Primary & Middle School Mathematics · Grades 1–9

63Fractions and Powers

Fractions and powers are the everyday tools of computation: sharing quantities, comparing proportions, writing very large or very small numbers. This chapter consolidates the rules for computing with them — with every intermediate step written out — and introduces scientific notation.

63.1 Computing with fractions

Definition 63.1 (Fraction)

A fraction $\dfrac ab$ (with $b \neq 0$ ) is the quotient of $a$ by $b$ . Two fractions are equal when one is obtained from the other by multiplying (or dividing) numerator and denominator by the same nonzero number:

\frac ab = \frac{a \times k}{b \times k} \qquad (k \neq 0).

Example 63.2

Simplify $\dfrac{42}{56}$ step by step:

\frac{42}{56} = \frac{42 \div 2}{56 \div 2} = \frac{21}{28} = \frac{21 \div 7}{28 \div 7} = \frac34 .

(In Chapter 64 the greatest common divisor will do this in one step.)

Method 63.3 (Adding and subtracting fractions)

Put both fractions over a common denominator (a common multiple of the two denominators);
add or subtract the numerators, keeping the denominator;
simplify the result if possible.

Example 63.4

Compute $\dfrac56 + \dfrac34$ . A common denominator of $6$ and $4$ is $12$ :

\frac56 + \frac34 = \frac{5 \times 2}{6 \times 2} + \frac{3 \times 3}{4 \times 3} = \frac{10}{12} + \frac{9}{12} = \frac{19}{12}.

Compute $2 - \dfrac37$ . Write $2$ as a fraction over $7$ :

2 - \frac37 = \frac{14}{7} - \frac37 = \frac{11}{7}.

Why a common denominator: cutting both wholes into 12 equal boxes makes the fractions comparable and addable — together, 10 + 9 = 19 twelfths. — Why a common denominator: cutting both wholes into $12$ equal boxes makes the fractions comparable and addable — together, $10 + 9 = 19$ twelfths.

Method 63.5 (Multiplying and dividing fractions)

To multiply, multiply numerators together and denominators together: $\dfrac ab \times \dfrac cd = \dfrac{ac}{bd}$ (simplify before multiplying whenever possible);
to divide, multiply by the inverse of the divisor: $\dfrac ab \div \dfrac cd = \dfrac ab \times \dfrac dc$ .

Example 63.6

\frac{7}{15} \times \frac{25}{14} = \frac{7 \times 25}{15 \times 14} = \frac{7 \times 25}{15 \times 2 \times 7} = \frac{25}{30} = \frac56,

where we simplified by the common factor $7$ , then by $5$ . And a division:

\frac{3}{4} \div \frac{9}{8} = \frac34 \times \frac89 = \frac{3 \times 8}{4 \times 9} = \frac{24}{36} = \frac23 .

63.2 Powers

Definition 63.7 (Power)

For a real number $a$ and a positive integer $n$ , the power $a^n$ is the product of $n$ factors equal to $a$ :

a^n = \underbrace{a \times a \times \dots \times a}_{n \text{ factors}} .

By convention $a^0 = 1$ (for $a \neq 0$ ), and negative exponents denote inverses:

a^{-n} = \frac{1}{a^n}.

Example 63.8

$2^4 = 2 \times 2 \times 2 \times 2 = 16$ ; $(-3)^2 = 9$ but $-3^2 = -(3^2) = -9$ (the exponent binds before the minus sign); $10^{-3} = \frac{1}{1000} = 0.001$ ; $5^1 = 5$ and $5^0 = 1$ .

Theorem 63.9 (Rules of exponents)

For all nonzero $a$ , $b$ and all integers $m$ , $n$ :

a^m \times a^n = a^{m+n}, \qquad \frac{a^m}{a^n} = a^{m-n}, \qquad \left(a^m\right)^n = a^{mn}, \qquad (ab)^n = a^n b^n .

Proof. For positive exponents, count the factors. For the first rule: $a^m \times a^n$ is $m$ factors $a$ followed by $n$ factors $a$ , in total $m + n$ factors. For the third: $\left(a^m\right)^n$ repeats $n$ times a block of $m$ factors, giving $mn$ factors. The other two rules are similar; the extension to zero and negative exponents is checked from $a^{-n} = \frac{1}{a^n}$ (and makes the convention $a^0 = 1$ the only consistent choice, since $a^n \times a^0$ must be $a^{n+0} = a^n$ ). ∎

Example 63.10

Simplify step by step:

\frac{3^7 \times 3^{-2}}{3^4} = \frac{3^{7 + (-2)}}{3^4} = \frac{3^5}{3^4} = 3^{5-4} = 3^1 = 3 .

