# 1.1: Our Number System

- Page ID
- 130901

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)When we are children, we learn to count objects using ordinary numbers like \(1,2,3,4,\) so we can describe quantities in the world around us. As we get older, we learn larger and larger numbers, so we can make sense of statements like "the population of the world is about 8 billion people."

We also learn about other types of numbers, such as \(0\), negative numbers, and non-whole numbers, which help us describe even more precise quantities, such as "I walked \(2.3\) miles today." In this section, we'll briefly review the meaning of these numbers and the multiple ways to express them.

- Accurately describe the meaning of the base \(10\) number system, including both whole numbers and non-whole numbers
- Correctly perform base \(10\) addition with any numbers and explain why it works
- Round numbers correctly to any place

## The Base 10 Number System

You're already an expert at addition — you've been adding numbers for at least a decade, if not much longer. But when's the last time you really thought about how addition works?

Many of us routinely rely on technology such as smartphones or calculators to perform basic addition. It's faster and less error-prone than using mental math or pencil and paper, particularly when dealing with large numbers. The purpose of this section is not to discourage use of technology, but rather to peek behind the curtain of how humans, the creators of technology, designed the number system we all use every day.

Below, you'll encounter the first example in this book. Take a few minutes to think about the answer to the question in the example before viewing the solution. This example, as well as the many others throughout this book, will be most impactful to your learning if you attempt to solve them before looking ahead at the solution. In fact, neuroscience has shown that the more you struggle initially, the more likely you'll be to remember the reasoning behind the solution, which in turn leads to durable and applicable mathematical skill.

Your seven-year-old cousin is learning to add numbers in school, and she asks you for help. She gives you the following problem:

\[\begin{array}{cccc}& 3& 7 & 4 \\ + & 2 & 5& 8\\ \hline &&&\\ \end{array} \]

How would you explain the solution to her?

###### Solution

The most common method used to solve this problem is explained below. This might not be how you learned how to do this, but for illustration, here it is:

- Add the two numbers in the rightmost column;
- If they add up to more than \(10\), put the ones digit of the sum in that column, and "carry" the \(1\) to the column to the left;
- Repeat this procedure in the next column to the left, remembering to add in any numbers that had been carried;
- Keep going from right to left until you finish.

In terms of what you'd tell your cousin, you might say: "First, add together the \(4\) and the \(8\) in the rightmost column. That answer is \(12\), so you write down the \(2\) underneath, and then carry the \(1\)." You could write it like this:

\[\begin{array}{cccc} & 3& 7 & \overset{1}{4} \\ + & 2 & 5& 8\\ \hline &&&2\\ \end{array}\]

From there, you could explain: "Next, add together all of the numbers in the next column. That is \(1 + 7 + 5\), which is \(13\). Just like before, write the \(3\) down below, and then carry the \(1\)." You could write it like this:

\[\begin{array}{cccc}& 3& \overset{1}{7} & 4 \\+ & 2 & 5& 8\\ \hline&&3&2\\ \end{array}\]

And then you could say: "Since you're out of columns, just add up the last numbers in that column, and put the answer underneath. That would give you \(1 + 3 + 2 = 6\). So the answer is \(632\)." Writing that would look like this:

\[\begin{array}{cccc}& \overset{1}{3}& 7 & 4 \\+ &2 & 5& 8\\ \hline&6&3&2\\ \end{array} \]

If you said that, your cousin would know exactly how to solve this sort of problem, and with practice she could likely solve many similar problems. It's a great explanation of *how* the process works.

But your cousin wants to know *why* the process works. She asks, "why do you carry the \(1\)?"

We use the **base \(10\)** or **decimal **number system. In this system, each digit is a single whole number between \(0\) and \(9\) that represents a particular power of \(10\), with the power increasing from right to left. We typically refer to these powers of ten as **place values.**

What does this mean when referring to a particular number? For example, consider the number \(4027\), which would be read as "four thousand and twenty-seven." If we break this down into its place values, we have that \(4\) is in the thousands place, \(0\) is in the hundreds place, \(2\) is in the tens place, and \(7\) is in the ones place. Therefore, we could choose to rewrite this number as follows:

\[4027 = (4 \times 1000) + (0 \times 100) + (2 \times 10) + (7 \times 1)\]

Writing numbers in this way is more time-consuming than writing them the normal way. However, it allows us to see a pattern that emerges. In order to see that pattern more fully, recall the following definition:

An** exponent** represents repeated multiplication a certain number of times. Exponents are written as superscripts. For example,

\[2^5 = 2 \times 2 \times 2 \times 2 \times 2 = 32\]

Notice that above, there are \(5\) copies of the number \(2\) being multiplied together. In general,

\[a^b = a \times a \times \cdots \times a \quad (b \; times)\]

A nonzero number to the \(0\) power is defined to equal \(1\); for example \(2^0 = 1\).

