0.4 Decimals
SLO 6.NS
The Number System. Divide fractions by fractions; compute fluently with
SLO 6.EE
Expressions and Equations. Write, read and evaluate expressions in which
SLO 7.NS
The Number System. Add, subtract, multiply and divide rational numbers,
SLO 7.EE
Expressions and Equations. Apply properties of operations to add, subtract,
SLO 8.NS
The Number System. Know that numbers which are not rational are irrational;
Learning Objectives
By the end of this section, you will be able to:
- read, write, compare, and round decimals using place value;
- round money amounts to the nearest cent;
- add, subtract, multiply, and divide decimals, and explain why each rule works;
- convert among decimals, fractions, and percents;
- simplify square roots and recognize when a square root is not a real number;
- classify a number as counting, whole, integer, rational, irrational, or real.
Almost every number you meet outside a math classroom is a decimal. Prices, batting averages, gas mileage, race times, interest rates — all decimals. They are so familiar that it is easy to treat them as their own species of number, with their own odd rules that you memorize and apply by reflex.
They are not a separate species. A decimal is a fraction whose denominator is a power of ten, written in a compressed form that reuses the place-value system from Section 0.1 instead of a fraction bar. Every rule in this section comes from that one fact. Line up the points to add, count the places to multiply, shift both points to divide — each of those is a consequence, not a separate law. Once you can see the fraction hiding behind the notation, you can rebuild a rule you have forgotten instead of guessing at it.
Students go wrong in two predictable places. The first is comparing: \(0.7\) and \(0.68\) look like a two-digit number losing to a three-digit number, and a habit built on whole numbers says the longer one wins. The second is bookkeeping in multiplication and division, where the digits come out right and the decimal point lands one spot off, turning $34.50 into $3.45. Both errors go away the same way — by thinking about what each place is worth instead of counting digits. The section ends with a bigger question that decimal notation finally lets us answer: exactly which numbers are out there, and how can we tell them apart?
0.4.1 Decimal Notation and Rounding
In Section 0.1 the place values ran ones, tens, hundreds, thousands, with each place worth ten times the place to its right. Nothing stops that pattern from running the other way. Moving right divides by ten each time, and the decimal point marks the spot where the whole-number part ends.
The decimal point is the symbol that separates the whole-number places from the fractional places. Every place to its right is worth one tenth of the place immediately before it.
Definition 0.4.1 — The places mirror around the ones place, and the point marks where the whole-number part ends.
Here are the places on both sides of the point:
| Place | Position | Value as a fraction | Value as a decimal |
|---|---|---|---|
| Tens | 2 left of the point | \(10\) | \(10\) |
| Ones | 1 left of the point | \(1\) | \(1\) |
| Tenths | 1 right of the point | \(\dfrac{1}{10}\) | \(0.1\) |
| Hundredths | 2 right of the point | \(\dfrac{1}{100}\) | \(0.01\) |
| Thousandths | 3 right of the point | \(\dfrac{1}{1{,}000}\) | \(0.001\) |
| Ten-thousandths | 4 right of the point | \(\dfrac{1}{10{,}000}\) | \(0.0001\) |
Two things in that table are worth naming. There is no "oneths" place, because the ones place is the pivot the pattern reflects around, so tenths sits opposite tens rather than opposite ones. And the "-th" ending is doing real work: ten thousand is a large number, but one ten-thousandth is a tiny one. Two letters are the entire difference between \(10{,}000\) and \(0.0001\).
Expanding a decimal makes the structure visible:
$$36.407 = 3(10) + 6(1) + 4\!\left(\tfrac{1}{10}\right) + 0\!\left(\tfrac{1}{100}\right) + 7\!\left(\tfrac{1}{1{,}000}\right)$$That zero in the hundredths place is a placeholder, doing the same job the zero in 4,072 did back in Section 0.1. Delete it and \(36.407\) collapses to \(36.47\), which is a different number.
Reading decimals aloud
There is a four-step routine for turning a decimal into words:
1. Name the whole-number part.
2. Say "and" for the decimal point.
3. Name the digits to the right of the point as if they were a whole number.
4. Name the place of the last digit.
So \(36.407\) is "thirty-six and four hundred seven thousandths." The last digit sits in the thousandths place, so the whole fractional part gets counted in thousandths.
Section 0.1 said 204 is "two hundred four," never "two hundred and four." Now you can see why. The word "and" is reserved for the decimal point, so "two hundred and four" would be indistinguishable from \(200.4\) when spoken aloud.
Here is the routine applied to several numbers at once. Watch what one extra placeholder zero does to the name.
| Decimal | Read as | Place of the last digit |
|---|---|---|
| \(0.9\) | nine tenths | tenths |
| \(0.53\) | fifty-three hundredths | hundredths |
| \(0.062\) | sixty-two thousandths | thousandths |
| \(7.4\) | seven and four tenths | tenths |
| \(7.04\) | seven and four hundredths | hundredths |
| \(12.385\) | twelve and three hundred eighty-five thousandths | thousandths |
| \(0.0009\) | nine ten-thousandths | ten-thousandths |
The \(7.4\) and \(7.04\) rows differ by one syllable when you say them and by one zero when you write them, and that zero costs the 4 a factor of ten: it is worth four tenths in the first number and only four hundredths in the second.
Going the other way, from words to standard form, the named place tells you where the last digit lands, and you fill any gaps with zeros. "Five and eighty-one thousandths" puts the 1 in the thousandths place, so the digits 81 take up hundredths and thousandths and the tenths place needs a placeholder: \(5.081\). Writing \(5.81\) instead is a common slip, and it multiplies the fractional part by ten.
Write "nine and forty-six thousandths" in standard form. Then read \(0.207\) aloud in words.
Solution
Step 1 — Place the last digit. The phrase names thousandths, so the last digit of the number goes in the thousandths place, three places right of the point.
Step 2 — Fill in the digits from the right. The fractional digits are 46. The 6 takes thousandths, the 4 takes hundredths, and nothing is left for tenths, so tenths gets a placeholder zero.
$$9.046$$Step 3 — Read \(0.207\). There is no whole-number part, so start with the fractional digits: 207. The last digit, 7, sits in the thousandths place.
Two hundred seven thousandths.
Answer: \(9.046\); "two hundred seven thousandths."
Comparing decimals
With whole numbers, more digits always means a bigger number. That rule is false for the fractional part of a decimal, and the false version is the most stubborn misconception in this section.
Compare \(0.7\) and \(0.68\). Counting digits suggests \(0.68\) should win. Place value says otherwise: \(0.7\) has 7 tenths while \(0.68\) has only 6 tenths, and no pile of hundredths can make up a whole missing tenth. So \(0.7 > 0.68\).
Think of tenths as dimes and hundredths as pennies. Someone with 7 dimes has more than someone with 6 dimes and 8 pennies, no matter how many coins each is holding. You compare the big places first and only look at the small ones to break a tie.
The reliable method is to compare place by place from the left and stop at the first place where the digits differ. Padding with trailing zeros makes the comparison easy to see, and padding is safe because adding zeros to the right of the last decimal digit does not change the value: \(0.7 = 0.70 = 0.700\), since \(\dfrac{7}{10} = \dfrac{70}{100} = \dfrac{700}{1{,}000}\).
| Comparison | Padded to match | First place they differ | Result |
|---|---|---|---|
| \(0.7\) and \(0.68\) | \(0.70\) and \(0.68\) | tenths: 7 beats 6 | \(0.7 > 0.68\) |
| \(0.325\) and \(0.33\) | \(0.325\) and \(0.330\) | hundredths: 2 loses to 3 | \(0.325 < 0.33\) |
| \(4.9\) and \(4.90\) | \(4.90\) and \(4.90\) | they never differ | \(4.9 = 4.90\) |
| \(0.08\) and \(0.1\) | \(0.08\) and \(0.10\) | tenths: 0 loses to 1 | \(0.08 < 0.1\) |
| \(-0.4\) and \(-0.35\) | \(-0.40\) and \(-0.35\) | see below | \(-0.4 < -0.35\) |
That last row brings in Section 0.2. Among positives, \(0.40\) is larger than \(0.35\). Making both negative flips their order on the number line, so \(-0.4\) sits to the left of \(-0.35\) and is therefore the smaller of the two. The rule from Section 0.2 still runs everything: a number is greater than whatever sits to its left, no matter how many decimal places it carries.
