Operations with Signed Rational Numbers

Fractions and decimals on both sides of zero

The fraction $\tfrac{3}{4}$ sits three quarters of a unit to the right of zero. Its opposite, $-\tfrac{3}{4}$, sits the same distance to the left, between $-1$ and $0$. In the same way $-2.5$ is the opposite of $2.5$, halfway between $-3$ and $-2$. A negative fraction or decimal is the mirror image of a positive one across zero, exactly as $-5$ mirrors $5$.

Every number in that picture can be written as one integer divided by another. The number $1.5$ is $\tfrac{3}{2}$, and the integer $-2$ is $\tfrac{-2}{1}$. For $-\tfrac{3}{4}$, ask what $(-3) \div 4$ is, which means asking what number times $4$ gives $-3$. Four copies of $-\tfrac{3}{4}$ are four steps of three quarters to the left, and together they land $3$ units left of zero. So $4 \times \left(-\tfrac{3}{4}\right) = -3$, and therefore

A rational number is a number that can be written as a fraction $\frac{a}{b}$, where $a$ and $b$ are integers and $b$ is not zero.

The denominator must not be zero because the bar means divide, and dividing by zero never gives one definite number. Many familiar numbers pass the test. Every integer is rational, since $-7 = \tfrac{-7}{1}$. Zero is rational too, as $0 = \tfrac{0}{1}$, and it is still neither positive nor negative and still its own opposite. Every decimal that ends is rational, because the decimals chapter wrote each one as a fraction over a power of ten, such as $10$, $100$ or $1000$. For example $-2.5 = -\tfrac{25}{10} = \tfrac{-5}{2}$.

A repeating decimal that came from dividing two integers is rational for the plainest reason: it is that fraction. The decimal $-0.\overline{3}$ is $-\tfrac{1}{3}$. (Algebra I shows that every repeating decimal is a fraction, even one you did not get by dividing.) A repeating decimal has no last digit, though, so writing out its digits has to stop somewhere, and the stopped decimal is no longer exact. When one turns up in a calculation, work with its fraction instead.

Comparing and ordering

The integers chapter compared numbers by where they sit: of two numbers on the line, the one farther right is the greater. That rule talked about positions, never about whole numbers, so it still decides every comparison. Every negative is less than zero, and less than every positive, so $-\tfrac{1}{8} < 0 < 0.01$.

Two negatives need more care. The absolute value of a number is still its distance from zero, its size without its sign, and now that distance can be a fraction of a unit: $\lvert -0.6 \rvert = 0.6$ and $\lvert -0.45 \rvert = 0.45$. Compare $-0.6$ and $-0.45$. The number $-0.6$ is farther from zero, and on the negative side farther from zero means farther left. So

even though $0.6 > 0.45$. That reversal is for two negatives only. Of two positives, the one farther from zero is the greater, and any negative is less than any positive, whatever their absolute values.

When the two numbers are written in different forms, put them in one form first. Compare $-\tfrac{2}{3}$ and $-0.6$. The fraction $\tfrac{2}{3}$ has a repeating decimal, so change the decimal instead: $0.6 = \tfrac{3}{5}$. Over the common denominator $15$, the sizes are $\tfrac{10}{15}$ and $\tfrac{9}{15}$. The number $-\tfrac{2}{3}$ is farther from zero, so it is farther left: $-\tfrac{2}{3} < -0.6$.

Adding and subtracting: the sign, then the size

The integers chapter added signed numbers by walking along the number line. A walk does not care whether its steps are whole units. Start at $\tfrac{1}{2}$ and add $-\tfrac{5}{6}$, which is a step of five sixths to the left. The first half unit of that step brings you to zero. The rest of it, $\tfrac{5}{6} - \tfrac{1}{2} = \tfrac{1}{3}$, carries you past zero, so you land at $-\tfrac{1}{3}$.

That walk split into two separate questions. Which side of zero do you finish on? And how far from zero are you? The first is a question about signs, and the integer rules answer it. The second is a question about sizes, and the fraction methods answer it.

So the rules from the integers chapter carry straight over, with fraction and decimal sizes in place of whole-number ones.

