Chapter 2 Ā· Operations with rational numbers

Ordering Rational Numbers and Finding Distance

Use the number line to put rational numbers in order and to find how far apart two of them are, using absolute values.

18 min read

Every rational number has exactly one place on the number line. In this lesson you use those places to answer two questions: which of two numbers is greater, and how far apart are they?

A quick reminder first. A rational number is a number you can write as a fraction mn\frac{m}{n}, where mm and nn are integers (whole numbers and their opposites) and nn is not 00. A negative sign can sit in front, on top, or on the bottom without changing the number: āˆ’mn=āˆ’mn=māˆ’n-\frac{m}{n} = \frac{-m}{n} = \frac{m}{-n}. For example, āˆ’34-\frac{3}{4}, āˆ’34\frac{-3}{4} and 3āˆ’4\frac{3}{-4} are one number at one place on the line.

In decimal form, a rational number either terminates (it ends, like āˆ’0.75-0.75) or repeats (like 0.3‾0.\overline{3}, where the bar marks the digit that repeats forever). When a decimal repeats, mark the repeat with the bar. Never cut it off.

Number line from -2 to 2, with a step of 0.25, showing 4 marked points. Markers: marker at -1.5, named -1.5; marker at -0.75, named -0.75; marker at 0.5, named 0.5; marker at 1.25, named 1.25.
The markers sit at āˆ’1.5-1.5, āˆ’0.75-0.75, 0.50.5 and 1.251.25. Left to right is least to greatest.

On the number lines in this lesson, numbers get greater as you move to the right. So less than means to the left of, and greater than means to the right of. This holds for every pair of rational numbers, whatever their signs.

The line above runs from āˆ’2-2 to 22 with a tick every 0.250.25. The marker at āˆ’0.75-0.75 is the place of āˆ’34-\frac{3}{4}. Read the four markers from left to right and you have them in order from least to greatest:

āˆ’1.5<āˆ’0.75<0.5<1.25-1.5 < -0.75 < 0.5 < 1.25

Now look at how far each marker is from 00. The absolute value of a number aa, written ∣a∣|a|, is its distance from 00 on the number line. It is also called the magnitude of aa. A distance is never negative, so ∣a∣|a| is never negative: āˆ£āˆ’1.5∣=1.5\left|-1.5\right| = 1.5 and ∣1.25∣=1.25|1.25| = 1.25.

Notice that āˆ’1.5-1.5 is farther from 00 than any other marker, yet it is the least of the four numbers. Order follows position on the line, not magnitude.

To compare or order rational numbers, you only need comparisons of positive fractions and decimals, which you already know.

  1. Sort by side. Negative numbers are left of 00 and positive numbers are right of 00. So every negative number is less than 00, and 00 is less than every positive number.
  2. Use one form. Write the absolute values in the same form: fractions with a common denominator, or decimals. If a decimal repeats, keep the repeat marked with its bar.
  3. Compare within each side. Of two positive numbers, the one with the larger absolute value is farther right, so it is greater. Of two negative numbers, the one with the larger absolute value is farther left, so it is less.

Worked example

Ordering fractions and decimals

Order āˆ’34\frac{-3}{4}, 0.50.5, āˆ’0.7-0.7 and 35\frac{3}{5} from least to greatest.

  1. Read the first number's sign. It is a fraction of two integers with the negative sign on top. That is the same number as the fraction with the sign in front, so it is a negative number.

    āˆ’34=āˆ’34\frac{-3}{4} = -\frac{3}{4}
  2. Sort by side of 0. The two negative numbers are left of 0 and the two positive numbers are right of 0. Every number left of 0 is less than every number right of 0, so the negative numbers come first.

    leftĀ ofĀ 0:Ā āˆ’34,Ā āˆ’0.7rightĀ ofĀ 0:Ā 0.5,Ā 35\text{left of } 0:\ -\frac{3}{4},\ -0.7 \qquad \text{right of } 0:\ 0.5,\ \frac{3}{5}
  3. Write the four absolute values in one form. Change each terminating decimal to a fraction, then rename all four with the denominator 20, because 20 is a multiple of 4, 10 and 5. Equivalent fractions name the same number, so no position changes.