And with two letters: $(2a^3)^2 \times a = 4 a^6 \times a = 4a^7$ .

63.3 Powers of ten and scientific notation

Proposition 63.11 (Powers of ten)

For a positive integer $n$ : $10^n$ is written “ $1$ followed by $n$ zeros”, and $10^{-n} = 0.0\dots01$ with the $1$ in the $n$ -th decimal place. Multiplying a decimal number by $10^n$ (resp. $10^{-n}$ ) shifts its decimal point $n$ places to the right (resp. left).

Definition 63.12 (Scientific notation)

The scientific notation of a positive decimal number is the unique way of writing it as

a \times 10^n, \qquad\text{with } 1 \leq a < 10 \text{ and } n \text{ an integer.}

Example 63.13

The Earth–Sun distance is about $149\,600\,000$ km $= 1.496 \times 10^8$ km. The size of a water molecule is about $0.000\,000\,000\,28$ m $= 2.8 \times 10^{-10}$ m. To compute with such numbers, group the decimal parts and the powers of ten:

(3 \times 10^5) \times (2.5 \times 10^{-2}) = (3 \times 2.5) \times 10^{5 + (-2)} = 7.5 \times 10^{3}.

Method 63.14 (Putting a number in scientific notation)

Move the decimal point just after the first nonzero digit; this gives the factor $a$ with $1 \leq a < 10$ ;
count the number of places the point moved: that count is $n$ , positive if the original number was $\geq 10$ , negative if it was $< 1$ ;
check: $a \times 10^n$ must reproduce the original number.

63.4 Exercises

Exercise 63.1 ★

Compute and give the result as a fully simplified fraction:

\frac13 + \frac15, \qquad \frac72 - \frac53, \qquad \frac{5}{12} + \frac38, \qquad 3 - \frac45 .

Solution

Solution of Exercise 63.1.

$\dfrac13 + \dfrac15 = \dfrac{5}{15} + \dfrac{3}{15} = \dfrac{8}{15}$ .

$\dfrac72 - \dfrac53 = \dfrac{21}{6} - \dfrac{10}{6} = \dfrac{11}{6}$ .

$\dfrac{5}{12} + \dfrac38 = \dfrac{10}{24} + \dfrac{9}{24} = \dfrac{19}{24}$ .

$3 - \dfrac45 = \dfrac{15}{5} - \dfrac45 = \dfrac{11}{5}$ .

Exercise 63.2 ★

Compute and simplify:

\frac59 \times \frac{27}{35}, \qquad \frac{8}{15} \div \frac{4}{5}, \qquad \frac23 \times \frac34 \times \frac45 .

Solution

Solution of Exercise 63.2.

$\dfrac59 \times \dfrac{27}{35} = \dfrac{5 \times 27}{9 \times 35} = \dfrac{27}{9} \times \dfrac{5}{35} = 3 \times \dfrac17 = \dfrac37$ (simplify by $9$ and by $5$ ).

$\dfrac{8}{15} \div \dfrac45 = \dfrac{8}{15} \times \dfrac54 = \dfrac{8 \times 5}{15 \times 4} = \dfrac{40}{60} = \dfrac23$ .

$\dfrac23 \times \dfrac34 \times \dfrac45 = \dfrac{2 \times 3 \times 4} {3 \times 4 \times 5} = \dfrac25$ (the $3$ s and $4$ s cancel).

Exercise 63.3 ★

Compute:

2^5, \qquad (-2)^4, \qquad -2^4, \qquad 4^{-2}, \qquad \left(\tfrac23\right)^3, \qquad 7^0 .

Solution

Solution of Exercise 63.3.

$2^5 = 32$ ; $(-2)^4 = 16$ ; $-2^4 = -16$ ; $4^{-2} = \frac{1}{16}$ ; $\left(\frac23\right)^3 = \frac{8}{27}$ ; $7^0 = 1$ .