We introduce this definition because it allows us to describe the pattern that occurs in base \(10\) numbers. Observe — or better yet, check with your phone or a calculator — that applying larger and larger exponents to \(10\) simply increases the number of \(0\)'s following the \(1\). For example, computing \(10^3\) gives us

\[10^3 = 10 \times 10 \times 10 = 1000\]

which has \(3\) zeros following the \(1\). This patterns occurs in general. The follow table shows the first several occurrences of this pattern:

Exponent |
Exponential Form |
Ordinary Form |

\(0\) | \(10^0\) | \(1\) |

\(1\) | \(10^1\) | \(10\) |

\(2\) | \(10^2\) | \(100\) |

\(3\) | \(10^3\) | \(1000\) |

\(4\) | \(10^4\) | \(10000\) |

In the United States, commas are used to separate groups of three digits in large numbers, starting from the ones place. That is, the number \(10000\) could be written \(10,000\), and both would be said as "ten thousand." In this text, we will typically omit commas unless it is helpful for comprehension.

Let's take this pattern and use it to further inspect the number \(4027\). We have

\[ \begin{array}{lll} 4027 &=& (4 \times 1000) + (0 \times 100) + (2 \times 10) + (7 \times 1) \\ 4027 &=& (4 \times 10^3) + (0 \times 10^2) + (2 \times 10^1) + (7 \times 10^0) \\ \end{array}\]

This is why our number system is called* base 10.* Each successive digit, going from right to left, corresponds to a power of \(10\). When we write a number as we did above on the right side of the equations, we call it the *base 10 expansion* of that number. The numbers \(4, 0, 2\) and \(7\) are called the* digits *of the number.

Why do we use \(10\) as the base for our number system?

###### Solution

The answer lies at the ends of your arms! Most humans have \(10\) fingers, so it's very convenient to count to \(10\) and then start over again. If we had \(8\) fingers, our number system would likely be very different, and the numeral \(9\) might not even exist! This is also why our fingers are sometimes called *digits *themselves.

Other base systems are used in particular contexts. For example, *binary numbers* are written in base 2, in which each successive place corresponds to another power of \(2\). This is the number system used in computer and electronics programming, though it is hidden from most device users. We also see base \(12\) and base \(60\) showing up in our computations of time: when counting seconds and minutes, we count to \(59\) and then start over again at \(0\).

Nevertheless, base \(10\) is the most common way to express numbers for the vast majority of people in our world today.

Now that we better understand the base \(10\) system, we can return to your cousin's question: why do we carry the 1 when we add? The answer is quite simple: because of our number system, we can only write a single digit in a given place. Remember your cousin's addition problem:

\[ \begin{array}{cccc} & 3& 7 & 4 \\+ & 2 & 5& 8\\ \hline &&&\\ \end{array}\]

In adding the \(4+8\), we get \(12\), but we can't put a \(12\) in the ones place, because \(12\) consists of two digits. Therefore, we have to remember what the number \(12\) actually *means.* We can rewrite \(12\) using the base \(10\) expansion to uncover that meaning:

\[ 12 = (1 \times 10^1) + (2 \times 10^0)\]

We can write the \(2\) in the ones place, since it's only one digit. However, the additional \(1\) in the tens place has to be added to the other values in the tens place, which in this case are \(7\) and \(5\). Since the sum of the digits in a single column exceed \(9\), the excess has to be "carried over" to the next column, which represents the next larger power of \(10\).

While these observations may seem very basic, these insights on how we represent numbers can have a big impact on how comfortable we feel working with more complicated types of numbers and making judgments about numerical questions.

## Decimals

Decimal numbers are written using the same logic and patterns that whole numbers do. While technically anything written in base 10 is a "decimal number," people often use the word "decimal" to describe numbers that are not whole numbers, such as \(98.7\) or \(510.106\). Non-whole numbers exist between whole numbers, and the section of the number after the decimal point describes precisely where the number lies on a number line.