Rounding
Rounding a decimal uses the same four steps you used on whole numbers, with one change at the end.
1. Find the digit in the place you are rounding to.
2. Look at the digit immediately to its right.
3. If that digit is 5 or more, add 1 to the rounding digit. Otherwise leave it alone.
4. Drop every digit to the right — do not replace them with zeros.
Step 4 is the change. When you rounded 6,473 to the nearest hundred, the zeros in 6,500 were required, because those places carry value. To the right of the decimal point a trailing zero carries no value, so you simply drop it.
Round \(8.4763\) to four different places and the answers land all over the map:
| Round to | Rounding digit | Digit to its right | Result |
|---|---|---|---|
| Tenths | 4 | 7 — round up | \(8.5\) |
| Hundredths | 7 | 6 — round up | \(8.48\) |
| Thousandths | 6 | 3 — leave it | \(8.476\) |
| Whole number | 8 | 4 — leave it | \(8\) |
Every one of those answers came from the original \(8.4763\), and that matters more than it looks. If you round to hundredths first and then round that result to tenths, you are rounding a rounded number, and the small error from the first step gets carried into the second. Sometimes you get lucky and the two routes agree; often you do not. Start over from the original number every time.
Rounding money
Money is the rounding job you will do most often, and it comes with a fixed convention: round to the nearest cent, which is the hundredths place. Amounts with more than two decimal places show up constantly. Sales tax on a purchase, a per-unit price at the grocery store, an interest calculation, and the price of gasoline are all computed to three, four, or five places before anyone rounds them.
When the rounding digit is 9 and it has to increase, it becomes 10 and carries into the next place. Rounding \(2.397\) to hundredths turns the 9 into 10, which carries into the tenths, giving \(2.40\). That trailing zero stays, because it reports how precise the answer is.
| Exact amount | Rounded to the nearest cent | Why |
|---|---|---|
| \(\$47.3826\) | \(\$47.38\) | thousandths digit is 2, so leave the 8 alone |
| \(\$47.3852\) | \(\$47.39\) | thousandths digit is 5, so the 8 becomes 9 |
| \(\$9.995\) | \(\$10.00\) | the nines roll over and carry into the dollars |
| \(\$0.4499\) | \(\$0.45\) | thousandths digit is 9, so the 4 becomes 5 |
Carry the exact value all the way through a calculation and round only at the very end. Rounding partway through is the same chained-rounding mistake as before, and on a long receipt those small errors add up into a total that visibly disagrees with the register.
A jacket costs $18.95 and the sales tax rate is \(7.25\%\), which gives an exact tax of $1.373875. What does the register charge for tax, and what is the total?
Solution
Step 1 — Find the hundredths digit of the tax. In $1.373875 the hundredths digit is 7, and the digit immediately to its right is 3.
Step 2 — Apply the rule. Since 3 is less than 5, the 7 stays and everything to the right is dropped.
$$\$1.373875 \rightarrow \$1.37$$Step 3 — Add to get the total. Line up the decimal points and add.
$$\$18.95 + \$1.37 = \$20.32$$Notice we rounded the tax once, at the end of the tax computation, and then worked with the rounded cents. That matches what a register does.
Answer: $1.37 in tax, for a total of $20.32.
Write "six and ninety-four hundredths" in standard form. Then decide which is larger, \(0.409\) or \(0.41\), and round \(0.409\) to the nearest hundredth.
Solution
Step 1 — Standard form. The phrase names hundredths, so the last digit lands two places right of the point. The fractional digits are 94, which fill tenths and hundredths exactly, with no placeholder needed.
$$6.94$$Step 2 — Compare \(0.409\) and \(0.41\). Pad the shorter one so both have three decimal places.
$$0.409 \quad \text{and} \quad 0.410$$The tenths match at 4. The hundredths are 0 and 1, and 0 is less than 1, so \(0.409 < 0.41\).
Step 3 — Round \(0.409\) to hundredths. The hundredths digit is 0, and the digit to its right is 9, which is 5 or more, so the 0 becomes 1.
$$0.409 \rightarrow 0.41$$Answer: \(6.94\); \(0.41\) is larger; \(0.409\) rounds to \(0.41\).
0.4.2 Operations with Decimals
You can only add quantities of the same kind. Tenths add to tenths and hundredths add to hundredths, but tenths never add directly to hundredths, any more than 3 hours can be added straight to 8 minutes. Lining up the decimal points is exactly what forces matching places into the same column.
A trailing zero is a zero at the right end of the fractional part of a decimal. It never changes the value, so you may add or remove trailing zeros freely to make places line up.
Add \(12.4 + 3.75 + 0.6\). Pad each addend with trailing zeros until all three have the same number of decimal places, then add column by column as you would with whole numbers, bringing the decimal point straight down:
$$\begin{array}{r} 12.40 \\ 3.75 \\ +\;0.60 \\ \hline 16.75 \end{array}$$Definition 0.4.2 — Appending zeros re-slices the same amount into finer pieces without moving its edge.
Subtraction works the same way. To subtract \(6.28\) from \(40\), remember that a whole number has an invisible decimal point at its right end, so \(40\) is \(40.00\) and the columns line up: \(40.00 - 6.28 = 33.72\).
The mistake this whole procedure exists to prevent is right-aligning the digits instead of the points. Right-aligning \(12.4\) and \(3.75\) would stack the 4 over the 5, adding tenths to hundredths, and the answer would be nonsense. Line up the points, not the edges.
Multiplying: count the decimal places
Multiplication does not need the points lined up. Multiply as though the points were not there, then place the point in the answer so that
$$\text{decimal places in the product} = \text{places in the first factor} + \text{places in the second factor}.$$Multiply \(1.6 \times 0.45\). Ignore the points and compute \(16 \times 45 = 720\). The factors carry 1 and 2 decimal places, for a total of 3, so the product needs three decimal places: \(0.720\), which we write as \(0.72\).
That rule looks like arbitrary bookkeeping until you rewrite the decimals as fractions, and then it turns into the fraction multiplication from Section 0.3:
$$0.7 \times 0.03 = \frac{7}{10} \times \frac{3}{100} = \frac{21}{1{,}000} = 0.021$$Multiplying the denominators takes \(10\) times \(100\) and gives \(1{,}000\), and the number of zeros in that denominator is exactly the number of decimal places. One place plus two places gives three places because \(10^1 \cdot 10^2 = 10^3\). The counting rule is a statement about exponents wearing a disguise. It also explains the placeholder in \(0.021\): the digits 21 have to be pushed out to the third decimal place, so the tenths place needs a zero to hold it open.
| Product | Digits, ignoring the points | Places needed | Answer |
|---|---|---|---|
| \(2.4 \times 3.5\) | \(24 \times 35 = 840\) | \(1 + 1 = 2\) | \(8.40 = 8.4\) |
| \(0.7 \times 0.03\) | \(7 \times 3 = 21\) | \(1 + 2 = 3\) | \(0.021\) |
| \(1.6 \times 0.45\) | \(16 \times 45 = 720\) | \(1 + 2 = 3\) | \(0.720 = 0.72\) |
| \(0.5 \times 0.5\) | \(5 \times 5 = 25\) | \(1 + 1 = 2\) | \(0.25\) |
| \(12 \times 0.08\) | \(12 \times 8 = 96\) | \(0 + 2 = 2\) | \(0.96\) |
The fourth row deserves a second look, because \(0.5 \times 0.5 = 0.25\) is smaller than either factor. Multiplying by a number between 0 and 1 shrinks things. That runs against an instinct built entirely on whole numbers, where multiplying never made anything smaller, and the instinct has to go before it causes trouble in Chapter 2.
