Part I

Algebra

Chapter 1

Fractions and So Forth

Holy Whole Numbers, Fractious Fractions

God created the integers. All the rest is the work of man.

  • Leopold Kronecker

Stories embedded deep in Western culture hint of an archaic past when counting was a form of magic, a divine prerogative too dangerous for mere humans. Zeus, the most powerful of the Greek gods, cruelly punishes Prometheus for giving humans the godlike knowledge of fire and, in Aeschylus’s version, of numbers. Yahweh, the ancient Hebrews’ “jealous god”, punishes King David by killing 70,000 of his men, prompting David’s impassioned cry, “I have sinned, and I have done wickedly: but these sheep, what have they done?” Never mind the sheep; what had David done? He had counted his men (2 Samuel 24).

Today, counting rarely provokes divine retribution, but the magical aura surrounding the whole numbers $ (1,2,3,4) $ remains potent: We count sheep (not David’s) as a charm to overcome insomnia, we “take a deep breath and count to 10” as a spell to allay anger or anxiety, we rate books, films, even pain, on pseudo-scientific scales of 1 to 10. Whole numbers speak to something deep within us. Fractions, alas, do not.

If whole numbers correspond to the divine activity of counting, then fractions – those broken, all-too-human numbers – correspond to the mundane, utilitarian activity of measurement. The gods do not seem particularly concerned to keep men from fractions, which is fortunate for the progress of science.

Far too many students fail calculus. They fail not because calculus is difficult (it isn’t, really), but because they struggle with algebra. Often, they have cracks in their algebraic foundations running all the way back to arithmetic, particularly the arithmetic of fractions. To understand calculus, one must understand algebra; to understand algebra, one must understand arithmetic. Accordingly, in this first chapter, we’ll review some basic algebra and arithmetic, emphasizing why the rules for manipulating numbers and algebraic symbols are what they are.

Understanding why such rules hold is every bit as important as understanding how to apply them. Pick up a 1000-page calculus book sometime. Feel its heft. Mere memorization of algorithms will not do, except as a temporary stopgap. The only way to truly learn mathematics is to understand it.

In Plato’s Meno, Socrates distinguishes knowledge from mere true opinions. For our purposes, a “true opinion” corresponds to a correctly memorized mathematical rule. “As long as they stay put,” says Socrates, “true opinions are fine things and do us much good. Only, they tend not to stay put for long. They’re always scampering away from a person’s soul. So they’re not very valuable until you shackle them by figuring out what makes them true.” $ ^{*} $

Do not settle for true opinions. Strive for knowledge.

In this first chapter, we will simultaneously review numerical and algebraic fractions. To lay some basic groundwork for the algebraic side of fractions, we’ll need to devote a few preliminary pages (before we reach fractions themselves) to some foundational algebraic ideas, all rooted in one core property: the so-called “distributive property”.

The Distributive Property

The distributive property (or more precisely, “the distributive property of multiplication over addition”) is the fact that multiplying something by a sum is equivalent to multiplying that same something by each individual term in the sum, then adding the results. For example, the distributive property tells us that

\[ 5(7+10)=(5\cdot7)+(5\cdot10)\quad and\quad(b+c+d)a=ab+ac+ad. \]

When describing such operations in words, we speak of “distributing the 5” (or the a) over the sum. Subtraction is just a special sort of addition (the addition of a negative), so we can distribute multiplication over subtraction, too. Thus, the distributive property guarantees that

\[ (x-y-w)z=xz-yz-wz\qquad and\qquad2a(a-b+c)=2a^{2}-2ab+2ac. \]

The distributive property is the bedrock on which much algebra is built, as you’ll see in the next few pages. It also justifies some simple techniques of mental arithmetic, as you’ll see in the exercises below.

Exercises

  1. To understand the distributive property visually, consider the figure at right.

The whole figure’s area equals the sum of the areas of the two rectangles it contains.

Rewrite the previous sentence as an algebraic equation involving a, b, and c.

[Hint: The whole figure’s height is a. What is whole figure’s width?]

  1. Consider the following technique for mental calculation:

What’s 32 times 7?

Let’s see, 32 sevens is 30 sevens and then 2 more sevens.

Well, 30 sevens is 210, and 2 sevens is 14.

Thus, 32 times 7 must be 210 plus 14, which is 224.

Explain how the distributive property makes a quiet appearance in that calculation.

  1. Using the technique from Exercise 2, compute the following in your head: $ 64 $ , $ 82 $ , $ 39 $ , $ 6 $ .

  2. One can mentally calculate a 15% tip as follows:

What’s 15% of $32?

Well, 10% of $32 is $3.20.

Half of that (i.e. 5% of the whole) is $1.60,

So a 15% tip would be $3.20 + $1.60 = $4.80.

Explain how the distributive property is at work here, too.

  1. Calculate 15% of the following amounts in your head: $28, $50, $72, $90.

  2. Students sometimes get overenthusiastic about distribution and try to distribute where distribution won’t work. For example, we can’t distribute multiplication over multiplication. To see why, find a counterexample. That is, find specific numbers a, b, c for which the expression $ a(b c) $ is not equal to $ ab ac $ .

  3. Now think of some specific counterexamples to demonstrate that

  1. Exponents do not distribute over addition. (That is, $ (a + b)^{n} a^{n} + b^{n} $ .)

  2. Square roots do not distribute over addition. (That is, $ + $ .)

The moral of the story: Multiplication distributes over addition, but not everything distributes over addition!

FOILed Again

(In Praise of Distribution, Part 1)

Early in your first algebra course, you learned how to multiply two “binomials”. That is, you learned how to expand $ (a + b)(c + d) $ out into the form $ ac + ad + bc + bd $ . Most algebra teachers summarize the steps in this process with the acronym FOIL (First, Outside, Inside, Last). The acronym is so common that it has become a verb, as in “when we FOIL this out, we obtain…”.

