Answer Choices

In many SAT math problems you can get the answer simply by trying each of the answer choices until you find the one that works. Unless you have some intuition as to what the correct answer might be, then you should always start with choice (C) as your first guess. The reason for this is simple. Answers are usually given in increasing or decreasing order. So very often if choice (C) fails you can eliminate two of the other choices as well.

There are a few exceptions to this rule. If the word least appears in the problem, then start with the smallest number as your first guess. Similarly, if the word greatest appears in the problem, then start with the largest number as your first guess.

Examples

Let’s take a look at two SAT math problems – one where we start with choice (C) and one where we do not.

1. When each side of a given square is lengthened by 3 inches, the area is increased by 45 square inches. What is the length, in inches, of a side of the original square?

               (A)   3
               (B)   4
               (C)   5
               (D)   6
               (E)   7

Let’s start with choice (C). If the original length of a side of the square is 5, then the length becomes 8 when we increase it by 3. The original square has an area of 5²= 25 and the new square has area 8² = 64. So the area is increased by 64 – 25 = 39 square inches. Thus, we can eliminate choice (C), and most likely (A) and (B) as well.

We next try choice (D). We have 6² = 36, 9² = 81 and 81 – 36 = 45. Thus, the answer is choice (D).

Here is an algebraic solution for those of you that really want to see it.

Thus, the answer is choice (D).

For this particular question, I prefer the solution by starting with choice (C) to the more tedious and confusing algebraic solution.

Now for the second example.

2. What is the largest positive integer value of k for which 3^k divides 18⁴?

               (A)   2
               (B)   4
               (C)   6
               (D)   7
               (E)   8

Pull out your calculator. Since the question has the word “largest” in it, we will start with the largest answer choice which is choice (E), and we will divide 18⁴ by 3⁸. We type 18^4 / 3^8 into our calculator and the output is 16. Since 16 is an integer, the answer is choice (E).

Note that all five answer choices give an integer, but 8 is the largest positive integer that works.

Here is a direct solution for those of you who really want to see it.

The prime factorization of 18 is 18 = 2·3². Therefore:

18⁴ = (2·3²)⁴ = 2⁴(3²)⁴ = 2⁴3⁸.

From this prime factorization it should be clear that 3⁸ divides 18⁴, but 3⁹ does not.

Again, for this particular question, most students will prefer the easier solution of starting with choice (E) (the largest answer choice) to the more confusing algebraic solution.

More SAT Math Practice Problems

More information on this extremely useful strategy, as well as many more problems to practice with, can be found in my 28 SAT Math Lessons Series. Click on the picture below for more information about these books.

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