Same signs: add the sizes and keep the shared sign. Different signs: subtract the smaller size from the larger, and take the sign of the number farther from zero. Subtraction: add the opposite, $a - b = a + (-b)$.

When the two sizes are equal and the signs differ, nothing is left over, and the sum is zero: $\tfrac{2}{5} + \left(-\tfrac{2}{5}\right) = 0$. That is the zero pair from the integers chapter, with a fraction in each half. Adding zero changes nothing, so zero needs no rule of its own.

This settles a question the fractions chapter left open. Adding and Subtracting Fractions kept every difference zero or positive by putting the larger fraction first. Now $\tfrac{1}{4} - \tfrac{2}{3}$ has an answer. Rewrite it as $\tfrac{1}{4} + \left(-\tfrac{2}{3}\right)$. The sizes are $\tfrac{3}{12}$ and $\tfrac{8}{12}$, so the difference of the sizes is $\tfrac{5}{12}$. The number $-\tfrac{2}{3}$ is farther from zero, so

Multiplying and dividing

Multiplying two negatives was where the integers chapter needed its most careful argument.

For nonzero signed rational numbers, multiply or divide the sizes as usual. The answer is positive when the two signs match and negative when they differ. With more factors, count the negative ones: an even count gives a positive answer, and an odd count a negative one. Each pair of negatives multiplies to a positive, so only an unpaired negative can leave the answer negative; and dividing is multiplying by the divisor’s reciprocal, which has the same sign as the divisor (shown below), so the count works for a chain of divisions too.

Dividing by a nonzero fraction still means multiplying by its reciprocal, the fraction turned upside down. The reciprocal of a negative fraction is negative. The reciprocal of $-\tfrac{2}{3}$ is $-\tfrac{3}{2}$, because two negatives make a positive and $\left(-\tfrac{2}{3}\right) \times \left(-\tfrac{3}{2}\right) = 1$. Turning a fraction upside down does not change its sign.

Zero follows the same rules it did for integers. If any factor is zero, the product is zero, as $\left(-\tfrac{5}{8}\right) \times 0 = 0$ shows. Zero divided by a nonzero number is zero, so $0 \div \left(-\tfrac{3}{4}\right) = 0$. But zero can never be the divisor. The quotient $(-2.5) \div 0$ would be a number that times $0$ gives $-2.5$. Every number times $0$ is $0$, so no such number exists, and $(-2.5) \div 0$ is undefined. And $0 \div 0$ fails the other way: every number times $0$ gives $0$, so no single number is the answer. That is the integers chapter’s argument, and it never needed $-2.5$ to be an integer. Zero also has no reciprocal, since no number times $0$ gives $1$.

Where the minus sign goes

So far $-\tfrac{3}{4}$ has been written with its minus sign in front, and also as $\tfrac{-3}{4}$. A third spelling puts the sign in the denominator, $\tfrac{3}{-4}$.

A minus sign on both the top and the bottom is different. In $\tfrac{-3}{-4}$ the signs match, so the quotient is positive: $\tfrac{-3}{-4} = \tfrac{3}{4}$. So a fraction is negative exactly when one of its numerator and denominator is negative and the other is positive. (A numerator of zero makes the fraction zero, which has no sign.) Writing the minus sign in front of the fraction, or in the numerator, with a positive denominator, is a convention for writing answers. It is not a rule of arithmetic, since all three spellings are equal.

Parentheses keep two signs from touching. A negative number that follows an operation sign is usually written in parentheses, as in $2.5 - (-0.75)$ or $3 \times \left(-\tfrac{1}{2}\right)$. A minus sign in front of parentheses means the opposite of everything inside, so $-\left(-\tfrac{3}{4}\right) = \tfrac{3}{4}$. Work out the inside first: $-(0.5 - 2) = -(-1.5) = 1.5$.

The same idea governs a negative mixed number. The sign belongs to the whole mixed number, so $-2\tfrac{1}{2}$ is the opposite of $2\tfrac{1}{2}$:

It does not mean $-2 + \tfrac{1}{2}$, which is $-1\tfrac{1}{2}$. On the number line, $-2\tfrac{1}{2}$ lies half a unit to the left of $-2$, not to its right.

Common mistakes