All algebra students know how to “FOIL”, but surprisingly few know why it works. This is a pity, since the explanation is so simple. It involves nothing but the distributive property. The key idea is that we can, when it’s convenient to do so, think of a binomial as a single entity to be distributed. To emphasize this idea, I’ll put a binomial in a grey box when I want to emphasize its unitary nature. Watch carefully:

\[ \begin{aligned}(a+b)(c+d)&=(a+b)(c+d)\\&=(a+b)c+(a+b)d\\&=ac+bc+ad+bd\\ \end{aligned} \]

\[ \begin{aligned}&(distributing the“box”,(a+b))\\&(distributing the c,and also the d)\end{aligned} \]

Thus, under the hood, the mysterious “FOIL” operation is just shorthand for carrying out the distributive property several times. If you understand this, you can easily figure out how to multiply two trinomials, three binomials, etc., without having to wait for someone to drill new acronyms into your head.

Exercises

  1. Explain to someone else why the “FOIL” rule is just a consequence of the distributive property.

  2. You probably know that the algebraic expression $ 3x^{2} + 4x^{2} $ can be rewritten as $ 7x^{2} $ . But why is this allowed? It might feel obvious, but a feeling isn’t an explanation. After all, someone else might feel that $ (3x{2})(4x{2}) $ should be $ 12x^{2} $ , which is entirely wrong. (Factors don’t care about your feelings.) In fact, $ 3x^{2} + 4x^{2} $ equals $ 7x^{2} $ because of the distributive property. Explain why.

[Hint: This requires some cleverness. Try “undistributing” something from $ 3x^{2} + 4x^{2} $ .]

  1. While reviewing some algebra, Esau encounters the expression $ x{2}x{3} $ . “Oh, I think I know how to simplify that,” says Esau. “You just multiply the exponents so it becomes $ x^{6} $ , right?” Looking up from the delicious red stew, he is cooking, Jacob flashes a grin at his brother and replies, “I don’t think so. You’re supposed to add the exponents, Esau. You should end up with $ x^{5} $ .” Esau begins to change his answer, but then hesitates and says, “You’re always tricking me, Jacob. Are you tricking me now?” To which Jacob replies, “Maybe. Maybe not.”

  1. Did Jacob give Esau the correct answer?

  2. Don’t settle for a true opinion. Turn it into knowledge: Explain why the correct answer is correct.

  3. Simplify each of the following: $ x{3}x{3} $ , $ xx^{4} $ , $ x{4}x{6} $ , $ (2x{2})(3x{4}) $ , $ (-3x^{3})(2x)(5x) $ .

  1. Use the distributive property (or equivalently, FOIL, when appropriate) to multiply the following polynomials.

\[ (3x-7)(2x+4) \]

\[ (-x^{4}+3)(-2x^{2}+6x-1) \]

\[ \begin{aligned}&b)\left(-x+2\right)\left(-2x-3\right)\\&d)\left(x^{2}-2x+3\right)\left(-x^{2}+2x-7\right)\\&f)\left(x+1\right)\left(x+2\right)\left(x+3\right)\\ \end{aligned} \]

\[ (x^{3}+x^{2}-x+1)(-x^{3}+x) \]

[Hint for Part F: One step at a time. First do $ (x + 1)(x + 2) $ . Then multiply the result by $ (x + 3) $ .]

\[ \mathsf{g})\left(x-1\right)(x^{3}+x^{2}+x+1) \]

\[ (x-1)(x^{5}+x^{4}+x^{3}+x^{2}+x+1) \]

  1. (x - 1)(x 99 + x 98 + x 97 + … + x 3 + x 2 + x + 1). [The “⋯” indicates that the pattern continues.]

Factoring Polynomials

(In Praise of Distribution, Part 2)

“Factoring” is nothing more than distribution in reverse (or “undistribution” as I called it in Exercise 9). For example, why is it true that $ 3x^{2} + 6x - 18 = 3(x^{2} + 2x - 6) $ ? If you read that equation from right to left, you’ll see why: It’s the good old distributive property that justifies that equals sign.

You may have met (and forgotten) the “difference of squares” identity, $ a{2}-b{2}=(a-b)(a+b) $ . Reading it from left to right, it is somewhat mysterious. What makes this strange factoring formula true? Why should we believe it? Well, if we read it from right to left, we see how we might prove that it’s true. We’ll just need to multiply the factors on the right and see if we end up with the expression on the left. I’ll do this now, putting a binomial in a grey box when we’re thinking of it as one block.

Claim. $ a{2}-b{2}=(a-b)(a+b) $ for all a and b.

\[ \begin{aligned}Proof.\quad(a-b)(a+b)&=(a-b)(a+b)\\&=(a-b)a+(a-b)b\quad&(distributing the(a-b))\\&=a^{2}-ab+ab-b^{2}\quad&(distributing the a,and also the b)\\&=a^{2}-b^{2},\text{as claimed.}\end{aligned} \]

Note how the proof boils down to three applications of distributive property – and nothing else. It is no exaggeration to say that the difference of squares formula holds because of the distributive property. In fact, the difference of squares formula is just the tip of the factoring iceberg. Every polynomial factorization can be justified by “reading right to left”, multiplying the factors, and verifying that the result is the original polynomial. Since multiplying those factors is, as we’ve seen, just a matter of repeatedly distributing, it follows that the entire topic of factoring polynomials is justified by the humble distributive property. There’s nothing deep here.

Most algebra textbooks make it seem as though factoring a polynomial is a complicated procedure. It isn’t. Almost any polynomial you’ll ever need to factor by hand will yield to one (or some combination) of three basic tricks:

  1. Pull out a factor common to all the terms.

  2. Use the difference of squares formula.

  3. Make (and check) educated guesses until you find the right combination.

The first two tricks are entirely mechanical, and require little comment. Here’s an example that uses both tricks in turn:

\[ 2x^{2}-32=2(x^{2}-16)=2(x-4)(x+4). \]

Nothing to it: We factored out the 2, noticed that one of the resulting factors was a difference of squares, and thus applied the difference of squares formula. You should commit the difference of squares formula to memory since it comes in handy so often.