    34=1520,0.5=510=1020,0.7=710=1420,35=1220\frac{3}{4}=\frac{15}{20},\quad 0.5=\frac{5}{10}=\frac{10}{20},\quad 0.7=\frac{7}{10}=\frac{14}{20},\quad \frac{3}{5}=\frac{12}{20}
  4. Compare the two negative numbers. Their absolute values are 15/20 and 14/20, and 15/20 is larger. So the negative fraction is farther from 0 than -0.7 is. Both are left of 0, so farther from 0 means farther left, and farther left means less.

    āˆ’34<āˆ’0.7-\frac{3}{4} < -0.7
  5. Compare the two positive numbers. Their absolute values are 10/20 and 12/20, and 12/20 is larger. So 3/5 is farther from 0 than 0.5 is. Both are right of 0, so farther from 0 means farther right, and farther right means greater.

    0.5<350.5 < \frac{3}{5}
  6. Join the two sides, with the negative numbers first because they are left of 0.

    āˆ’34<āˆ’0.7<0.5<35-\frac{3}{4} < -0.7 < 0.5 < \frac{3}{5}

Answer

From least to greatest, the numbers are āˆ’34\frac{-3}{4}, āˆ’0.7-0.7, 0.50.5 and 35\frac{3}{5}.

Now the second question: how far apart are two numbers? The distance between two numbers is the length of the piece of number line between them. It is how far you travel to get from one to the other, so it is the same in both directions and it is never negative.

Absolute values do the measuring, because each one is a distance from 00. The line below marks āˆ’1.5-1.5, āˆ’0.5-0.5 and 0.750.75, with a tick every 0.250.25.

  • Opposite sides of zero. Going from āˆ’1.5-1.5 to 0.750.75, you pass through 00. You travel 1.51.5 to reach 00, then 0.750.75 more. Add: 1.5+0.75=2.251.5 + 0.75 = 2.25.
  • Same side of zero. Going from āˆ’1.5-1.5 to āˆ’0.5-0.5, you stop before 00. The whole trip from āˆ’1.5-1.5 to 00 would be 1.51.5 long, and you skip its last 0.50.5. Subtract: 1.5āˆ’0.5=11.5 - 0.5 = 1.

Check by counting the gaps between ticks: 99 gaps of 0.250.25 make 2.252.25, and 44 gaps of 0.250.25 make 11.

Number line from -2 to 2, with a step of 0.25, showing 3 marked points. Markers: marker at -1.5, named -1.5; marker at -0.5, named -0.5; marker at 0.75, named 0.75.
From āˆ’1.5-1.5 to 0.750.75 is 99 gaps of 0.250.25. From āˆ’1.5-1.5 to āˆ’0.5-0.5 is 44 gaps of 0.250.25.

The distance between two rational numbers pp and qq is written ∣pāˆ’q∣|p-q|, the absolute value of their difference. Here is why that fits.

A difference tells you what to add: 7āˆ’3=47-3=4 because 3+4=73+4=7. In the same way, pāˆ’qp-q is the number you add to qq to land on pp. On the number line it describes the move from qq to pp: how far, and in which direction (right is positive, left is negative). The absolute value keeps the how far and drops the direction. So ∣pāˆ’q∣|p-q| is the length of the move, which is the distance between the two points.

Try it with q=āˆ’1.5q=-1.5 and p=0.75p=0.75 on the line above. The move from qq to pp is 2.252.25 to the right, and the move back is 2.252.25 to the left. The length is 2.252.25 either way, so ∣pāˆ’q∣|p-q| and ∣qāˆ’p∣|q-p| are both 2.252.25, the same number you found by adding absolute values. So adding absolute values (opposite sides of 00) or subtracting them (same side of 00) gives ∣pāˆ’q∣|p-q| using only arithmetic with positive numbers.

Key concept

Position gives order and distance

Every rational number has one position on the number line.

Order. The number farther left is less. Order follows position, not absolute value.