Exercise 63.4 ★

Write as a single power:

5^3 \times 5^6, \qquad \frac{7^9}{7^5}, \qquad \left(2^3\right)^4, \qquad \frac{3^2 \times 3^7}{3^5}, \qquad 2^5 \times 4 .

Solution

Solution of Exercise 63.4.

$5^3 \times 5^6 = 5^9$ ; $\dfrac{7^9}{7^5} = 7^4$ ; $\left(2^3\right)^4 = 2^{12}$ ; $\dfrac{3^2 \times 3^7}{3^5} = 3^{9-5} = 3^4$ ; $2^5 \times 4 = 2^5 \times 2^2 = 2^7$ .

Exercise 63.5 ★

Write in scientific notation: $45\,000$ ; $0.0072$ ; $302.5$ ; $0.000\,000\,9$ ; twelve million.

Solution

Solution of Exercise 63.5.

$45\,000 = 4.5 \times 10^4$ ; $0.0072 = 7.2 \times 10^{-3}$ ; $302.5 = 3.025 \times 10^2$ ; $0.000\,000\,9 = 9 \times 10^{-7}$ ; twelve million $= 1.2 \times 10^7$ .

Exercise 63.6 ★★

Compute, giving the result in scientific notation:

(2 \times 10^7) \times (4.5 \times 10^{-3}), \qquad \frac{9 \times 10^5}{3 \times 10^{-2}}, \qquad (5 \times 10^3)^2 .

Solution

Solution of Exercise 63.6.

$(2 \times 10^7) \times (4.5 \times 10^{-3}) = 9 \times 10^{4}$ .

$\dfrac{9 \times 10^5}{3 \times 10^{-2}} = 3 \times 10^{5 - (-2)} = 3 \times 10^7$ .

$(5 \times 10^3)^2 = 25 \times 10^6 = 2.5 \times 10^7$ .

Exercise 63.7 ★★

A tank is $\frac35$ full. After adding $30$ liters it is $\frac34$ full. What is the capacity of the tank? (Express the added fraction of the tank first.)

Solution

Solution of Exercise 63.7.

The added water represents $\dfrac34 - \dfrac35 = \dfrac{15}{20} - \dfrac{12}{20} = \dfrac{3}{20}$ of the tank. So $\frac{3}{20}$ of the capacity is $30$ liters: $\frac{1}{20}$ is $10$ liters, and the capacity is $200$ liters.

Exercise 63.8 ★★

Alice eats $\frac14$ of a cake, then Ben eats $\frac13$ of what remains, then Carla eats $\frac12$ of what still remains. What fraction of the cake is left at the end? (Compute the remainder after each step.)

Solution

Solution of Exercise 63.8.

After Alice: $1 - \frac14 = \frac34$ of the cake remains. Ben eats $\frac13 \times \frac34 = \frac14$ , leaving $\frac34 - \frac14 = \frac12$ . Carla eats half of that, $\frac14$ , leaving $\frac14$ of the cake.

Exercise 63.9 ★★

Light travels $3 \times 10^8$ meters per second. The Sun is about $1.5 \times 10^{11}$ m away. How long does sunlight take to reach us? Give the answer in seconds (scientific notation not needed), then in minutes and seconds.

Solution

Solution of Exercise 63.9.

Time $= \dfrac{\text{distance}}{\text{speed}} = \dfrac{1.5 \times 10^{11}}{3 \times 10^8} = 0.5 \times 10^3 = 500$ seconds, i.e. $8$ minutes and $20$ seconds.

Exercise 63.10 ★★★

Which is larger, $2^{30}$ or $3^{20}$ ? (Hint: write both as powers with exponent $10$ using $\left(a^m\right)^n = a^{mn}$ , and compare $2^3$ with $3^2$ .)

Solution

Solution of Exercise 63.10.

Write both numbers as powers with exponent $10$ :

2^{30} = \left(2^3\right)^{10} = 8^{10}, \qquad 3^{20} = \left(3^2\right)^{10} = 9^{10}.

Since $8 < 9$ , multiplying ten copies of each: $8^{10} < 9^{10}$ , so $2^{30} < 3^{20}$ .