The third trick, making educated guesses, works best for polynomials of the form $ ax^{2} + bx + c $ . For example, suppose we wish to factor $ x^{2} + x - 12 $ . Well, if it factors, the result might look like this: $ x^{2} + x - 12 = (x - 12)(x - 1) $ . (We’ll leave the constant terms blank to keep some play in the joints.) Note that putting those two x’s on the right was an educated guess; we did it because their product (FOIL’s “F”) is $ x^{2} $ , which matches one term of our “target polynomial” on the left. To continue our educated guessing, we observe that whatever the factors’ constant terms are, their product (FOIL’s “L”) must be

−12 to match the target polynomial’s constant term. There are many possibilities here, so let’s pick one pair of numbers whose product is 12 and try it out: How about $ (x - 4)(x + 3) $ ? Will this work? We can check by multiplying it out in our heads; $ ^{*} $ doing so, we find that we end up with the wrong x term: we end up with -x instead of +x. Close, but no cigar. What if we just swapped the locations of the positive and negative, like this: $ (x + 4)(x - 3) $ ? A quick check shows that this does work, so our factoring is complete: $ x^{2} + x - 12 = (x + 4)(x - 3) $ . And that’s all there is to the third trick. With a little practice, which you’ll in the exercises below, you’ll develop an intuition for making good guesses.

As you can see, factoring efficiently depends upon being able to multiply binomials in your head… which is just FOILing… which is just using the distributive property. Yes, it requires practice and patience, but everything here should be entirely comprehensible.

Exercises

  1. Prove that $ (a+b){2}=a{2}+b^{2}+2ab $

  2. Prove that $ (a-b){2}=a{2}+b^{2}-2ab $

  3. Prove that $ (a+b){3}=a{3}+b{3}+3a{2}b+3ab^{2} $

  4. Prove the “difference of cubes” identity: $ a{3}-b{3}=(a-b)(a{2}+ab+b{2}) $ .

  5. Factor the following polynomials as much as possible.

  1. $ x^{2} + 6x + 8 $

  2. $ x^{2} - 100 $

  3. $ 10x^{2} + 5x $

  4. $ x^{2} - 7x + 10 $

  5. $ 3x^{2} - 3x - 6 $

  6. $ x^{2} - 25 $

  7. $ -4x^{2} - 32x - 64 $

  8. $ -15x^{2} - 30x + 45 $

  9. $ x^{4} - 16 $ [Hint: $ x^{4} = (x{2}){2} $ ]

  10. $ 9x^{2} - 4 $ [Hint: The first term is a square.]

  11. $ 81x^{8} - 1 $

  12. $ 2x^{2} + 5x + 2 $ [Hint: $ (2x - 3)(x - 3) $ ]

  13. $ 3x^{2} - 8x - 3 $

  14. $ 4x^{2} - 4x - 3 $

  1. The algebraic identities that you proved in exercises 12 and 13 are used so frequently that you should commit them to memory. From now on, whenever you need to square a binomial, you should apply these identities directly; don’t reinvent the wheel each time. For example, to expand $ (x + 3)^{2} $ , you should not “FOIL it out” (even though doing so will, of course, yield the correct expansion). Rather, you should just mentally apply the identity, which allows you to write $ (x + 3)^{2} = x^{2} + 9 + 6x $ without any further ado.

In this spirit, quickly expand the following:

  1. $ (x + 5)^{2} $

  2. $ (x - 5)^{2} $

\[ (x+11)^{2} \]

\[ (2x+1)^{2} \]

\[ \mathsf{d})(x-12)^{2} \]

\[ (3x-2)^{2} \]

  1. $ (a+)^{2} $ [Hint: $ $ is, by definition, the number whose square is 2, so the value of $ ()^{2} $ must be…]

  2. $ (a-)^{2} $ i) $ (2x+1/2)^{2} $ j) $ (2a+3b)^{2} $ k) $ (-1)^{2} $

Why Minus Times Minus is Plus (In Praise of Distribution, Part 3)

Minus times minus is plus.

The reasons for this we need not discuss.

– Ms. Anonymous

Most rules for working with negatives are easy to understand if we think in terms of debits and credits. It’s obvious why $ (-1)(1) $ should be -1: A debit of 1 incurred one time is obviously a debit of 1. Similarly, it is obvious why $ -1 + 1 = 0 $ : Adding a credit of 1 to a debit of 1 results in a net value of 0.

But why should $ (-1)(-1) $ be 1? A “debits and credit justification” exists, but not a very good one. $ ^{*} $ A much better argument – and certainly a more artistic argument – rests on the distributive property.

Claim 1. $ (-1)(-1)=1 $ .

Proof. We’ll begin the proof in godlike fashion, by creating something from nothing.

\[ \mathbf{0}=(-1)(0)=(-1)(-1+1)=(-1)(-1)+(-1)(1)=(-\mathbf{1})(-\mathbf{1})-\mathbf{1}.^{\dagger} \]

We’ve established that $ 0 = (-1)(-1) - 1 $ . Thus, whatever $ (-1)(-1) $ may be, we’ve deduced that subtracting 1 from it leaves us with zero. Obviously, the only number with that property is 1, so it follows that $ (-1)(-1) = 1 $ , as claimed.

Since every negative number can be written as $ (-1) $ times a positive number, we see that, for instance,

\[ (-2)(-3)=(-1)(2)(-1)(3)=(2)(3)(-1)(-1)=(2)(3), \]

where that last equals sign is justified by Claim 1. And if we wish to write up a formal proof covering the product of any two negatives, we could do so as follows.

Claim 2. (Minus times minus is always plus.) $ (-a)(-b)=ab $ for any a and b.

Proof. Let -a and -b represent any two negative numbers. Then

\[ (-a)(-b)=(-1)(a)(-1)(b)=ab(-1)(-1)=ab. \]

The last equals sign is justified by Claim 1.