Let's investigate a number between \(0\) and \(1\) first: the number \(0.439\). The \(0\) to the left of the decimal point tells us that this number is between \(0\) and \(1\), and the part after the decimal point describes its value. The place values of this number work just like they do for whole numbers, except that instead of counting in groups of \(10\), \(100\), \(1000\), and so on, they stand for the *reciprocals *of those values. We use negative exponents to denote these reciprocals since it is shorter to write.

A **negative exponent** on a number indicates that you should find the reciprocal of that number. That is,

\[2^{-5} = \frac{1}{2^5} = \frac{1}{32}\]

In general,

\[a^{-b} = \frac{1}{a \times a \times \cdots \times a} \quad (b \; times)\]

Let's use this definition in context to help us understand how decimals can be expressed using base 10 notation.

Rewrite \(10^{-1}\), \(10^{-2}\), and \(10^{-3}\) without any exponents, and convert them to decimal form.

###### Solution

Using the definition of negative exponents along with the original definition of exponents, we see that

\[10^{-1} = \frac{1}{10^1} = \frac{1}{10}\]

\[ 10^{-2} = \frac{1}{10^2} = \frac{1}{100}\]

\[ 10^{-3} = \frac{1}{10^3} = \frac{1}{1000}\]

These numbers are valid fractions, but it's often easier to work with decimals in calculations. Each of these fractions corresponds to a decimal value in the following way. The pronunciation is included as well.

\[\frac{1}{10} = 0.1 = \text{one tenth} \]

\[\frac{1}{100} = 0.01 = \text{one hundredth} \]

\[\frac{1}{1000} = 0.001 = \text{one thousandth} \]

Note that the reciprocal of a number is named similarly to the number itself, with a "th" afterwards.

Of course, we could keep going with this pattern to ten thousandths, hundred thousandths, and so on.

These negative powers of \(10\) form the basis for how we write decimals, just as the positive powers of \(10\) give us a way to write any whole number. A final example is given next. Remember, you'll learn better if you try it first before looking at the answer!

Write \(0.439\) in expanded form, using negative powers of \(10\).

###### Solution

Each successive place in this number corresponds to a negative power of \(10\) as we saw in the previous example. The \(4\) following the decimal point is in the \(0.1\) place, also known as the \(\frac{1}{10}\) place or the \(10^{-1}\) place. Next, the \(3\) is in the \(0.01\) place, also known as the \(\frac{1}{100}\) place or the \(10^{-2}\) place. And last, the \(9\) is in the \(0.001\) place, also known as the \(\frac{1}{1000}\) place or the \(10^{-3}\) place.

We can therefore expand this number in the following way:

\[0.439 = 4 \times 0.1 + 3 \times 0.01 + 9 \times 0.001 = 4 \times 10^{-1} + 3 \times 10^{-2} + 9 \times 10^{-3}\]

Note the similarities and differences with a base \(10\) expansion of a whole number.

Now that we've got some tools to describe and compare whole numbers and decimals, we'll introduce one last important concept for the chapter.

## Rounding

Rounding is a skill you've likely seen before, but may need to practice. Let's get started with two examples.

The population of Bluffington is \(57,489\) people. Express this fact as a sentence in which the population is rounded to the nearest thousand.

###### Solution

To round the number \(57,489\) to the nearest thousand, the first step is to locate the thousands place. After rounding, every number after the thousands place will be a \(0\).

In this case, the \(7\) is in the thousands place, because it stands for \(7 \times 1000\) in the base 10 expansion of \(57,489\). Since the number in the thousands place is a \(7\), it tells us that the number \(57,489\) is between \(57,000\) and \(58,000\), and will be rounded to one of those two values.

The question is: which of \(57,000\) and \(58,000\) is \(57,489\) closer to? The answer is \(57,000\), because the number \(57,489\) is less than \(57,500\), which is the halfway point between \(57,000\) and \(58,000\).

So, as a sentence, we could write, "The population of Bluffington is approximately \(57,000\) people."

Try the next exercise on your own before looking at the answer.