63.5 Problem: The secret code of repeating decimals

Problem 63.1

Weekend problem — every fraction has a decimal writing that stops or repeats, every repeating decimal is a fraction, and $0.999\ldots$ equals exactly $1$

Divide $3$ by $11$ and the digits fall into a chant: $0.272727\ldots$ forever. Divide $1$ by $4$ and the digits stop dead: $0.25$ . This problem proves that these are the only two possible behaviors for a fraction — and then reverses the machine: given any repeating decimal, it rebuilds the fraction hiding behind it. On the way it settles, once and for all, the most argued-about equality in school mathematics.

Part I — From fraction to decimal.

Compute by long division the decimal writings of $\frac14$ , $\frac13$ , $\frac56$ and $\frac{3}{11}$ . Which stop, which repeat, and with what repeating block?
Explain why the decimal writing of any fraction $\frac ab$ must either stop or eventually repeat. (In the long division by $b$ , which values can the successive remainders take? What happens the moment a remainder shows up for the second time?)
Push the division of $1$ by $7$ far enough to find the repeating block. How long is it, and how does that length compare with the bound promised by question 2?
Some fractions stop because their denominator can be blown up to a power of ten: $\frac14 = \frac{25}{100}$ , $\frac{3}{8} = \frac{375}{1000}$ . Explain the criterion, and why no whole-number multiplier can ever turn $3$ , $6$ or $11$ into $10$ , $100$ , $1000, \dots$ (what digit would the product end in?).
Without dividing, sort $\frac{9}{40}$ , $\frac{5}{12}$ and $\frac{33}{50}$ into “stops” and “repeats”, then give the decimal writing of the two that stop.

Part II — From repeating decimal back to the fraction.

The shift trick. Let $x = 0.7777\ldots$ Compute $10x$ , then $10x - x$ , and deduce $x$ as a fraction. Check by dividing.
Let $x = 0.727272\ldots$ Which power of ten shifts $x$ by exactly one repeating block? Use it to write $x$ as a fraction in lowest terms.
A mixed case: $x = 0.58333\ldots$ (the $3$ alone repeats). Compute $1000x - 100x$ and deduce $x$ as a fraction in lowest terms.
The famous one: apply the shift trick to $x = 0.9999\ldots$ What fraction — what number — do you find? Reconcile the result with your intuition, remembering the phantom neighbour of Problem 38.1 and the shrinking remainder of Problem 39.1: is $0.999\ldots$ “just below” $1$ , or another name for $1$ ?
Convert $2.454545\ldots$ and $0.123123123\ldots$ into fractions in lowest terms.

Part III — The dictionary completed.

Questions 6–10 suggest a dictionary: a block of $k$ digits repeating from the decimal point equals that block over $k$ nines ( $0.727272\ldots = \frac{72}{99}$ ). Use the dictionary to predict $0.037037\ldots$ as a fraction, simplify — and marvel: which surprisingly simple fraction chants “ $037$ ”? Check it by division.
Combine the dictionary with a shift: write $0.0272727\ldots$ as a fraction in lowest terms.
The number $0.101001000100001\ldots$ (a $1$ , then one $0$ , then a $1$ , then two $0$ s, then a $1$ , then three $0$ s, and so on) never stops. Is it a repeating decimal? What does Part I then say about it — can it be a fraction? (You have just met your first provably irrational number; the most famous one is caught in Chapter 65.)
Roughly how many digits does $2^{30}$ have? Use $2^{10} = 1\,024 \approx 1.024 \times 10^3$ and the exponent rules (Theorem 63.9) to write $2^{30}$ in (approximate) scientific notation, and conclude. (Compare Exercise 63.10.)
Finale: write $0.20262026\ldots$ as a fraction, and state the two-way moral of the whole problem: which decimal writings are fractions, and which fractions have which decimal writings?

Solution

Solution of Problem 63.1.

1. $\frac14 = 0.25$ (stops); $\frac13 = 0.333\ldots$ (block $3$ ); $\frac56 = 0.8333\ldots$ (block $3$ after the $8$ ); $\frac{3}{11} = 0.272727\ldots$ (block $27$ ).

2. At each step of the division by $b$ , the remainder is one of the $b$ values $0, 1, \dots, b - 1$ (Theorem 37.7). If a remainder $0$ appears, the division stops. Otherwise, after at most $b$ steps some remainder must appear a second time — there are more steps than possible remainders — and from that moment the division repeats exactly the sequence of digits it produced after the first appearance: the decimal writing cycles forever. Stop or repeat: no third behavior exists.