Congratulations. You are now among the select few people who know why minus times minus minus is plus! By now, I trust that you’ve been convinced of the fundamental importance of the distributive property, which, among other things, lets us make rapid mental calculations, explains why we can multiply and factor polynomials, and helps us understand why minus times minus is plus. I’ll not dwell on it any longer. It will always be there, functioning smoothly under the hood. Part of the process of learning to think mathematically is to develop an appreciation for how even simple mathematical ideas can have enormous logical ramifications. You’ll have another opportunity shortly, when we turn our attention to fractions. But first, some exercises.

Exercises

  1. You can now understand how to divide with negatives, provided you understand division itself.

Quick review: $ 10 $ is the number of times 5 “goes into” 10. Phrased differently, $ 10 $ asks the question, “5 times what is 10?” The answer, obviously, is 2. Similarly, $ 8 (-4) $ asks how many times $ (-4) $ goes into 8; in other words, “-4 times what is 8?” Of course, the answer is -2.

  1. Explain why $ (-8) (-4) $ is 2.

  2. Convince yourself that minus divided by minus must always be plus.

  3. Explain why $ 9 (-3) $ is -3.

  4. Convince yourself that plus divided by minus must always be minus.

  5. What about minus divided by plus?

  1. Explain why $ 10 (1/3) = 30 $ .

  2. The expressions -8/2, 8/-2, and -(8/2) are all equal, right? (“Of course!” you shout out, “They’re all -4!”) And how about 3/-5, -3/5, and -(3/5)? They’re all equal, too? (“Naturally,” I hear you cry, “Each is -0.6.”) Let’s cut to the chase and express this algebraically:

The expressions -a/b, a/-b, and -(a/b) are all equal, regardless of what a and b are.

Convince yourself that this is true.

  1. Can you divide a nonzero number by zero? If so, what is the result? If not, why not?

  2. Can you divide zero by a nonzero number? If so, what is the result? If not, why not?

  3. What about 0/0?

  4. Explain to a friend why minus times minus is plus.

Fractions: Two Intuitive Rules That Lead to All the Others

Any fool can know. The point is to understand.

  • Albert Einstein

We’ll begin our study of fractions with an observation: You get the same amount of dessert by taking two fifths of a pie as you do by taking one fifth two times. Or, translating this statement into symbols,

\[ \frac{2}{5}=2\left(\frac{1}{5}\right). \]

Of course, there was nothing special about 2 or 5 in that last example. We might just as well have noted that seven eighths of a pie is the same as one eighth taken seven times, so $ 7/8 = 7(1/8) $ . Algebra, the science of patterns, allows us to describe the pattern we are seeing here as follows.

Intuitive Fraction Rule 1:

\[ \frac{a}{b}=a\left(\frac{1}{b}\right) \]

Now for a second observation. Since 10 tenths make a whole, cutting each tenth into thirds would give us 30 equal parts. Thus, each third of a tenth is exactly 1/30 of the whole. Or, in symbols,

\[ \left(\frac{1}{3}\right)\left(\frac{1}{10}\right)=\frac{1}{30}. \]

Similarly, an eighth of a half must be a sixteenth, so $ (1/8)(1/2)=1/16 $ . The algebraic pattern here is:

Intuitive Fraction Rule 2:

\[ \left(\frac{1}{a}\right)\left(\frac{1}{b}\right)=\frac{1}{ab} \]

We shall take it as an axiom that our two “intuitive fraction rules” hold for all values of a and b.

Now for a surprise: Over the next few pages, we’ll see that every aspect of fractional arithmetic – that subject that confuses so many people – follows logically from those two simple rules! This should encourage those who find fractions confusing. Along the way, we’ll clarify quite a bit of algebra too.

Exercise

  1. The Queen of Sheba tests Solomon with a riddle: “O Great King, we all know the rule for multiplying fractions: Multiply the tops and multiply the bottoms. For example, $ (2/3)(4/5) = 8/15 $ .” Solomon nods sagely. “But why is this so? The wise men of my land say only that it is the gods’ will, but I am told that you, a mortal man, know the explanation. What is it, good King?”

Solomon begins by explaining the two intuitive fraction rules, which the Queen allows are quite intuitive. He then proceeds to show her why $ (2/3)(4/5) $ must be 8/15 because of the two rules.

Your problem: Explain how Solomon did it.

Multiplying and Reducing Fractions

From the two intuitive fraction rules, we can fully explain the well-known rule for multiplying fractions. (And if you’ve solved exercise 24, you’ve basically discovered this on your own already.)

Multiplication Rule for Fractions.

To multiply fractions, we multiply the tops and multiply the bottoms. Or in symbols,

\[ \frac{\boldsymbol{a}}{\boldsymbol{b}}\cdot\frac{\boldsymbol{c}}{\boldsymbol{d}}=\frac{\boldsymbol{a}\boldsymbol{c}}{\boldsymbol{b}\boldsymbol{d}} \]

\[ \begin{aligned}Proof.\frac{a}{b}\cdot\frac{c}{d}&=a\left(\frac{1}{b}\right)c\left(\frac{1}{d}\right)\quad&(applying Intuitive Fraction Rule1)\\&=ac\left(\frac{1}{b}\right)\left(\frac{1}{d}\right)\quad&(reordering the multiplication)\\&=ac\left(\frac{1}{bd}\right)\quad&(Intuitive Fraction Rule2)\\&=\frac{ac}{bd}\quad&(Intuitive Fraction Rule1^{*})\end{aligned} \]

Thus, for example, $ (6/7)(2/3) = 12/21 $ . You probably also know that 12/21 can be reduced to 4/7. But can you explain why we are allowed to remove that common factor of 3 from its top and bottom? This is no idle question. People who don’t understand why numerical fractions can be reduced (even if they know how to do it) tend to botch the analogous algebraic operation, which we’ll discuss in the next section.

We “reduce” a fraction by throwing out factors that are common to its numerator and denominator. An example will show exactly why we can do this:

\[ \begin{array}{l}\frac{12}{21}=\frac{4\cdot3}{7\cdot3}=\frac{4}{7}\cdot\frac{3}{3}=\frac{4}{7}\cdot\mathbf{1}=\frac{4}{7}.\end{array}^{\dagger} \]

Thus, reducing a fraction is really nothing more than eliminating a hidden factor of 1. Very simple.