Compute \(0.24 \times 3.5\).
Solution
Step 1 — Multiply as whole numbers. Drop the points and multiply 24 by 35.
$$24 \times 35 = 840$$Step 2 — Count the decimal places in the factors. There are 2 places in \(0.24\) and 1 place in \(3.5\), so the product needs \(2 + 1 = 3\) places.
Step 3 — Place the point. Starting from the right end of 840, count three places to the left. That requires writing the digits as 0.840.
$$0.840$$Step 4 — Drop the trailing zero. A zero at the right end of the fractional part changes nothing.
$$0.84$$A quick sanity check: \(0.24\) is about a quarter, and a quarter of 3.5 is a little less than 1, so 0.84 is believable.
Answer: \(0.84\).
Dividing by a whole number
Put the decimal point in the quotient directly above the point in the dividend, then divide the way you would with whole numbers:
$$19.5 \div 6 = 3.25$$If the division does not come out even, append trailing zeros to the dividend and keep going, since \(19.5\) and \(19.500\) are the same number. Check any division by multiplying back: \(6 \times 3.25 = 19.5\).
Dividing by a decimal
A decimal divisor is awkward, so we get rid of it. Write the division as a fraction and multiply the top and the bottom by whatever power of ten turns the divisor into a whole number. Multiplying both parts by the same thing leaves the value unchanged — that is the equivalent-fractions idea from Section 0.3:
$$\frac{8.61}{0.3} = \frac{8.61 \times 10}{0.3 \times 10} = \frac{86.1}{3} = 28.7$$In practice people describe this as "move both decimal points the same number of places to the right, enough places to make the divisor whole." The word doing the work is both. Moving only the divisor's point changes the answer by a factor of ten.
| Problem | Shift both by | Rewritten | Answer |
|---|---|---|---|
| \(8.61 \div 0.3\) | 1 place | \(86.1 \div 3\) | \(28.7\) |
| \(0.144 \div 0.06\) | 2 places | \(14.4 \div 6\) | \(2.4\) |
| \(7 \div 0.25\) | 2 places | \(700 \div 25\) | \(28\) |
| \(5.5 \div 1.1\) | 1 place | \(55 \div 11\) | \(5\) |
The third row shows the other instinct that needs retraining. Dividing 7 by 0.25 gives 28, which is much larger than the number we started with. That is not a mistake. Dividing by a number between 0 and 1 makes things bigger, because the question "how many quarter-units fit inside 7 units?" has a large answer.
Multiplying and dividing by powers of ten
This case gets its own rule because it comes up constantly and takes no computation at all. Multiplying by 10 bumps every digit up one place, which is the same as sliding the decimal point one place to the right. Dividing by 10 slides it one place to the left.
| Operation | Point moves | Example |
|---|---|---|
| \(\times 10\) | 1 place right | \(4.62 \times 10 = 46.2\) |
| \(\times 100\) | 2 places right | \(4.62 \times 100 = 462\) |
| \(\times 1{,}000\) | 3 places right | \(4.62 \times 1{,}000 = 4{,}620\) |
| \(\div 10\) | 1 place left | \(4.62 \div 10 = 0.462\) |
| \(\div 100\) | 2 places left | \(4.62 \div 100 = 0.0462\) |
| \(\div 1{,}000\) | 3 places left | \(4.62 \div 1{,}000 = 0.00462\) |
Count the zeros in the power of ten and that is how many places the point moves, appending placeholder zeros whenever the digits run out. In \(4.62 \times 1{,}000\) the point has to move three places, but only two digits follow it, so a zero fills the ones place: \(4{,}620\). This shortcut is behind every metric-unit conversion you will ever do, and Section 1.3 comes back to it as a statement about exponents.
Nothing about the order of operations from Section 0.1 changes when decimals show up. Grouping symbols still go first, then exponents, then multiplication and division from left to right, then addition and subtraction from left to right. The only difference is that each individual computation now needs the decimal point tracked, and the point is where the errors hide. In a problem like \(0.6(4.5 - 1.25) + 0.8\), almost everyone gets the digits right; what separates a correct answer from a wrong one is whether the point in the product landed in the right spot. Count the places before you write the product down, every time.
Simplify \(\;0.6(4.5 - 1.25) + 0.8\).
Solution
Step 1 — Grouping symbols first. Pad \(4.5\) to \(4.50\) so the places line up, then subtract.
$$4.50 - 1.25 = 3.25$$The expression is now \(0.6(3.25) + 0.8\).
Step 2 — Multiply. Ignore the points: \(6 \times 325 = 1950\). There is 1 decimal place in \(0.6\) and 2 in \(3.25\), so the product needs 3 places.
$$1.950 = 1.95$$Step 3 — Add. Pad \(0.8\) to \(0.80\) and add.
$$1.95 + 0.80 = 2.75$$Answer: \(2.75\).
Compute \(0.9 \times 0.04\), then compute \(6.3 \div 0.7\), then simplify \(\;2.5 + 0.4(6 - 1.5)\).
Solution
Part 1 — \(0.9 \times 0.04\). Ignore the points: \(9 \times 4 = 36\). There is 1 place in \(0.9\) and 2 in \(0.04\), so the product needs 3 places, which means a placeholder zero in the tenths spot.
$$0.036$$Part 2 — \(6.3 \div 0.7\). Shift both points one place right so the divisor becomes a whole number.
$$\frac{6.3}{0.7} = \frac{63}{7} = 9$$Part 3 — \(2.5 + 0.4(6 - 1.5)\). Grouping first:
$$6.0 - 1.5 = 4.5$$Then multiply. Ignore the points: \(4 \times 45 = 180\), and \(1 + 1 = 2\) places are needed.
$$1.80 = 1.8$$Then add:
$$2.5 + 1.8 = 4.3$$Answer: \(0.036\); \(9\); \(4.3\).
0.4.3 Decimals, Fractions, and Percents
Decimals, fractions, and percents are three notations for the same quantities. Each one is convenient somewhere and clumsy somewhere else. Fractions are exact and are what algebra prefers. Decimals are easiest to compare and are what a calculator hands you. Percents are the clearest way to report a proportion to another person. Being fluent means converting between them without stopping to think, because a problem almost never arrives in the notation you want to work in.
Turning a decimal into a fraction
Read the decimal aloud. The name is the fraction. Since \(0.36\) is "thirty-six hundredths," it is \(\dfrac{36}{100}\), which reduces to \(\dfrac{9}{25}\).
| Decimal | Its name | Fraction | Lowest terms |
|---|---|---|---|
| \(0.4\) | four tenths | \(\dfrac{4}{10}\) | \(\dfrac{2}{5}\) |
| \(0.36\) | thirty-six hundredths | \(\dfrac{36}{100}\) | \(\dfrac{9}{25}\) |
| \(0.125\) | one hundred twenty-five thousandths | \(\dfrac{125}{1{,}000}\) | \(\dfrac{1}{8}\) |
| \(0.06\) | six hundredths | \(\dfrac{6}{100}\) | \(\dfrac{3}{50}\) |
| \(2.05\) | two and five hundredths | \(2\dfrac{5}{100}\) | \(\dfrac{41}{20}\) |
The denominator is always the place value of the last digit, so the number of decimal places tells you the power of ten before you write anything down. Reducing afterward uses the greatest-common-factor method from Section 0.3.
Turning a fraction into a decimal
A fraction bar means division, so \(\dfrac{a}{b}\) is \(a \div b\). Divide the numerator by the denominator, appending trailing zeros to the numerator as needed:
$$\frac{3}{8} = 3 \div 8 = 0.375$$Every fraction converts, but the division can end in one of two ways, and each way has a name.
