Beautiful, Simple, Exact, Crazy: Mathematics in the Real World

By Apoorva Khare, Anna Lachowska

Yale UNIVERSITY PRESS

Copyright © 2015 Yale University
All rights reserved.
ISBN: 978-0-300-19089-2

Contents

Preface, vii,
Acknowledgments, xv,
1 Algebra: The art and craft of computation, 1,
2 Velocity: On the road, 19,
3 Acceleration: After the apple falls, 39,
4 Irrational: The golden mean and other roots, 59,
5 Exponents: How much would you pay for the island of Manhattan?, 79,
6 Logarithms I: Money grows on trees, but it takes time, 97,
7 Logarithms II: Rescaling the world, 107,
8 e: The queen of growth and decay, 129,
9 Finite series: Summing up your mortgage, geometrically, 149,
10 Infinite series: Fractals and the myth of forever, 167,
11 Estimation: What is your first guess?, 193,
12 Modular arithmetic I: Around the clock and the calendar, 213,
13 Modular arithmetic II: How to keep (and break) secrets, 237,
14 Probability: Dice, coins, cards, and winning streaks, 261,
15 Permutations and combinations: Counting your choices, 277,
16 Bayes’ law: How to win a car … or a goat, 297,
17 Statistics: Babe Ruth and Barry Bonds, 325,
18 Regression: Chasing connections in big data, 341,
The bigger story, 353,
Solutions to odd-numbered exercises, 357,
Practice exams, 423,
Index, 457,

CHAPTER 1

Algebra: The art and craft of computation

In this book, we hope to show you that a wide variety of real-world problems and applications can be tackled systematically, comprehensively, and relatively simply by using just a few mathematical formulas and techniques. In order to introduce the mathematics and then to apply it to the real world, it is essential to be able to work with mathematical expressions and quantities in a systematic manner. Thus, we first need to be comfortable with basic operations like adding or multiplying polynomials; solving equations and systems of (linear or other) equations; and choosing an optimal way to simplify an algebraic expression. Developing these techniques is the goal of this chapter.

Sometimes these techniques produce unexpected results which some of you may have seen as “magic tricks.” For instance, you can check that multiplying two consecutive odd numbers (or consecutive even numbers) yields one less than a perfect square (e.g., 5 · 7 + 1 = 36 = 62, 10 · 12 + 1 = 121 = 112). Is this always the case, or can we find two consecutive odd or even numbers for which this phenomenon does not occur? Note that it is impossible – even for the biggest computer – to verify this for all integers in finite time, because there are infinitely many numbers. But, as we will see, there is a simple way to perform just one calculation – and it will do the job for every single case.

Similarly, multiplying three successive integers and adding the middle integer to this product always yields a perfect cube! Why? Once again, we will see in the exercises in this chapter how one calculation reveals the answer for all possible cases.

Thus, the purpose of this chapter is to discuss mathematical techniques which will then be used throughout the remainder of this book. We will see real-world applications of linear and quadratic equations in Chapters 2, 3, and 4.

The distributive law

We start by discussing one of the most important principles used in simplifying and manipulating mathematical expressions. As a first example, consider the following elementary computation:

2 · 39 + 2 · 61 = ?

One way to solve this equation is to explicitly calculate that 2 · 39 = 78, 2 · 61 = 122, and then to add them and obtain 200. This solution involves carrying out two multiplications (neither of which seems trivial enough to do mentally) followed by one addition. However, here is a simpler approach, which involves only one addition and then one trivial multiplication: we first add: 39 + 61 = 100, and then multiply the result by 2:

2 · 39 + 2 · 61 = 2 · (39 + 61) = 2 · 100 = 200.

Why does the first equality hold in the preceding equation? This is the principle of distributivity. It says that 2 · 39 + 2 · 61 = 2 · (39 + 61). More generally, the distributive law says that given any three real numbers a, b, c,

a · b + a · c = a · (b + c), a · b – a · c = a · (b – c).

The distributive law is at the heart of many algebraic manipulations and simplifications which will be used throughout the book. Thus it is important that we understand it – as well as its implications – thoroughly. Here are further applications of the distributive law.

Example 1.1. Compute (without using a calculator):

1. 17 · 3 + 3 · 283.

2. 3 · 3 + 3 · 17 + 20 · 17.

Solution: While the problem itself is elementary, our goal here is to see how to solve it using the distributive law, and make it easy enough to perform the computation mentally. For the first part, we compute using the distributive law:

17 · 3 + 3 · 283 = 3 · 17 + 3 · 283 = 3 · (17 + 283) = 3 · 300 = 900.

For the second part, we group terms two at a time, using the distributive law:

(3 · 3 + 3 · 17) + 20 · 17 = 3 · (3 + 17) + 20 · 17 = 3 · 20 + 20 · 17 = 20 · (3 + 17) = 20 · 20 = 400.