Exercises

  1. Reduce the following fractions: $ $ , $ $ , $ $ .

  2. Simplify $ $ by hand, without first computing $ 208 $ or $ 12 $ . Justify each step in your work.

[Hint: Break some of the big numbers on top down to simpler factors; when you do, you’ll find the numerator and denominator have some common factors. Once you remove them, you’ll be able to proceed more comfortably.]

  1. Explain why $ a()= $ . (This is another algebraic fact that we use all the time without thinking about it.)

Cancelling Above and Below the Bar

And may there be no moaning of the bar,

When I put out to sea.

  • Alfred Lord Tennyson, Crossing the Bar

Everyone knows that we can simplify $ (ab + a)/a $ by cancelling some $ a^{}s $ , but not everyone knows why. Consequently, many students produce incorrect “simplifications” such as ab or $ b + a $ . Those who truly understand algebra never make such mistakes, for they know that cancelling is nothing more than discarding a hidden factor of 1. The logic here is the same as with reducing ordinary numerical fractions.

Example. Simplify $ $ .

Solution.

\[ \frac{ab+a}{a}=\frac{a(b+1)}{a} \]

(factoring a from the numerator)

\[ =\frac{a}{a}\cdot\frac{b+1}{1} \]

(by the multiplication rule for fractions)

\[ =b+1. \]

\[ (because a/a=1) \]

I repeat: “Cancelling above and below the bar” is merely shorthand for discarding a hidden factor of 1. It is essential that you understand this thoroughly. Whenever you are tempted to “cancel” something from above and below a fraction bar, just ask yourself if you can separate it off as a factor of 1. If you can, you can cancel; if not, you can’t cancel. That’s all there is to it.

“Cancelling” above and below the fraction bar

If the top and bottom of a fraction have a common factor, you can “cancel” it from both places.

You cannot cancel above and below a fraction bar under any other circumstance.

You cannot cancel in other circumstances simply because there is no logical justification for doing so. Mathematics is justified by logic, even if “math” is all too often justified by the teacher’s authority.

Exercises

  1. The crucial word in the box above is factor. Lest there be any confusion at all, please recall that

k is called a factor of an algebraic expression if the expression can be written in the form $ k () $ . [For example, $ 3x^{2} $ is a factor of $ 6x{3}y{2} $ because we can write the latter as $ 3x{2}(2xy{2}) $ .]

  1. Is $ 2a{2}b{2} $ a factor of $ 4a{2}b{2}-18a{5}b{3} $ ? If so, explain why. If not, explain why not.

  2. Provide a definition for a term of an algebraic expression similar to the definition of a factor above.

  3. Can we cancel common terms from a fraction’s top and bottom? If so, why? If not, give a counterexample.

  4. Simplify the following expressions as much as possible:

\[ \frac{a^{2}b}{ab^{2}}, \]

\[ \frac{3x+3xy}{6xyz}, \]

\[ \frac{5a}{5a+10b-15c}, \]

\[ \frac{c^{2}-d^{2}}{c-d}. \]

  1. A good mathematical joke: $ = $ . Discuss.

The Old Multiply By 1 Trick

To reduce a fraction, we remove a factor of 1. Surprisingly, introducing a factor of 1 can also be useful. Multiplying a fraction by a judiciously disguised 1 preserves the fraction’s value but changes its form to something more convenient. I call this “the old multiply by 1 trick”. It nicely captures the spirit of algebra. $ ^{*} $

Here’s an easy example of it in action: 3/5 is how many twentieths? To answer this question, we observe that multiplying our fraction’s bottom by 4 would change it to 20, which is just what we want. Good news: We can get what we want if we also multiply the top by 4, because the net effect will be to multiply the fraction by 1, which preserves its value. Thus,

\[ \frac{3}{5}=\frac{3}{5}\cdot\frac{4}{4}=\frac{12}{20} \]

Such is the old multiply by 1 trick. Now let’s see it in a more substantial application.

Dividing Fractions

It’s not yours to wonder why,

Just invert and multiply.

  • Mr. Anonymous

Everyone knows the rule for dividing fractions, but few know why it works. The explanation, however, is simple if you understand the old multiply by 1 trick. Watch for it in the proof below.

Division Rule for Fractions.

To divide one fraction by another, we “invert and multiply.” Or in symbols,

\[ \frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a}{b}\cdot\frac{d}{c} \]

Proof.

(the old multiply by 1 trick)

\[ \begin{aligned}\frac{\frac{a}{b}}{\frac{c}{d}}&=\frac{\frac{a}{b}}{\frac{c}{d}}\cdot\frac{\frac{d}{c}}{\frac{d}{c}}\\&=\frac{\frac{ad}{bc}}{1}\\&=\frac{ad}{bc}\\&=\frac{a}{b}\cdot\frac{d}{c}\end{aligned} \]

(multiplication rule for fractions)

(dividing by 1 does nothing)

(multiplication rule for fractions)

Having proved the theorem, let’s consider a typical algebraic example in which we can apply it.

Problem. Simplify $ $ .

Solution.

\[ \begin{aligned}\frac{\frac{a^{2}-b^{2}}{c}}{\frac{a-b}{c^{2}}}&=\frac{a^{2}-b^{2}}{c}\cdot\frac{c^{2}}{a-b}\quad&(invert and multiply)\\&=\frac{(a^{2}-b^{2})c^{2}}{c(a-b)}\quad&(multiplication rule for fractions)\\&=\frac{(a^{2}-b^{2})c}{(a-b)}\quad&(cancelling a factor of c from top and bottom)\\&=\frac{(a-b)(a+b)c}{(a-b)}\quad&(factoring a difference of squares)\\&=(a+b)c\quad&(cancelling a factor of(a-b)from top and bottom).\end{aligned} \]

Note how each step in our solution was justified by something whose validity we established earlier. Thus, if you’ve understood all we’ve done so far, such problems should pose no real difficulties.