A terminating decimal is a decimal whose digits stop. The division reaches a remainder of zero.
Definition 0.4.3 — A terminating decimal stops because the division reaches a remainder of zero.
A repeating decimal is a decimal in which a block of digits repeats forever. The division never reaches a remainder of zero; instead a remainder comes back around, and from that point the quotient digits cycle.
Definition 0.4.4 — A remainder that comes back around is what makes the quotient digits cycle forever.
We write a repeating decimal with a bar over the repeating block. The bar is not decoration. It states that the block goes on without end, which writing \(0.8333\ldots\) only hints at.
| Fraction | Division | Decimal | Type |
|---|---|---|---|
| \(\dfrac{3}{8}\) | \(3 \div 8\) | \(0.375\) | terminating |
| \(\dfrac{7}{20}\) | \(7 \div 20\) | \(0.35\) | terminating |
| \(\dfrac{5}{6}\) | \(5 \div 6\) | \(0.8\overline{3}\) | repeating |
| \(\dfrac{4}{11}\) | \(4 \div 11\) | \(0.\overline{36}\) | repeating |
| \(\dfrac{2}{3}\) | \(2 \div 3\) | \(0.\overline{6}\) | repeating |
Look closely at where each bar starts. In \(0.8\overline{3}\) only the 3 repeats and the 8 appears once, so the number is \(0.8333\ldots\). In \(0.\overline{36}\) the bar covers both digits, so the pair repeats as a block: \(0.363636\ldots\).
Predicting which fractions terminate
You can tell which ending you will get before you divide. Reduce the fraction to lowest terms and factor the denominator. The decimal terminates exactly when that denominator's only prime factors are 2 and 5, the two primes that divide 10. Any other prime factor forces the division to repeat.
| Fraction | Denominator factored | Only 2s and 5s? | Prediction | Actual |
|---|---|---|---|---|
| \(\dfrac{3}{8}\) | \(2 \cdot 2 \cdot 2\) | yes | terminates | \(0.375\) |
| \(\dfrac{7}{20}\) | \(2 \cdot 2 \cdot 5\) | yes | terminates | \(0.35\) |
| \(\dfrac{5}{6}\) | \(2 \cdot 3\) | no, there is a 3 | repeats | \(0.8\overline{3}\) |
| \(\dfrac{4}{11}\) | \(11\) | no, there is an 11 | repeats | \(0.\overline{36}\) |
Reducing first is not optional here. The fraction \(\dfrac{6}{15}\) has a denominator with a factor of 3, which predicts a repeating decimal, but in lowest terms it is \(\dfrac{2}{5} = 0.4\), which terminates. Test the fraction in lowest terms, not whatever form you were handed.
Percent
A percent is a fraction whose denominator is 100, written with the symbol \(\%\) in place of the denominator.
If you got 17 out of 20 on a quiz, nobody compares that to 43 out of 50 in their head. Rewriting both as scores out of 100 makes them line up instantly, and that rewriting is all a percent is.
Definition 0.4.5 — The percent sign stands in for a denominator of 100.
So \(42\%\) means \(\dfrac{42}{100}\), which is \(0.42\). Since dividing by 100 moves the decimal point two places to the left, both conversions are mechanical. To go from a percent to a decimal, drop the \(\%\) and move the point two places left. To go from a decimal to a percent, move the point two places right and attach the \(\%\).
| Percent | Decimal | Fraction in lowest terms |
|---|---|---|
| \(75\%\) | \(0.75\) | \(\dfrac{3}{4}\) |
| \(8\%\) | \(0.08\) | \(\dfrac{2}{25}\) |
| \(125\%\) | \(1.25\) | \(\dfrac{5}{4}\) |
| \(0.8\%\) | \(0.008\) | \(\dfrac{1}{125}\) |
| \(250\%\) | \(2.5\) | \(\dfrac{5}{2}\) |
Two of those rows contradict a widespread belief that a percent has to land between 0 and 100. It does not. A percent of \(125\%\) is a perfectly good number — it is \(1.25\), more than the whole — and it turns up any time something grows past its original size. At the other end, \(0.8\%\) is less than one percent, and the temptation to write it as \(0.8\) is an error by a factor of one hundred. Read the symbol as "per hundred" every time and \(0.8\%\) becomes \(\dfrac{0.8}{100} = 0.008\) without any drama.
Some equivalents come up so often that recognizing them on sight saves real time, especially in the data work of Chapter 4 where the same handful of proportions keeps reappearing. Table 0.4.1 collects them.
| Fraction | Decimal | Percent |
|---|---|---|
| \(\dfrac{1}{2}\) | \(0.5\) | \(50\%\) |
| \(\dfrac{1}{3}\) | \(0.\overline{3}\) | \(33\dfrac{1}{3}\%\) |
| \(\dfrac{2}{3}\) | \(0.\overline{6}\) | \(66\dfrac{2}{3}\%\) |
| \(\dfrac{1}{4}\) | \(0.25\) | \(25\%\) |
| \(\dfrac{3}{4}\) | \(0.75\) | \(75\%\) |
| \(\dfrac{1}{5}\) | \(0.2\) | \(20\%\) |
| \(\dfrac{2}{5}\) | \(0.4\) | \(40\%\) |
| \(\dfrac{1}{8}\) | \(0.125\) | \(12.5\%\) |
| \(\dfrac{3}{8}\) | \(0.375\) | \(37.5\%\) |
| \(\dfrac{1}{10}\) | \(0.1\) | \(10\%\) |
| \(\dfrac{1}{100}\) | \(0.01\) | \(1\%\) |
| \(1\) | \(1.0\) | \(100\%\) |
The \(\dfrac{1}{3}\) row explains why you usually see \(33\dfrac{1}{3}\%\) rather than \(33.3\%\). The decimal repeats forever, so any decimal percent you write down is rounded, while the fraction form is exact. When exactness matters, keep the fraction.
Convert \(\dfrac{3}{8}\) to a decimal and then to a percent. Then convert \(24\%\) to a decimal and to a fraction in lowest terms.
Solution
Step 1 — Fraction to decimal. The bar means division, so divide 3 by 8.
$$3 \div 8 = 0.375$$Step 2 — Decimal to percent. Move the point two places right and attach the symbol.
$$0.375 \rightarrow 37.5\%$$Step 3 — Percent to decimal. Drop the symbol and move the point two places left.
$$24\% \rightarrow 0.24$$Step 4 — Decimal to fraction. Read it: "twenty-four hundredths." The denominator is the place value of the last digit.
$$\frac{24}{100} = \frac{6}{25}$$Both 24 and 100 are divisible by 4, which reduces the fraction in one step.
Answer: \(\dfrac{3}{8} = 0.375 = 37.5\%\); \(24\% = 0.24 = \dfrac{6}{25}\).
Convert \(0.45\) to a fraction in lowest terms and to a percent. Then decide which is largest: \(\dfrac{5}{8}\), \(0.63\), or \(61\%\).
Solution
Step 1 — Decimal to fraction. Read \(0.45\) as "forty-five hundredths."
$$\frac{45}{100} = \frac{9}{20}$$Both 45 and 100 are divisible by 5.
Step 2 — Decimal to percent. Move the point two places right.
$$0.45 \rightarrow 45\%$$Step 3 — Put all three candidates in one notation. Decimals are easiest to rank, because they sort by place value with no common denominator to hunt for.
$$\frac{5}{8} = 0.625 \qquad 0.63 = 0.630 \qquad 61\% = 0.610$$Step 4 — Compare place by place. All three have 6 tenths, so move to hundredths: 2, 3, and 1. The largest hundredths digit is 3.
Answer: \(0.45 = \dfrac{9}{20} = 45\%\); the largest of the three is \(0.63\).