A more advanced version of this rule is when both factors are sums or differences. For instance, we can multiply out 101 · 201 without using a calculator, as follows:

101 · 201 = (100 + 1) · (200 + 1) = (100 + 1) · 200 + (100 + 1) · 1

= 200 · (100 + 1) + 101 = 200 · 100 + 200 · 1 + 101

= 20, 000 + 200 + 101 = 20, 301.

You may have seen a variation of this method:

101 · 201 = (100 + 1) · (200 + 1) = 100 · 200 + 100 · 1 + 1 · 200 + 1 · 1 = 20, 301.

This is sometimes called the “FOIL” method – essentially, it is simply the distributive law applied twice – once to the terms in the first factor, and once to the terms in the second.

Variables

A useful feature of the distributive law is that it applies equally well to variables. Thus if x, y are any unknown real numbers, then we once again have (for instance): 2 · (x + y) = 2x + 2y. The distributive law can also be used to “contract” expressions into more compact forms. Here is an example: the expression

(x2 – 2)(x + 2) – 3x2 + 6

looks somewhat unwieldy. However, look closely: the last two terms can be written using the distributive law as -3(x2 – 2). Now applying the distributive law again, we get

(x2 – 2)(x + 2) – 3×2 + 6 = (x2 – 2)(x + 2) + (x2 – 2) · (-3)

= (x2 – 2)(x + 2 – 3) = (x2 – 2)(x – 1),

and so just like long expressions involving numbers (as in Example 1.1), variable expressions can also be simplified.

An advantage of using variables is that once verified, a statement involving variables automatically holds for every real value assumed by these variables. For instance, consider the following “magic trick.”

Example 1.2.Get your number back! Here’s a magic trick: start with any number, add 2 to it, multiply the sum by 5, subtract 10 from the product, and divide the difference by 5 – and lo and behold! You get your original number back.

Is it possible to explain what is going on without having to verify whether or not this works for every single starting number?

Solution: Indeed it is. Suppose we start with the number x – as opposed to a specific value, we use a variable because it can be set equal to any number. Then the effect of the given operations on the initial value can be explicitly computed:

x -> x + 2, x + 2 -> 5(x + 2) = 5x + 10, 5x + 10 -> (5x + 10) – 10 = 5x, 5x -> (5x)/5 = x,

and indeed, we get the original number back as the trick claims. (Note that the second operation uses the distributive law.) Thus we have explained the magic trick for every starting number x. Of course, if you want to repeat the calculations corresponding to a specific starting number x, simply set x everywhere above equal to that starting number.

The above example shows that it is indeed possible to use variables to explain general phenomena, or to solve “word problems” using simple mathematics. We now mention an example which will help us answer the question posed at the beginning of this chapter.

Example 1.3.The sum times the difference. One very useful identity involves taking two numbers a, b, and multiplying their sum times their difference. Let us compute the result using the distributive law:

(a + b) · (a – b) = a2 + ba – ab – b2 = a2 – b2.

This means that the sum and the difference of any two real numbers a and b multiply to yield the difference between their squares.

Conversely, this identity can help compute the product of any two numbers: take the square of the number exactly in the middle, and subtract the square of the half of the difference between them. Indeed, if the numbers to multiply are (a + b) and (a – b), then the number in the middle is a, and half of the difference between them is b, and our statement reads

(a + b) · (a – b) = a2 – b2,

which is the sum times the difference formula. It works especially well for two integers that are equidistant from a number whose square is easy to compute.

For instance, this identity enables us to compute 297 · 303 without too much fuss (or a calculator):

297 · 303 = (300 – 3) · (300 + 3) = 3002 – 32 = 90, 000 – 9 = 89, 991.

Example 1.4.Multiplying consecutive odd or even numbers. We now return to the example mentioned at the beginning of the chapter: multiplying two consecutive odd (or even) numbers always seems to yield one less than a perfect square. To check this for every pair of consecutive odd/even numbers at once, we replace these numbers by variables: denote the number between the two consecutive odd/even numbers as n. Then the two numbers that we are to multiply, are (n + 1) and (n – 1). Their product, by the previous example, is

(n + 1) · (n – 1) = n2 – 12 = n2 – 1.

This is precisely one less than the square of the middle number n, as we claimed.

This general calculation also provides a method to compute such products: take the square of the number in between, and subtract 1. For instance, if we want to compute the product of the two successive odd numbers 101 · 99, then we simply use n = 100 in the above calculation, to get 1002 – 12 = 9, 999.

Solving equations

Often in the real world, we see that the same quantity can be expressed in different ways using different physical units. For instance, the same distance can be given in yards, miles, or meters. Or, the same temperature can be expressed in Celsius and in Fahrenheit (or even in Kelvin) degrees. In solving questions involving such changes, a simple tool that is useful is a linear equation. Here is an example.