Exercises

  1. True or False? Explain why each true statement is true:

  1. $ =1+a^{2} $ b) $ = $ c) $ =a+3 $ d) $ = $

  2. $ = $ f) $ = $ g) $ = $ h) $ = $

  3. $ = $ j) $ = $ k) $ =1 $ l) $ = $

  1. The expression $ n! $ (“n factorial”) represents the product of the first n numbers. (E.g. $ 4! = 4 = 24 $ .) Your problem: Simplify the expression $ 98!/100! $ .

  2. True or false: $ = $ . [Moral: In such fractions, it’s crucial to draw one fraction bar longer than the other.]

  3. Given any number a, its reciprocal is defined to be the number 1/a. (For example, the reciprocal of 8 is 1/8.) Useful fact: To take the reciprocal of a fraction, we just flip the top and bottom. (E.g. the reciprocal of 2/3 is 3/2.)

  1. Prove the preceding “useful fact”, remember it, and use it whenever such expressions arise from now on.

  2. Simplify the following: $ $ , $ $ , $ $ , $ $ . [Careful with that last one. Remember exercise 33.]

  1. Show how to use the multiply by 1 trick to write $ $ in the cleaner form $ $ .

  2. Simplify the following expressions as much as possible:

\[ \begin{aligned}&a)\frac{x^{2}-4}{(x-4)(x+4)}\quad b)\frac{x^{2}-16}{(x-4)(x+4)^{2}}\quad c)\frac{\frac{x}{y}}{z}\quad d)\frac{\frac{a+b}{c}}{d}\quad e)\frac{\frac{a+b}{c}}{cd}\quad f)\frac{\frac{a}{b}}{bd}\cdot\frac{b^{2}}{a^{2}}\quad g)\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10}{4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10}\\&h)\frac{3a+2b}{9a^{2}-4b^{2}}\quad i)\frac{3x^{2}-15x}{15x-3x^{2}}\quad j)\left[\left(\frac{ab}{cd}\cdot\frac{ac}{bd}\right)\div\frac{d^{2}}{a^{2}}\right]\frac{a^{4}}{d^{2}}\quad k)\frac{10x^{2}-10x-60}{5x+10}\quad l)\frac{\frac{2x^{2}-3x-2}{2x+1}}{\frac{x-2}{5}}.\end{aligned} \]

(A Parenthetical Aside (on Parentheses))

Algebra is generalized arithmetic: Algebraic expressions represent unspecified numbers. Often, we can view the same expression in multiple ways. Consider 3d - c. We can view this as a representation of one number $ (3d - c) $ , or as the difference of two numbers $ (3d $ and c), or as a combination of three numbers $ (3, d, $ and c). Whenever we wish to emphasize that we are thinking of an algebraic expression as a single number – a single package – we do so by enclosing it in parentheses.

This brings us to this brief section’s central idea:

When we combine algebraic expressions, we think of each expression as representing a single number. Hence, we initially enclose each individual expression in parentheses. We can remove the parentheses in subsequent steps, provided we distribute any negatives or constant factors preceding them.

For example, if we wish to subtract 3d - c from $ 2d + c $ , we form the difference $ (2d + c) - (3d - c) $ . The first set of parentheses isn’t preceded by anything that needs to be distributed, so we can drop them when we simplify. But before we remove the second set, we must distribute that pesky negative. Carrying out these simplifications, we obtain:

\[ (2d+c)-(3d-c)=2d+c-3d+c=2c-d. \]

Before too long, you will reach the stage in your studies at which you never mistakenly omit parentheses. At that point, you’ll be able to do much of this mentally, but until then, for your own sake, write it out.

Given a slightly different subtraction such as, say, $ (2d + c) - 5(3d - c) $ , we’d need to distribute not just the negative (i.e. -1) but -5 to the terms in the second set of parentheses. Thus, we’d have

\[ (2d+c)-5(3d-c)=2d+c-15d+5c. \]

Naturally, all that I’ve written in this section about parentheses applies to other grouping symbols, such as brackets, which we use to mitigate clutter in expressions such as $ [3a-(a+b)][(a-b)(a+b)] $ .

Exercises

  1. Remove all grouping symbols and simplify:

\[ a)-(a-b)+(a+2b) \]

\[ \mathsf{b})\left(2a+b+3c\right)-2(c-a+b) \]

\[ \mathsf{c})3(a-b)-3(b-a) \]

\[ \mathsf{d})\left(-2a-b\right)-\left[5a-\left(3a+3b\right)-\left(a-b\right)\right] \]

\[ \mathsf{e})x-(x-y+z)+[x-y-(z+y+x)] \]

\[ \mathfrak{f})\ a-\left[a+\left(a-(a+a)\right)\right]-a \]

  1. Subtract $ 2x^{2} + x - 1 $ from $ -x^{2} + 5x + 1 $ .

Adding and Subtracting Fractions

To add (or subtract) fractions with the same denominator, we just add (or subtract) their numerators, putting the result over their common denominator. This much is clear even to pizza-mad schoolchildren: Suppose a pizza is sliced into 10 equal pieces. A child with one piece (1/10 of the pie) who filches two more (2/10 of the whole) from inattentive classmates now obviously has 3/10 of the whole pie.

But what if the fractions have different denominators? Mathematicians, being lazy by nature, like to solve new problems by transforming them into old problems that we’ve already figured out how to solve. Let’s be lazy: To add or subtract fractions with different denominators, we’ll use the “multiply by 1 trick” to transform them into fractions with the same denominator. This common denominator will need to be a multiple of both original denominators. $ ^{*} $ Here’s a typical example:

Problem. \(\frac{5}{6}-\frac{1}{10}\).

Solution. We need a denominator that is a multiple of both 6 and 10. The smallest such number is 30, so we’ll use the “multiply by 1 trick” to change those sixths and tenths into thirtieths:

\[ \begin{aligned}\frac{5}{6}-\frac{1}{10}&=\left(\frac{5}{6}\cdot\frac{5}{5}\right)-\left(\frac{1}{10}\cdot\frac{3}{3}\right)\quad&(the“multiply by1trick”)\\&=\frac{25}{30}-\frac{3}{30}\quad&(bythemultiplicationruleforfractions)\\&=\frac{22}{30}\quad&(subtractingfractionswiththesamedenominator)\\&=\frac{11}{15}\quad&(reducingthefraction).\\ \end{aligned} \]

If you understand that one numerical example, you understand them all. I won’t belabor the point.