0.4.4 Square Roots and the Real Numbers
Squaring a number multiplies it by itself, so \(9^2 = 81\). A square root asks the reverse question: given 81, what number was squared to produce it?
A square root of \(m\) is a number whose square is \(m\). It is written \(\sqrt{m}\), and the symbol \(\sqrt{\;\;}\) is called the radical sign.
$$\sqrt{81} = 9 \quad \text{because} \quad 9^2 = 81$$Definition 0.4.6 — Squaring asks what 9 rows of 9 come to; the square root asks the reverse.
There is a subtlety hiding in that line. Both 9 and \(-9\) square to 81, since like signs multiply to a positive back in Section 0.2. If \(\sqrt{81}\) meant "either one," it would not name a single number, and every expression containing it would be ambiguous. So the radical sign is defined to hand back only the nonnegative one.
The principal square root of a nonnegative number is its nonnegative square root. The symbol \(\sqrt{m}\) always means the principal root, so \(\sqrt{81} = 9\) and never \(-9\).
To ask for the negative root, put the sign outside the radical: \(-\sqrt{81} = -9\). That is the same "negate afterward" structure as the absolute-value expressions in Section 0.2. The radical sign groups, so you finish everything underneath it first and apply the outside sign last.
Definition 0.4.7 — Two numbers square to 81, so the radical is defined to return only the nonnegative one.
A perfect square is a number that is the square of an integer. Its square root is an integer.
Definition 0.4.8 — A perfect square fills a complete square of dots with nothing left over.
Knowing the first fifteen perfect squares on sight is worth the small effort. It makes simplifying radicals fast, and it is the recognition step behind factoring quadratics in Chapter 7.
| \(n\) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| \(n^2\) | 1 | 4 | 9 | 16 | 25 | 36 | 49 | 64 | 81 | 100 | 121 | 144 | 169 | 196 | 225 |
When a square root is not a real number
What would \(\sqrt{-49}\) be? It would have to be a number whose square is \(-49\). But a positive squared is positive, a negative squared is positive, and zero squared is zero, so no real number squares to a negative.
$$\sqrt{-49} \text{ is not a real number.}$$This is a real gap in our number system rather than an oversight, and it is the reason a later course introduces imaginary numbers. For now, treat the square root of a negative the way you treat division by zero from Section 0.2: an expression with no value among the numbers we have.
Do not confuse it with \(-\sqrt{49} = -7\), which is perfectly fine. The negative sits inside the radical in the first case and outside it in the second, and the two mean completely different things.
| Expression | Value | Reason |
|---|---|---|
| \(\sqrt{49}\) | \(7\) | the principal, nonnegative root |
| \(-\sqrt{49}\) | \(-7\) | take the root first, then negate |
| \(\sqrt{-49}\) | not a real number | nothing real squares to a negative |
| \(\sqrt{0}\) | \(0\) | \(0^2 = 0\) |
| \((-7)^2\) | \(49\) | like signs give a positive product |
Simplify what is underneath the radical completely before taking the root. Here \(\sqrt{25 - 9} = \sqrt{16} = 4\), while \(\sqrt{25} - \sqrt{9} = 5 - 3 = 2\). Different answers, so the grouping is not optional.
Simplify \(\;2\sqrt{36} - \sqrt{25 - 9}\).
Solution
Step 1 — Grouping first. The radical groups everything under it, so do the subtraction underneath before taking that root.
$$2\sqrt{36} - \sqrt{16}$$Step 2 — Take both roots. Both radicands are perfect squares.
$$2(6) - 4$$Step 3 — Multiplication before subtraction.
$$12 - 4$$Step 4 — Subtract.
$$8$$Answer: \(8\).
Estimating a root that is not a perfect square
Most numbers are not perfect squares, and their roots are not integers. You can still pin such a root between two consecutive integers by finding the perfect squares on either side of it. To estimate \(\sqrt{55}\), notice that the nearest perfect squares are 49 and 64:
$$7 = \sqrt{49} \; < \; \sqrt{55} \; < \; \sqrt{64} = 8$$So \(\sqrt{55}\) lands between 7 and 8. Since 55 is closer to 49 than to 64, the root is closer to 7. Squaring candidates closes in on it: \(7.4^2 = 54.76\) is too small, \(7.41^2 = 54.9081\) is still too small, and \(7.42^2 = 55.0564\) overshoots. The root therefore sits between 7.41 and 7.42, nearer the upper end, so \(\sqrt{55} \approx 7.42\) to two decimal places.
| Root | Perfect squares around it | Between | Closer to |
|---|---|---|---|
| \(\sqrt{55}\) | 49 and 64 | 7 and 8 | 7 |
| \(\sqrt{30}\) | 25 and 36 | 5 and 6 | 5 |
| \(\sqrt{90}\) | 81 and 100 | 9 and 10 | 9 |
| \(\sqrt{140}\) | 121 and 144 | 11 and 12 | 12 |
Estimating is not a lesser substitute for a calculator; it is how you catch a mistyped key. If a calculator reports \(\sqrt{55} \approx 24.1\), the estimate tells you instantly that something went in wrong.
Sorting numbers into sets
The decimal work from Section 0.4.3 turns out to answer a structural question about which numbers exist and how they relate.
A rational number is a number that can be written as a ratio \(\dfrac{p}{q}\) of two integers, where \(q \neq 0\).
The name comes from ratio, not from reasonable. The definition is broader than it looks, and it swallows nearly every number you have met so far: \(\dfrac{3}{4}\) is already a ratio; \(12\) is \(\dfrac{12}{1}\); \(-7\) is \(\dfrac{-7}{1}\); \(2.6\) is \(\dfrac{13}{5}\); and \(0.\overline{6}\) is \(\dfrac{2}{3}\).
That last pair of examples points at a test you can run by eye. A number is rational exactly when its decimal form terminates or repeats. Both directions hold. Any ratio of integers, divided out, has to terminate or repeat, because the remainders along the way can only take finitely many values, so eventually one repeats and the quotient digits start cycling. And any terminating or repeating decimal can be written back as a ratio of integers.
That leaves room for a third kind of decimal: one that runs forever without ever settling into a repeating block.
Definition 0.4.9 — Five familiar numbers, each rewritten as a ratio of two integers.
An irrational number is a real number that is not rational. Its decimal form neither terminates nor repeats.
Definition 0.4.10 — One decimal stops, one repeats, and one does neither.
Two families supply most of the examples. The first is roots of numbers that are not perfect squares: \(\sqrt{55} = 7.4161984\ldots\) runs forever with no repeating block, and in fact the square root of any whole number that is not a perfect square is irrational. The second is \(\pi\), the ratio of a circle's circumference to its diameter, which is \(3.14159265\ldots\) and also irrational. The familiar \(3.14\) and \(\dfrac{22}{7}\) are approximations, and both of them are rational, which is exactly why neither one equals \(\pi\).
Notice the asymmetry: \(\sqrt{36} = 6\) is rational, while \(\sqrt{35}\) is irrational. The radical sign is not what makes a number irrational. Whether the number underneath it is a perfect square is.
A real number is any number that is either rational or irrational. Equivalently, a real number is any number with a location on the number line.
Definition 0.4.11 — The sets nest outward, the irrationals sit beside them, and together they fill the line.
These sets nest inside one another, so a number in an inner set automatically belongs to every set outside it:
$$\text{counting} \subset \text{whole} \subset \text{integer} \subset \text{rational} \subset \text{real}$$The counting numbers are \(1, 2, 3, 4, \dots\). The whole numbers add zero to that list. The integers add the opposites, giving \(\dots, -2, -1, 0, 1, 2, \dots\) as in Section 0.2. The rational numbers add every ratio of integers, which is every terminating or repeating decimal. The irrational numbers are everything real that is left over, and this is the one set in the list that does not nest — it sits beside the rationals rather than inside them. Together the rationals and irrationals make up the real numbers.