Distance. For any two rational numbers pp and qq, the distance between them is ∣pāˆ’q∣|p-q|, and it is never negative. Find it from ∣p∣|p| and ∣q∣|q|, their distances from 00:

  • Opposite sides of 00: add the absolute values.
  • Same side of 00: subtract the smaller absolute value from the larger. Equal numbers are 00 apart.
  • One number is 00: the distance is the other number's absolute value.

Each rule gives ∣pāˆ’q∣|p-q|, because ∣pāˆ’q∣|p-q| is the length of the move from one number to the other.

distanceĀ betweenĀ pĀ andĀ q=∣pāˆ’q∣\text{distance between } p \text{ and } q = |p - q|

Worked example

Distance across zero on a trail

A straight trail runs east and west. A park gate on the trail is at position 00. Positions are measured in miles from the gate: east of the gate is positive and west is negative. A picnic table is at āˆ’114-1\frac{1}{4} and a lookout is at 2.52.5. How far apart are the picnic table and the lookout?

  1. Picture the trail as a number line with west on the left, and find each place's side of 0. The picnic table's position is negative, so it is west of the gate, left of 0. The lookout's position is positive, so it is east of the gate, right of 0. They are on opposite sides of 0, so the path between them passes through the gate.

  2. Write each absolute value. It is that place's distance from the gate in miles, and a distance is never negative.

    āˆ£āˆ’114∣=114∣2.5∣=2.5\left|-1\frac{1}{4}\right| = 1\frac{1}{4} \qquad |2.5| = 2.5
  3. Use one form so you can add. Change the terminating decimal to a mixed number, then rename its fraction part in fourths. Equivalent fractions name the same number, so the distance does not change.

    2.5=2510=212=2242.5 = 2\frac{5}{10} = 2\frac{1}{2} = 2\frac{2}{4}
  4. The places are on opposite sides of 0, so add the absolute values: the picnic table's distance to the gate plus the gate's distance to the lookout.

    114+224=3341\frac{1}{4} + 2\frac{2}{4} = 3\frac{3}{4}
  5. Connect this to the formula. Call the lookout's position p and the picnic table's position q. The length you found is the absolute value of their difference. You found it by adding two positive numbers, with no negative number to subtract.

    ∣pāˆ’q∣=114+212=334|p-q| = 1\frac{1}{4} + 2\frac{1}{2} = 3\frac{3}{4}

Answer

The picnic table and the lookout are 3343\frac{3}{4} miles apart.

Worked example

Distance on the same side of zero

Find the distance between āˆ’56-\frac{5}{6} and āˆ’14-\frac{1}{4} on a number line.

  1. Both numbers are negative, so both are left of 0. They are on the same side, and the path between them does not pass through 0. So you subtract absolute values.

  2. Write the absolute values and rename them with the common denominator 12, because 12 is a multiple of 6 and 4. Multiply the top and bottom of 5/6 by 2, and of 1/4 by 3.

    āˆ£āˆ’56∣=56=1012āˆ£āˆ’14∣=14=312\left|-\frac{5}{6}\right| = \frac{5}{6} = \frac{10}{12} \qquad \left|-\frac{1}{4}\right| = \frac{1}{4} = \frac{3}{12}
  3. Subtract the smaller absolute value from the larger. The larger, 10/12, is the far point's distance from 0. Taking away 3/12, the near point's distance from 0, leaves the gap between the two points.

    1012āˆ’312=712\frac{10}{12} - \frac{3}{12} = \frac{7}{12}
  4. Keep the answer as a fraction. The decimal form of 7/12 repeats, so no terminating decimal equals it, and a cut-off decimal would be a different number.

    712=0.583‾\frac{7}{12} = 0.58\overline{3}

Answer

The distance between āˆ’56-\frac{5}{6} and āˆ’14-\frac{1}{4} is 712\frac{7}{12}.

Takeaway. Position on the number line answers both questions. To order rational numbers, ask which is farther left: every negative number is less than every positive number, and of two negative numbers, the one with the larger absolute value is less. The distance between two rational numbers pp and qq is ∣pāˆ’q∣|p-q|. Find it with absolute values: add them when the numbers are on opposite sides of 00, and subtract the smaller from the larger when they are on the same side. Next, you will look for errors in work that converts between fractions and decimals.