Algebra is generalized arithmetic, so the rules for numerical fractions work for algebraic fractions, too. For example, to subtract algebraic fractions with the same denominator, we just subtract their numerators. We must, however, take care with parentheses!

Example 1. Subtract \(\frac{3d-c}{c\sqrt{b}}\) from \(\frac{14d^{2}+2d+c}{c\sqrt{b}}\) and simplify the result (if possible).

Solution.

\[ \begin{aligned}\frac{14d^{2}+2d+c}{c\sqrt{b}}-\frac{3d-c}{c\sqrt{b}}&=\frac{\left(14d^{2}+2d+c\right)-\left(3d-c\right)}{c\sqrt{b}}\quad(note the parentheses!)\\&=\frac{14d^{2}+2d+c-3d+c}{c\sqrt{b}}\\&=\frac{14d^{2}-d+2c}{c\sqrt{b}}.\end{aligned} \]

Be sure you understand those parentheses in the first step: When subtracting two fractions with a common denominator, the new numerator is the difference of the two original numerators; each of these was an algebraic expression, so to subtract them, we had to think of each as a single package. Consequently, we had to enclose each of them in parentheses. Had we omitted the parentheses, we would have ended up with the wrong numerator, and hence the wrong answer.

To add or subtract algebraic fractions with different denominators, we’ll use – just as you’d expect – the “multiply by 1 trick” to give the fractions a common denominator. For example,

Example 2. Add and simplify: $ + $ .

Solution. For a common denominator, we need a multiple of $ 2a^{2}b $ and 6ab. The “smallest” such denominator is $ 6a^{2}b $ . Using the “multiply by 1 trick,” we’ll convert to this new denominator.

\[ \begin{aligned}\frac{3}{2a^{2}b}+\frac{7}{6ab}&=\left(\frac{3}{2a^{2}b}\cdot\frac{3}{3}\right)+\left(\frac{7}{6ab}\cdot\frac{a}{a}\right)\quad&(the“multiply by1trick”)\\&=\frac{9}{6a^{2}b}+\frac{7a}{6a^{2}b}\\&=\frac{9+7a}{6a^{2}b}\end{aligned} \]

After you’ve become comfortable adding and subtracting fractions with different denominators, you need not write out every step. In practice, we usually condense the process as follows:

Addition and Subtraction Rule for Fractions. We find the denominator (a so-called “common denominator”) as follows: Take any multiple of the given denominators.* We find the numerator as follows: Multiply each given numerator by the factor that will turn its denominator into the common denominator. These products will be the terms in the new numerator.

If we re-do the previous example using this shortcut addition rule, we’ll have somewhat less to write. Taking $ 6a^{2}b $ as our common denominator, the addition rule quickly tells us that

\[ \frac{3}{2a^{2}b}+\frac{7}{6ab}=\frac{3(3)+7(\boldsymbol{a})}{6a^{2}b}=\frac{9+7a}{6a^{2}b}. \]

Beginning students frequently make mistakes when subtracting fractions – usually because they omit necessary parentheses. Be especially careful when subtracting.

Example 3. Subtract and simplify: $ - $ .

\[ \begin{aligned}Solution.\ \frac{3}{x-5}-\frac{2x-1}{x+5}&=\frac{[3(x+5)]-[2x-1)(x-5)]}{(x-5)(x+5)}\quad&(subtraction rule)\\&=\frac{[3x+15]-[2x^{2}-11x+5]}{(x-5)(x+5)}\quad&(distributing within the brackets)\\&=\frac{3x+15-2x^{2}+11x-5}{x^{2}-25}\quad&(distributing a minus;difference of squares)\\&=\frac{-2x^{2}+14x+10}{x^{2}-25}\quad&(combining like terms)\quad\diamond\end{aligned} \]

Exercises

  1. Carefully explain why $ + = $ .

  2. Simplify the following expressions.

  1. $ (+) $

  2. $ --(-) $

  3. $ 1 - $

  1. In Example 2 above, we took $ 6a^{2}b $ as our common denominator. Suppose we had used $ 12a{3}b{2} $ instead. Would this have changed the result? Work it out that way and see.

  2. Express each of the following as a single fraction, and simplify as much as possible:

  1. $ + $ b) $ ()() $ c) $ - $ d) $ + $

  2. $ + $ f) $ - $ g) $ + + $ h) $ - + - $

  3. $ - $ j) $ - $ k) $ 3 - $ l) $ 1 - $

  4. $ - $ n) $ - + - $

Last Words and One Nasty Example

You can now add, subtract, multiply, or divide any two algebraic fractions. No lingering mysteries remain; you are fully initiated. Should you ever need to combine seventeen fractions linked in all sorts of intricate arithmetical ways, you have all the knowledge necessary to do it. All you must do is slow down and take each piece in order. It may be tedious, but it shouldn’t be difficult.

To illustrate my point, I’ll offer one last example. If you have understood everything so far in this chapter, it should pose no conceptual difficulties, even though it is rather involved.

Problem. Simplify the following ugly expression as much as possible: $ + $

Solution. Before we begin hacking away at this overgrown mess, let’s consider it from a distance to convince ourselves that, despite its ugliness, this is something we can handle. Begin by observing that the expression has two terms. True, the first one looks nasty (“Ugh! Fractions in a fraction!”), until we realize that when we subtract the two fractions in its numerator, they’ll obviously combine into a single fraction, which we can then divide by the fraction in the left term’s denominator. The result of that division will obviously be… a single fraction. Thus, when the dust settles, we’ll have rewritten the whole ugly first term as one fraction. All the “fractions within fractions” will be gone. As for the second term, you can probably simplify it in your head, turning it into one fraction, too. Then, all that will remain is to add two ordinary algebraic fractions – an easy task.