Table 0.4.2 sorts ten sample numbers into those sets.
| Number | Counting | Whole | Integer | Rational | Irrational | Real |
|---|---|---|---|---|---|---|
| \(12\) | yes | yes | yes | yes | yes | |
| \(0\) | yes | yes | yes | yes | ||
| \(-7\) | yes | yes | yes | |||
| \(\dfrac{3}{4}\) | yes | yes | ||||
| \(-2.6\) | yes | yes | ||||
| \(0.\overline{45}\) | yes | yes | ||||
| \(\sqrt{9}\) | yes | yes | yes | yes | yes | |
| \(\sqrt{35}\) | yes | yes | ||||
| \(\pi\) | yes | yes | ||||
| \(\sqrt{-16}\) | not real |
Three rows of that table carry the whole idea. The \(\sqrt{9}\) row looks exotic and is simply 3, a counting number, so always simplify before you classify — the notation can disguise an ordinary number. The \(\sqrt{35}\) row sits one line away and belongs to none of the inner sets. And \(\sqrt{-16}\) belongs to nothing on the chart at all, because it is outside the real numbers entirely.
For each number, list every set it belongs to among counting, whole, integer, rational, irrational, and real: \(\sqrt{64}\), \(-\dfrac{5}{2}\), and \(\sqrt{20}\).
Solution
Step 1 — Simplify \(\sqrt{64}\) first. Since \(8^2 = 64\), this number is just 8.
An 8 is a counting number, so it is also whole, an integer, rational (it is \(\dfrac{8}{1}\)), and real.
Step 2 — Classify \(-\dfrac{5}{2}\). It is already a ratio of two integers with a nonzero denominator, so it is rational and therefore real. It is negative and not a whole amount, so it is none of counting, whole, or integer.
As a decimal it is \(-2.5\), which terminates, confirming that it is rational.
Step 3 — Classify \(\sqrt{20}\). The number 20 sits between the perfect squares 16 and 25, so it is not a perfect square and its root is not an integer.
$$4 < \sqrt{20} < 5$$A square root of a whole number that is not a perfect square is irrational, so \(\sqrt{20}\) is irrational and real.
Answer: \(\sqrt{64} = 8\) is counting, whole, integer, rational, and real; \(-\dfrac{5}{2}\) is rational and real; \(\sqrt{20}\) is irrational and real.
The number line from Section 0.2 held the integers as evenly spaced ticks. Fractions and decimals fill the space between those ticks, so \(2.6\) sits six tenths of the way from 2 to 3, and \(-\dfrac{3}{4}\) sits three quarters of the way from 0 toward \(-1\). The irrationals take the positions that are left over, and \(\sqrt{55}\) has a definite spot just past 7.4 even though no finite decimal names it exactly. The result is that the real numbers fill the line completely, with no gaps anywhere. That completeness is what lets us talk about the graph of a line in Chapter 5 as one solid unbroken object instead of a dotted trail of points, and it is why the properties collected in Section 0.5 can be stated for all real numbers at once.
Simplify \(\;\sqrt{100} - 3\sqrt{4}\). Then say whether \(\sqrt{-9}\) is a real number and why. Then name every set that \(\sqrt{49}\) belongs to.
Solution
Part 1 — Simplify \(\sqrt{100} - 3\sqrt{4}\). Take each principal root first.
$$\sqrt{100} = 10 \qquad \sqrt{4} = 2$$Multiplication comes before subtraction:
$$10 - 3(2) = 10 - 6 = 4$$Part 2 — Is \(\sqrt{-9}\) real? It would have to be a number whose square is \(-9\). Positives square to positives, negatives square to positives, and zero squares to zero, so no real number works. It is not a real number.
Part 3 — Classify \(\sqrt{49}\). Simplify before classifying: \(\sqrt{49} = 7\).
Seven is a counting number, so it is also whole, an integer, rational (it is \(\dfrac{7}{1}\)), and real.
Answer: \(4\); \(\sqrt{-9}\) is not a real number; \(\sqrt{49} = 7\) is counting, whole, integer, rational, and real.
Problem Set 0.4
Problem 1. Write \(0.078\) in words.
Solution
Step 1 — Locate the last digit's place: In \(0.078\) the digits after the point are 0, 7, 8, filling tenths, hundredths, and thousandths, so the last digit sits in the thousandths place, and that place name is what the whole phrase gets read in.
Step 2 — Name the fractional digits as a whole number: Reading 078 as a count gives seventy-eight, and attaching the place of the last digit turns that count into "seventy-eight thousandths," since a decimal is just a fraction over a power of ten wearing compressed notation.
Answer: Seventy-eight thousandths.
Problem 2. Write "twenty-three and five tenths" in standard form.
Solution
Step 1 — Place the last digit: The phrase names tenths, so the last digit of the number belongs one place right of the point, because the fractional part is being counted in tenths.
Step 2 — Fill in the whole and fractional parts: "Twenty-three" is the whole-number part, and "five tenths" is a single digit, 5, sitting in the tenths place with no placeholder needed.
$$23.5$$Answer: \(23.5\).
Problem 3. Write "four and sixty-two thousandths" in standard form.
Solution
Step 1 — Place the last digit: The phrase names thousandths, so the last digit of the number lands three places right of the point, since each place is worth one tenth of the place before it.
Step 2 — Fill in the digits from the right: The fractional digits are 62. The 2 takes thousandths and the 6 takes hundredths, leaving nothing for tenths, so tenths needs a placeholder zero.
$$4.062$$Answer: \(4.062\).
Problem 4. In the number \(5.3096\), which digit is in the thousandths place, and what is that digit worth?
Solution
Step 1 — Locate the thousandths place: In \(5.3096\) the places after the point run tenths (3), hundredths (0), thousandths (9), ten-thousandths (6), so the digit in the thousandths place is 9.
Step 2 — State its value: The thousandths place is worth \(\dfrac{1}{1{,}000}\), so that digit is worth nine thousandths.
$$9 \times \frac{1}{1{,}000} = 0.009$$Answer: The digit \(9\) is in the thousandths place, worth \(0.009\).
Problem 5. Which is larger, \(0.5\) or \(0.48\)? Explain your reasoning in one sentence.
Solution
Step 1 — Compare by place value: Padding \(0.5\) to \(0.50\) shows it has 5 tenths while \(0.48\) has only 4 tenths, and since tenths are compared before hundredths, no amount of hundredths in \(0.48\) can make up a whole missing tenth.
Answer: \(0.5\) is larger than \(0.48\).
Problem 6. Order from smallest to largest: \(0.207\), \(0.27\), \(0.027\).
Solution
Step 1 — Pad to the same number of places: Writing all three with three decimal places gives \(0.207\), \(0.270\), and \(0.027\), so every digit can be compared column by column.
Step 2 — Compare tenths first: \(0.027\) has 0 tenths while \(0.207\) and \(0.270\) each have 2 tenths, so \(0.027\) is the smallest of the three, since the tenths place is compared before any smaller place.
Step 3 — Break the tie in hundredths: Between \(0.207\) and \(0.270\), the hundredths digits are 0 and 7, and 0 is less than 7, so \(0.207\) is smaller than \(0.270\).
Answer: \(0.027 < 0.207 < 0.27\).
Problem 7. Order from smallest to largest: \(-0.6\), \(-0.56\), \(-0.65\).
Solution
Step 1 — Pad to matching places: Writing all three with two decimal places gives \(-0.60\), \(-0.56\), and \(-0.65\), so the digits line up by place value.
Step 2 — Order the positive versions first: Comparing \(0.60\), \(0.56\), and \(0.65\) by tenths and then hundredths gives \(0.56 < 0.60 < 0.65\).