In performing a calculation, you need to enter \(45 \div 40\) into your calculator. This gives you the answer \(1.125\). However, question you are answering with this calculation asks for an amount of money, in dollars and cents. Round the number \(1.125\) to the correct decimal place and value so it expressed as a number of dollars and cents.

###### Solution

To answer this question, you'll need to know that money amounts in the United States are typically expressed as a number of dollars followed by a number of cents, where a dollar is composed of one hundred cents. That means that each cent is *one hundredth* of a dollar, so we will need to round to the hundredths place.

The hundredths place refers to the second number after a decimal point in a decimal number. In the number \(1.125\), the current value of the hundredths place is a \(2\). This tells us the number \(1.125\) is between \(1.12\) and \(1.13\).

However, we still need to ask what \(1.125\) is closest to: either \(1.12\) or \(1.13\). In reality, it's exactly halfway between these two values, but we adopt (as do most textbooks) the habit of "rounding up" from the halfway point. Therefore, the number \(1.125\) rounded to the nearest hundredth is \(1.13\). As a dollar amount, you would write \(\$1.13\).

These two examples illustrate the general principles of rounding. The process in general is summed up in the Procedure box below.

To round a number to a given place value, follow these steps:

- Identify the the original digit in the place value you're rounding to.
- Look at the digit to the
*right*of the original digit.- If the digit to the right is \(4\) or less, keep the original digit in the place to which you are rounding.
- If the digit to the right is \(5\) or more, increase the original digit by \(1\).

- All digits to the right of the original digit either become \(0\) (if you are rounding to a whole number) or go away (if you are rounding to a decimal value).

One more example will help to clarify. Consider the number \(34.5718\). We label the place values of this number in the table below:

3 | 4 | . | 5 | 7 | 1 | 8 |

Tens | Ones | Decimal Point | Tenths | Hundredths | Thousandths | Ten Thousandths |

Let's practice rounding this number in various ways:

**Nearest ten or "tens place":**The digit in the tens place currently is a \(3\). The digit to the right is a \(4\). Since this \(4\) is smaller than \(5\), keep the \(3\) as it is, and replace the \(4\) with a \(0\). We also get rid of everything after the decimal point. Therefore, the answer in this case is \(30\).**Nearest whole or "ones place":**This means "round to the ones place." The ones digit is currently a \(4\). The digit to the right of the ones place is currently a \(5\). This means that we need to "round up," so the \(4\) will increase by \(1\) to become a \(5\). The rest of the digits after the decimal point go away. Therefore, the answer in this case is \(35\).**Tenths place or "one decimal place":**The digit currently in the tenths place is a \(5\). The digit to the right is a \(7\), which is greater than \(5\). Therefore, we increase the value in the tenths place from \(5\) to \(6\). We then delete everything following the digit in the tenths place. Therefore, the answer in this case is \(34.6\).**Hundredths place or "two decimal places":**The digit currently in the hundredths place is a \(7\). The digit to the right is a \(1\). Since \(1\) is less than \(5\), we keep the \(7\) as is, and delete everything after. Therefore, the answer is \(34.57\).**Thousandths place or "three decimal places":**The answer is \(34.572\). Do you see why?

Throughout this book, there will be specific rounding directions on many of the exercises. Two rules also apply throughout, and will be reinforced through examples:

- When computing a number of living things, round to the nearest whole.
- When computing an amount of money, always round to the hundredths place (two digits after the decimal point)

Other rounding directions will be provided as needed.

## Exercises

- Write the number \(35,023\) using its base \(10\) expansion, explicitly writing the power of ten correspond to each place.
- Round the number \(67.192\) to the following places.
- Tens
- Ones
- Tenths
- Hundredths

- Consider the following subtraction problem:
\[ \begin{array}{cccc}

& 3& 7 & 4 \\

- & 2 & 5& 8\\ \hline

&&&\\

\end{array} \]Pretend you're explaining to your cousin how to do this by hand, showing her work. What would you say to her? Can you explain what happens when you "borrow?" You should write at least 5-6 sentences and include each step in the process.

- Go to the link at the start of this section that describes the world population. You will see a website with two tallies, that are counting the World Population and the US population.
- Write down the number that the US population counter says when you look at the website for the first time. It may be slowly increasing, write down the closest number that you can.
- Round that number to the millions place.
- Write a sentence that describes the number you found in the previous part. Your sentence should answer the question "about how many people live in the United States right now?