3. $\frac17 = 0.142857\,142857\ldots$ : the block $142857$ has length $6$ . Question 2 promised a repetition within at most $7$ steps — here the remainders $1, 3, 2, 6, 4, 5$ all appear before the $1$ returns: the longest possible chant for a division by $7$ .

4. $\frac ab$ stops exactly when some $\frac{a \times k}{b \times k}$ has denominator $10$ , $100$ , $1000, \dots$ — that is, when the denominator can be multiplied up to a power of ten, as $4 \times 25 = 100$ or $8 \times 125 = 1\,000$ . For $3$ and $6$ this is hopeless: a power of ten has digit sum $1$ , and every multiple of $3$ has a digit sum that is a multiple of $3$ — so no multiple of $3$ (hence none of $6$ ) is ever $10$ , $100$ , $1\,000, \dots$ For $11$ : dividing $10$ , $100$ , $1\,000$ , $10\,000$ by $11$ always leaves a remainder ( $10$ , $1$ , $10$ , $1, \dots$ ), never $0$ . Denominators that reach a power of ten stop; all others repeat.

5. $\frac{9}{40}$ : $40 \times 25 = 1\,000$ , stops: $\frac{225}{1000} = 0.225$ . $\frac{33}{50}$ : $50 \times 2 = 100$ , stops: $\frac{66}{100} = 0.66$ . $\frac{5}{12}$ : $12$ is a multiple of $3$ , repeats ( $0.41666\ldots$ ).

6. $10x = 7.777\ldots$ , so $10x - x = 7$ (the tails cancel perfectly), $9x = 7$ and $x = \frac79$ . Division check: $7 \div 9 = 0.777\ldots$

7. The block has two digits, so shift by $100$ : $100x - x = 72.7272\ldots - 0.7272\ldots = 72$ , hence $99x = 72$ and $x = \frac{72}{99} = \frac{8}{11}$ .

8. $1000x = 583.333\ldots$ and $100x = 58.333\ldots$ ; subtracting, $900x = 525$ , so $x = \frac{525}{900} = \frac{7}{12}$ (divide by $75$ ). Indeed $\frac{7}{12} = 0.58333\ldots$

9. $10x - x = 9.999\ldots - 0.999\ldots = 9$ , so $9x = 9$ and $x = 1$ . Not “just below” $1$ : $0.999\ldots$ is the number $1$ , written in a second costume. Any number strictly below $1$ leaves a gap, and Problem 38.1 showed a gap always contains further numbers — while nothing fits between $0.999\ldots$ and $1$ . The endless nines are the chocolate remainder of Problem 39.1 shrunk below every positive amount: nothing is left.

10. $2.4545\ldots = 2 + \frac{45}{99} = 2 + \frac{5}{11} = \frac{27}{11}$ . And $0.123123\ldots = \frac{123}{999} = \frac{41}{333}$ (divide by $3$ ).

11. Dictionary: $0.037037\ldots = \frac{37}{999}$ . Since $999 = 27 \times 37$ , this simplifies to $\frac{1}{27}$ : the fraction $\frac{1}{27}$ chants “ $037$ ”. Division check: $1 \div 27 = 0.037037\ldots$

12. $0.0272727\ldots = \frac{1}{10} \times 0.272727\ldots = \frac{1}{10} \times \frac{27}{99} = \frac{27}{990} = \frac{3}{110}$ (divide by $9$ ).

13. The blocks of zeros grow without end, so no fixed block can repeat forever: it is not a repeating decimal (and it never stops). By Part I, every fraction stops or repeats — so this number is no fraction at all: it is irrational, built to order. The superstar of irrationality, $\sqrt2$ , is unmasked in Chapter 65.

14. $2^{30} = \left(2^{10}\right)^3 \approx (1.024)^3 \times 10^9 \approx 1.07 \times 10^9$ : a number slightly above $10^9$ , hence with $10$ digits. (Exactly: $2^{30} = 1\,073\,741\,824$ .)

15. $0.20262026\ldots = \frac{2026}{9999}$ , already in lowest terms ( $9999 = 3 \times 3 \times 11 \times 101$ shares no factor with $2026 = 2 \times 1013$ ). The completed dictionary: the fractions are exactly the decimals that stop or repeat — denominators of tens’ friends stop, all others chant a block; and conversely every chant, even your graduation year, is a fraction in disguise.