Step 3 — Flip the order for negatives: Making each number negative reverses its position on the number line, since a number is greater than whatever sits to its left, so the negative with the largest magnitude becomes the smallest value.
$$-0.65 < -0.6 < -0.56$$Answer: \(-0.65 < -0.6 < -0.56\).
Problem 8. Round \(3.7482\) to the nearest tenth.
Solution
Step 1 — Find the rounding digit and its neighbor: In \(3.7482\) the tenths digit is 7, and the digit immediately to its right is 4.
Step 2 — Apply the rounding rule: Since 4 is less than 5, the tenths digit stays the same, and everything to its right is dropped rather than replaced with zeros, because those fractional places carry no value once removed.
$$3.7482 \rightarrow 3.7$$Answer: \(3.7\).
Problem 9. Round \(3.7482\) to the nearest hundredth.
Solution
Step 1 — Find the rounding digit and its neighbor: Rounding again from the original number \(3.7482\), the hundredths digit is 4, and the digit immediately to its right is 8.
Step 2 — Apply the rounding rule: Since 8 is 5 or more, the hundredths digit increases by 1, from 4 to 5, and the remaining digits are dropped.
$$3.7482 \rightarrow 3.75$$Answer: \(3.75\).
Problem 10. Round \(0.0961\) to the nearest thousandth.
Solution
Step 1 — Find the rounding digit and its neighbor: In \(0.0961\) the thousandths digit is 6, and the digit immediately to its right, in the ten-thousandths place, is 1.
Step 2 — Apply the rounding rule: Since 1 is less than 5, the thousandths digit stays at 6 and the trailing digit is dropped, because a place past the rounding digit carries no value once removed.
$$0.0961 \rightarrow 0.096$$Answer: \(0.096\).
Problem 11. A phone case costs $14.99 and the tax comes to $1.086775. Round the tax to the nearest cent and find the total.
Solution
Step 1 — Find the hundredths digit of the tax. In $1.086775 the hundredths digit is 8, and the digit immediately to its right is 6.
Step 2 — Apply the rounding rule. Since 6 is 5 or more, the hundredths digit rounds up from 8 to 9, and everything past it is dropped.
$$\$1.086775 \rightarrow \$1.09$$Step 3 — Add to find the total. Line up the decimal points and add the rounded tax to the price, since only matching places can be added directly.
$$\$14.99 + \$1.09 = \$16.08$$Answer: $1.09 in tax, for a total of $16.08.
Problem 12. Add: \(7.9 + 0.46 + 12\).
Solution
Step 1 — Line up the points: Addition only combines matching places, so pad each addend with trailing zeros until all three show two decimal places — this does not change any of their values.
$$\begin{array}{r} 7.90 \\ 0.46 \\ +\;12.00 \\ \hline 20.36 \end{array}$$Step 2 — Add column by column: With the points aligned, tenths sit over tenths and hundredths sit over hundredths, so the columns add exactly like whole-number addition, and the point drops straight down into the answer.
Answer: \(20.36\).
Problem 13. Subtract: \(25 - 8.37\).
Solution
Step 1 — Give 25 its invisible decimal point: A whole number has an unwritten decimal point at its right end, so \(25\) is \(25.00\), padded to match the two decimal places in \(8.37\).
$$\begin{array}{r} 25.00 \\ -\;8.37 \\ \hline 16.63 \end{array}$$Step 2 — Subtract column by column: With the points lined up, each column subtracts hundredths from hundredths and tenths from tenths, borrowing across columns exactly as with whole numbers.
Answer: \(16.63\).
Problem 14. Multiply: \(0.8 \times 0.06\).
Solution
Step 1 — Multiply as whole numbers: Multiplication does not need the points aligned, so drop them and multiply \(8 \times 6\).
$$8 \times 6 = 48$$Step 2 — Count the decimal places in the factors: \(0.8\) has 1 decimal place and \(0.06\) has 2, so the product needs \(1 + 2 = 3\) places — this is really \(10^1 \cdot 10^2 = 10^3\) in disguise.
Step 3 — Place the point: Starting from the right end of 48, three places is more digits than 48 has, so a placeholder zero fills the tenths spot.
$$0.048$$Answer: \(0.048\).
Problem 15. Multiply: \(3.2 \times 4.5\).
Solution
Step 1 — Multiply as whole numbers: Ignore the points and multiply \(32 \times 45\).
$$32 \times 45 = 1{,}440$$Step 2 — Count the decimal places: \(3.2\) carries 1 place and \(4.5\) carries 1 place, so the product needs \(1 + 1 = 2\) places.
Step 3 — Place the point and drop the trailing zero: Counting two places from the right of 1440 gives \(14.40\), and a trailing zero at the right end of the fractional part changes nothing.
$$14.40 = 14.4$$Answer: \(14.4\).
Problem 16. Divide: \(14.4 \div 8\).
Solution
Step 1 — Set the quotient's point: Dividing by a whole number keeps the point in the quotient directly above the point in the dividend, so no places need to shift first.
Step 2 — Divide as with whole numbers: \(8\) goes into \(14\) once with \(6\) left over; bringing down the \(4\) makes \(64\), and \(8\) goes into \(64\) exactly \(8\) times.
$$14.4 \div 8 = 1.8$$Step 3 — Check by multiplying back: \(8 \times 1.8 = 14.4\), which matches the dividend, so the quotient is correct.
Answer: \(1.8\).
Problem 17. Divide: \(2.16 \div 0.4\).
Solution
Step 1 — Clear the decimal divisor: A decimal divisor is awkward, so multiply both the dividend and the divisor by the same power of ten — here \(10\), since \(0.4\) needs only one place to become whole. Multiplying both parts by the same thing leaves the value unchanged.
$$\frac{2.16}{0.4} = \frac{2.16 \times 10}{0.4 \times 10} = \frac{21.6}{4}$$Step 2 — Divide by the now-whole divisor: \(4\) goes into \(21\) five times with \(1\) left over; bringing down the \(6\) makes \(16\), and \(4\) goes into \(16\) exactly \(4\) times.
$$21.6 \div 4 = 5.4$$Answer: \(5.4\).
Problem 18. Compute without long multiplication: \(0.735 \times 1{,}000\).
Solution
Step 1 — Use the power-of-ten shortcut instead of multiplying it out: Multiplying by \(1{,}000\) has three zeros, and each zero slides the decimal point one place to the right — no digit-by-digit multiplication is needed.
Step 2 — Slide the point three places right: Starting from \(0.735\), moving the point three places right lands exactly on the last digit, since the number has exactly three decimal places.
$$0.735 \times 1{,}000 = 735$$Answer: \(735\).
Problem 19. Compute without long division: \(58.2 \div 100\).
Solution
Step 1 — Use the power-of-ten shortcut instead of dividing it out: Dividing by \(100\) has two zeros, and each zero slides the decimal point one place to the left — no long division is needed.
Step 2 — Slide the point two places left: Starting from \(58.2\), moving the point two places left passes the \(8\) and the \(2\), landing just past a placeholder zero in the tenths spot of the new whole-number part.
$$58.2 \div 100 = 0.582$$Answer: \(0.582\).
Problem 20. Simplify: \(\;1.5 + 0.2(8 - 3.5)\).
Solution
Step 1 — Grouping symbols first: Pad \(8\) to \(8.0\) so the points line up, then subtract.
$$8.0 - 3.5 = 4.5$$The expression is now \(1.5 + 0.2(4.5)\).
Step 2 — Multiply: Ignore the points and multiply \(2 \times 45 = 90\). There is 1 decimal place in \(0.2\) and 1 in \(4.5\), so the product needs \(1 + 1 = 2\) places.
$$0.90 = 0.9$$Step 3 — Add: With the points lined up, tenths add to tenths.
$$1.5 + 0.9 = 2.4$$Answer: \(2.4\).
Problem 21. Convert \(0.85\) to a fraction in lowest terms.