Example 1.5. What temperature has the same numerical value when measured in Fahrenheit and Celsius?

Solution: Denote the unknown temperature in Celsius, say, by T. Then this same temperature in Fahrenheit is given by: 32 + 9/5T. The conditions of the problem imply that we are to equate these two expressions. In formulas, we have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

We conclude that -40°F = -40°C (and no other temperature has this property).

The above equation T = 32 + 9/5T is an example of a linear equation. It models a linear relationship, or dependency, between two varying quantities (in this case, the temperature in Celsius and in Fahrenheit). A linear equation is characterized by the way the variable appears in it: it can either stand alone, or be multiplied by a numeric coefficient. The variable can be denoted by any letter, say, x, T, w, y, A, and so on. All other terms of the equation are numbers.

Such dependencies are ubiquitous in the real world – for instance, one can use linear equations to model production costs (involving an overhead amount and manufacturing cost per unit), simple interest in banking, distance traveled by a vehicle or a jogger with constant velocity, or conversions between different units of temperature, money, weight, distance, and so on. For example, one can determine which temperature in Fahrenheit is twice (and which is half) the number that it equals in Celsius. Here is another real-world example which can be solved using linear equations.

Example 1.6.Taxicab fares. Suppose in a given city, cab drivers charge an initial fare of $3, followed by an additional charge of $2 per mile. What is the fare if the passenger travels for 10 miles? How many miles has the passenger traveled if the fare is $11? Answer: $23 and 4 miles, respectively. The linear equation here is: fare = 3 + 2 · miles.

Another class of equations that is easy to solve involves equating a product to zero. For instance: find all possible values of the variables a, b, c such that a · b · c = 0. The answer is that if three numbers are nonzero, their product cannot be zero; hence at least one of the variables a, b, c must equal zero in this case. Thus, the solution is that a = 0 and b, c are arbitrary real numbers; or b = 0 and a, c are arbitrary; or c = 0 and a, b are arbitrary.

Example 1.7. Using the above idea, find all solutions to the following equations.

1. (x – 1)(x + 2) = 0.

2. (x – 1)2(x + 3)(y – 2) = 0.

3. y2 – 1 = 99.

Solution: The first two equations are easy to solve, using the previous reasoning. Thus for the first equation, either x – 1 = 0 or x + 2 = 0, whence x = 1, or x = -2. For the second equation, we write the equation as (x – 1) · (x – 1) · (x + 3) · (y – 2) = 0. Hence at least one of the four factors is zero, which leads to: x = 1, and y is any number, or x = -3 and y is any number, or y = 2 and x is any number. (We allow for both x = 1 and y = 2 to occur simultaneously.)

Finally for the last equation, we add 1 to both sides to get: y2 = 100. Hence y = 10 or -10. The other way to see this is to use Example 1.3. To do so, subtract 99 from both sides to get:

y2 – 100 = 0 [??] y2 – 102 = 0 [??] (y + 10)(y – 10) = 0.

Hence we obtain the complete set of solutions: y = -10 or y = 10. For convenience we write y = ±10.

Polynomials: Solving quadratic equations

Now we move from simple dependencies to more involved ones. Using variables allows us to define a useful class of expressions called polynomials. A polynomial is an expression in which several (finitely many) powers of x can appear, either alone or multiplied by numbers. For instance, f(x) = 3x2 – 1 is equal to f(x) = 3 · x2 + (-1) · x0. The number next to each power of x (or of the one variable that is used) is called its coefficient. For instance, the coefficients of x0, x1, x2, and x3 in 3x2 – 1 are -1, 0, 3, 0, respectively.

The highest power of x whose coefficient in a polynomial p(x) is nonzero is called the degree of the polynomial, and is denoted by deg(p). A polynomial is said to be constant, linear, quadratic, cubic, quartic, and so on, if its degree is, respectively, 0, 1, 2, 3, 4, and so on. A linear equation can be expressed as the condition that a linear polynomial in the variable equals zero: ax + b = 0 for some real numbers a, b. In general, given a polynomial p(x), one is often interested in determining the set of x such that p(x) equals a given value – in other words, solving a polynomial equation.

In the remainder of this chapter, we will learn how to solve some polynomial equations which are more involved than linear equations. For instance, using the distributive law (or the FOIL method), you can check that the first of the equations in Example 1.7 was an example of a second-degree polynomial equation in x – i.e., a quadratic equation: x2 + x – 2 = 0. Other examples of quadratic equations are x2 – 3x+2 = 0, or 2x2 -8x+8 = 0, or more generally,

ax2 + bx + c = 0

for some real numbers a, b, c with a ≠ 0. (Note that if a = 0 then the equation becomes a linear equation.)

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