Having watched the preliminary dumbshow, let’s proceed to the actual details of calculation, confident that, at least in outline, we already know how matters will work out.

The first term’s numerator is

\[ \begin{aligned}\frac{2}{x-2}-\frac{x+1}{x+2}&=\frac{\left[2(x+2)\right]-\left[(x-2)(x+1)\right]}{(x-2)(x+2)}\\&=\frac{\left[2(x+2)\right]-\left[(x-2)(x+1)\right]}{x^{2}-4}\\&=\frac{\left[2x+4\right]-\left[x^{2}-x-2\right]}{x^{2}-4}\\&=\frac{2x+4-x^{2}+x+2}{x^{2}-4}\\&=\frac{-x^{2}+3x+6}{x^{2}-4}\end{aligned} \]

(note the brackets!)

(difference of squares)

(distributing within the brackets)

(removing brackets, distributing the negative)

(combining like terms).

Now that we’ve simplified the first term’s numerator, we’ll divide it by the first term’s denominator. Doing so, we find that the entire first term simplifies to

\[ \begin{aligned}\frac{\frac{-x^{2}+3x+6}{x^{2}-4}}{\frac{3x-7}{x^{2}-4}}&=\frac{-x^{2}+3x+6}{x^{2}-4}\cdot\frac{x^{2}-4}{3x-7}\quad&(division rule for fractions)\\&=\frac{\left(-x^{2}+3x+6\right)\left(x^{2}-4\right)}{(x^{2}-4)(3x-7)}\quad&(multiplication rule for fractions)\\&=\frac{-x^{2}+3x+6}{3x-7}\quad&(cancelling(x^{2}-4)above and below the bar).\end{aligned} \]

Having reduced the first term to something manageable, we can tackle the original problem:

\[ \begin{aligned}\frac{\frac{2}{x-2}-\frac{x+1}{x+2}}{\frac{3x-7}{x^{2}-4}}+\frac{\frac{2}{x^{2}}}{-5}&=\frac{-x^{2}+3x+6}{3x-7}+\frac{\frac{2}{x^{2}}}{-5}\\&=\frac{-x^{2}+3x+6}{3x-7}-\frac{2}{5x^{2}}\\&=\frac{[\left(-x^{2}+3x+6\right)\left(5x^{2}\right)]-\left[2(3x-7)\right]}{(3x-7)\left(5x^{2}\right)}\\&=\frac{\left[-5x^{4}+15x^{3}+30x^{2}\right]-\left[6x-14\right]}{15x^{3}-35x^{2}}\\&=\frac{-5x^{4}+15x^{3}+30x^{2}-6x+14}{15x^{3}-35x^{2}}.\end{aligned} \]

(by our work on the first term above)

(division of fractions, second term)

(subtraction rule for fractions)

(a great flurry of multiplication)

(removing brackets upstairs)

There were many steps in that last problem, but each was simple. Occasional mistakes in such problems are inevitable, but bear in mind that there are mistakes and mistakes. Accidentally writing $ 3 = 6 $ in the midst of a larger problem is wrong, but it presumable isn’t a conceptual error. On the other hand, omitting crucial parentheses, forgetting to distribute negatives, or botching the multiply by 1 trick are serious mistakes that most likely stem from conceptual misunderstandings. If you intend to take further mathematics courses, you need to clear up any and all such misunderstandings immediately. Those courses will give you plenty of new material to think about; if you are still struggling with basic algebra at that stage, you won’t see the forest for the trees.

Exercises

  1. Simplify as much as possible:
  1. $ $ b) $ $ c) $ ()(b+c) $ d) $ ()b+c $
  1. True or false (explain your answers):

  1. $ -3^{2} = 9 $

  2. -x always represents a negative number.

  3. $ (-3)^{2} = -9 $

  4. a - b = -(b - a).

  5. $ = $

  6. $ = $

  7. $ = $

  1. Express as a single fraction – and simplify as much as possible:

  1. $ + $ [Hint: You may find Exercise 44d useful.]

  2. $ $

  3. $ $

  4. $ $

\[ \mathrm{e})\left[\frac{5x+4}{x+1}-\frac{-3x^{2}+9x+4}{(x+1)^{2}}\right]\div\left(\frac{4x^{3}}{(x+1)^{2}}\right) \]

  1. $ 1 + $

  2. $ $

  3. $ $

  4. $ $

\[ \mathrm{j})\left(\frac{x^{2}}{1+x^{2}}\cdot\frac{\left(1+x^{2}\right)\left(-x^{2}\right)}{x^{4}}\right)(a-b) \]

  1. Early in the chapter, I claimed that all of fractional arithmetic follows logically from “two intuitive rules”:

\[ \frac{a}{b}=a\left(\frac{1}{b}\right)\text{and}\left(\frac{1}{a}\right)\left(\frac{1}{b}\right)=\frac{1}{ab}. \]

Now that you know all the rules of fractional arithmetic, it’s worth reexamining my claim. I explicitly pointed out in the text how the multiplication rule for fractions follows directly from the two intuitive rules. What about cancelling factors above and below the bar? Well, we saw that this operation is really just a matter of detaching a hidden factor of 1, like so:

\[ \frac{6ab}{3bc}=\frac{3b\cdot2a}{3b\cdot c}=\left(\frac{3b}{3b}\right)\left(\frac{2a}{c}\right)=1\left(\frac{2a}{c}\right)=\frac{2a}{c}. \]

The “detachment” (at the second equals sign) justified by the multiplication rule… which was built on the two intuitive rules. (The other steps are justified by completely obvious facts such as “multiplying by 1 doesn’t change anything” or “anything divided by itself is 1”.) Thus cancellation above and below the bar is ultimately a logical consequence of the two intuitive rules.

  1. Convince yourself that the “multiply by 1 trick” is ultimately justified by the two intuitive rules.

  2. Do the same for the division rule for fractions.

  3. Do the same for the addition and subtraction rule for fractions.

  4. Congratulate yourself: You’ve now seen that the whole vexed subject of fractions is actually quite simple, built up from just a few intuitive rules and the logical consequences thereof.