Solution
Step 1 — Read the decimal to name its fraction: \(0.85\) is "eighty-five hundredths," and the denominator is always the place value of the last digit — here, hundredths.
$$0.85 = \frac{85}{100}$$Step 2 — Reduce to lowest terms: Both \(85\) and \(100\) share a greatest common factor of \(5\), so divide top and bottom by \(5\).
$$\frac{85}{100} = \frac{17}{20}$$Answer: \(\dfrac{17}{20}\).
Problem 22. Convert \(\dfrac{7}{8}\) to a decimal.
Solution
Step 1 — Read the fraction bar as division: A fraction bar means the numerator divided by the denominator, so \(\dfrac{7}{8} = 7 \div 8\).
Step 2 — Divide, appending trailing zeros to the numerator as needed: \(8\) does not go into \(7\), so place a decimal point and continue: \(8\) into \(70\) goes \(8\) times with \(6\) left over, \(8\) into \(60\) goes \(7\) times with \(4\) left over, and \(8\) into \(40\) goes exactly \(5\) times, leaving no remainder — so the decimal terminates.
$$7 \div 8 = 0.875$$Answer: \(0.875\).
Problem 23. Convert \(\dfrac{2}{9}\) to a decimal and write it using bar notation.
Solution
Step 1 — Divide. The fraction bar means division, so divide 2 by 9. The division never reaches a remainder of zero — the remainder 2 keeps coming back, so the digit 2 keeps coming back with it.
$$2 \div 9 = 0.2222\ldots$$Step 2 — Write it with bar notation. A single digit repeats forever, so the bar covers just that digit.
$$0.\overline{2}$$Answer: \(\dfrac{2}{9} = 0.\overline{2}\).
Problem 24. Without dividing, predict whether \(\dfrac{9}{40}\) gives a terminating or a repeating decimal, and explain how you know.
Solution
Step 1 — Factor the denominator. The fraction \(\dfrac{9}{40}\) is already in lowest terms, and \(40 = 2^3 \cdot 5\) has only 2s and 5s as prime factors, which is exactly the condition for a terminating decimal.
Answer: Terminating, because \(40\)'s only prime factors are 2 and 5.
Problem 25. Convert \(0.6\) to a percent, and convert \(140\%\) to a decimal.
Solution
Step 1 — Decimal to percent. Move the point of \(0.6\) two places right and attach the symbol, padding with a trailing zero to make the shift visible.
$$0.60 \rightarrow 60\%$$Step 2 — Percent to decimal. Drop the \(\%\) on \(140\%\) and move the point two places left.
$$140\% \rightarrow 1.40$$The trailing zero drops, since it carries no value.
Answer: \(0.6 = 60\%\); \(140\% = 1.4\).
Problem 26. Which is largest: \(\dfrac{7}{10}\), \(0.71\), or \(69\%\)?
Solution
Step 1 — Put all three candidates in one notation. Decimals sort by place value with no common denominator to hunt for.
$$\frac{7}{10} = 0.70 \qquad 0.71 \qquad 69\% = 0.69$$Step 2 — Compare place by place. The tenths digits are 7, 7, and 6, so \(0.69\) drops out first. Between \(0.70\) and \(0.71\), the hundredths digits are 0 and 1, and 1 wins.
Answer: \(0.71\) is the largest.
Problem 27. Simplify: \(\;\sqrt{144}\).
Solution
Step 1 — Find the perfect square. The radical sign asks for the nonnegative number whose square is 144, and \(12^2 = 144\).
Answer: \(\sqrt{144} = 12\).
Problem 28. Simplify: \(\;-\sqrt{25}\).
Solution
Step 1 — Take the root first. The radical sign always means the principal, nonnegative root, so \(\sqrt{25} = 5\) regardless of the sign sitting outside it.
Step 2 — Apply the outside negative. The minus sign is outside the radical, so it negates the root only after the root is taken.
$$-\sqrt{25} = -(5)$$Answer: \(-\sqrt{25} = -5\).
Problem 29. Explain in one sentence why \(\sqrt{-36}\) is not a real number.
Solution
Step 1 — Consider what squares to a negative. A positive number squared is positive, a negative number squared is positive, and zero squared is zero, so no real number squares to \(-36\).
Answer: \(\sqrt{-36}\) is not a real number because nothing real squares to a negative.
Problem 30. Between which two consecutive whole numbers does \(\sqrt{72}\) lie?
Solution
Step 1 — Find the nearest perfect squares. The perfect squares on either side of 72 are \(64 = 8^2\) and \(81 = 9^2\), and \(64 < 72 < 81\).
Step 2 — Take the roots of the bracketing squares. Since square roots preserve order, the same inequality holds after taking the root of each part.
$$8 = \sqrt{64} \; < \; \sqrt{72} \; < \; \sqrt{81} = 9$$Answer: \(\sqrt{72}\) lies between 8 and 9.
Problem 31. Simplify: \(\;\sqrt{81} + \sqrt{16 - 7}\).
Solution
Step 1 — Grouping first. The radical sign groups everything underneath it, so the subtraction inside the second radical has to be simplified before either root is taken.
$$\sqrt{81} + \sqrt{16 - 7} = \sqrt{81} + \sqrt{9}$$Step 2 — Take both roots. Both radicands are perfect squares.
$$9 + 3$$Step 3 — Add.
$$12$$Answer: \(12\).
Problem 32. List every set that \(\sqrt{16}\) belongs to among counting, whole, integer, rational, irrational, and real.
Solution
Step 1 — Simplify before classifying. The radical sign can disguise an ordinary number, so simplify first: \(\sqrt{16} = 4\).
Step 2 — Sort the simplified number. Four is a counting number, which automatically makes it whole, an integer, and rational (it is \(\dfrac{4}{1}\)), and every rational number is real.
Answer: \(\sqrt{16} = 4\) belongs to counting, whole, integer, rational, and real.
Problem 33. List every set that \(-4.75\) belongs to among counting, whole, integer, rational, irrational, and real.
Solution
Step 1 — Classify the decimal. \(-4.75\) is a terminating decimal, and every terminating decimal is a ratio of integers, so it is rational and therefore real.
Step 2 — Rule out the smaller sets. It is negative and not a whole amount, so it is none of counting, whole, or integer.
Answer: \(-4.75\) belongs to rational and real only.
Problem 34. Explain in one sentence why \(\sqrt{10}\) is irrational but \(\sqrt{100}\) is not.
Solution
Step 1 — Check whether each radicand is a perfect square. 10 is not a perfect square, and the square root of a whole number that is not a perfect square is irrational; 100 is a perfect square, since \(10^2 = 100\), so its root is the whole number 10.
Answer: \(\sqrt{10}\) is irrational because 10 is not a perfect square, but \(\sqrt{100} = 10\) is rational because 100 is.
Key Terms
decimal point — the symbol separating the whole-number places from the fractional places, with each place to its right worth one tenth of the place before it.
trailing zero — a zero at the right end of the fractional part of a decimal; it never changes the value.
terminating decimal — a decimal whose digits stop, because the division reaches a remainder of zero.
repeating decimal — a decimal in which a block of digits repeats forever, written with a bar over the repeating block.
percent — a fraction whose denominator is 100, written with the \(\%\) symbol in place of the denominator.
square root — a number whose square is the given number; \(\sqrt{m}\) is the square root of \(m\).
radical sign — the symbol \(\sqrt{\;\;}\), which groups everything underneath it.
principal square root — the nonnegative square root, which is what \(\sqrt{m}\) always means.
perfect square — a number that is the square of an integer, so its square root is an integer.
rational number — a number that can be written as a ratio of two integers with a nonzero denominator; equivalently, a number whose decimal terminates or repeats.
irrational number — a real number that is not rational; its decimal neither terminates nor repeats.
real number — any number that is rational or irrational, which is any number with a location on the number line.