Locating Irrational Numbers on a Number Line | Grade 8 Math | LevelKite

A number line gives every number its own point. Positive numbers are to the right of $0$ , negative numbers are to the left, and a number's distance from $0$ shows its size. To locate a number means to say where its point is.

The short marks on the line are ticks. They are evenly spaced, and the distance from one tick to the next is the step. A number that matches a tick is easy to locate: you point at the tick. But some numbers never match a tick, however finely you mark the line in decimals. In this lesson you will learn to say where those numbers are by using the ticks around them.

Every tick on the number lines in this lesson is an integer or a terminating decimal, such as $-2$ , $1.4$ or $1.41$ . Each of these is a rational number. For example, $1.41 = \frac{141}{100}$ .

Some square roots are rational, and they land exactly on a tick. Because $3$ is nonnegative and $3^2 = 9$ , you know that $\sqrt{9} = 3$ . The point for $\sqrt{9}$ is exactly the tick at $3$ .

An irrational number such as $\sqrt{2}$ can never do that. It is not equal to any rational number, so its point is not on any of these ticks. When its point is on the part of the line you can see, it is in a gap between two neighboring ticks, which are ticks one step apart. The tick just left of it is a lower bound, a value less than the number. The tick just right of it is an upper bound, a value greater than the number. To locate an irrational number, you find these two bounds. You can use the same squaring you used when you approximated square roots.

Key concept

Trap the number between two ticks

An irrational number has exactly one point on the number line. When every tick is a rational number, that point is never on a tick. So you locate it by naming the two neighboring ticks it lies between.

For a square root, let $n \ge 0$ be the number under the root sign, and let $a$ and $b$ be tick values with $0 \le a < b$ . If $a^2 < n < b^2$ , then $a < \sqrt{n} < b$ . This works because, among nonnegative numbers, the larger number always has the larger square, so the order of the squares matches the order of the numbers.

The ticks $a$ and $b$ are rational bounds. A tick can approximate the irrational number, and you show that with the symbol $\approx$ (read is approximately equal to). A rational tick is never equal to an irrational number.

0 \le a < b \text{ and } a^2 < n < b^2 \;\Longrightarrow\; a < \sqrt{n} < b

Worked example

Locate the square root of 2 with tenths ticks

A number line runs from $1$ to $2$ with a step of $0.1$ between ticks. Between which two neighboring ticks does $\sqrt{2}$ lie, and which of those two ticks is it nearer to?

Check that the point is on this line. The ends of the line, 1 and 2, are nonnegative, and their squares are 1 and 4. The number 2 is between those squares. Among nonnegative numbers, the larger number has the larger square, so the square root of 2 is between 1 and 2.
$1^2 = 1 < 2 < 4 = 2^2 \;\Rightarrow\; 1 < \sqrt{2} < 2$
Test the tenths ticks by squaring them, as you did when you approximated square roots. The square of 1.3 is 1.69 and the square of 1.4 is 1.96, both less than 2. The square of 1.5 is 2.25, the first square greater than 2. The ticks 1.4 and 1.5 are nonnegative, so the square root of 2 is between them.
$1.4^2 = 1.96 < 2 < 2.25 = 1.5^2 \;\Rightarrow\; 1.4 < \sqrt{2} < 1.5$
Find the midpoint of the gap, the value halfway between the two ticks. It splits the gap into a left half, nearer to 1.4, and a right half, nearer to 1.5. To find it, add the two ticks and divide by 2.
$\frac{1.4 + 1.5}{2} = \frac{2.9}{2} = 1.45$
Square the midpoint and compare the result with 2. The square of 1.45 is 2.1025, which is greater than 2. Both 1.45 and the square root of 2 are nonnegative, so the one with the larger square is the larger number. The square root of 2 is less than 1.45, so its point is in the left half of the gap.
$1.45^2 = 2.1025 > 2 \;\Rightarrow\; \sqrt{2} < 1.45$
Read the number line below. Its two markers sit on the ticks 1.4 and 1.5, the bounds from step 2. The square root of 2 has no marker of its own: its point is in the gap between the markers, in the half nearer to 1.4. The squares you computed are the evidence for this, not the look of the picture.
$\sqrt{2} \approx 1.4$

Answer

$\sqrt{2}$ lies between the ticks at $1.4$ and $1.5$ and is nearer to $1.4$ , so $\sqrt{2} \approx 1.4$ to the nearest tenth.

The markers at

1.4

and

1.5

are rational bounds.

\sqrt{2}

is in the gap between them, in the half nearer to

1.4

.

Worked example

Zoom in to hundredths

Zoom in on the gap from $1.4$ to $1.5$ and cut it into steps of $0.01$ . Between which two neighboring ticks does $\sqrt{2}$ lie now, and which of those two ticks is it nearer to?

Square the hundredths ticks in order, starting just after 1.4, until a square is greater than 2. The square of 1.41 is 1.9881, which is less than 2. The square of 1.42 is 2.0164, which is greater than 2. Both ticks are nonnegative, so the square root of 2 is between them.
$1.41^2 = 1.9881 < 2 < 2.0164 = 1.42^2 \;\Rightarrow\; 1.41 < \sqrt{2} < 1.42$
Use the midpoint test. Add the two ticks and divide by 2 to get the midpoint, 1.415. Its square is 2.002225, which is greater than 2. Both 1.415 and the square root of 2 are nonnegative, so the larger square belongs to the larger number: the square root of 2 is less than 1.415 and nearer to 1.41.
$\frac{1.41 + 1.42}{2} = \frac{2.83}{2} = 1.415, \qquad 1.415^2 = 2.002225 > 2 \;\Rightarrow\; \sqrt{2} < 1.415$
Read the number line below. It shows only the part of the first line from 1.4 to 1.5. The markers sit on the new bounds, 1.41 and 1.42. This gap is 0.01 wide, one tenth as wide as the gap in the first example, but the point for the square root of 2 is still inside a gap, not on a tick.
$\sqrt{2} \approx 1.41$

Answer

$\sqrt{2}$ lies between the ticks at $1.41$ and $1.42$ and is nearer to $1.41$ , so $\sqrt{2} \approx 1.41$ to the nearest hundredth. On the first number line, that puts the point just to the right of the tick at $1.4$ .

This window shows only the part of the line from

1.4

to

1.5

, in steps of

0.01

.

\sqrt{2}

is in the gap between the markers at

1.41

and

1.42

.

Watch out: a marker is not the number. It is tempting to write $\sqrt{2} = 1.41$ and say that the marker at $1.41$ is the point for $\sqrt{2}$ . On the first number line you could not even see a difference, because the two points are less than $0.01$ apart. But looks are not evidence. Test the claim by squaring: $1.41^2 = 1.9881$ , and $1.9881$ is not $2$ , so $1.41$ is not $\sqrt{2}$ . Zooming in again cannot fix this. Every new tick is still a terminating decimal, which is a rational number, and the irrational number $\sqrt{2}$ cannot equal a rational number. The marker at $1.41$ is only a lower bound. Write $\sqrt{2} \approx 1.41$ , not $\sqrt{2} = 1.41$ , and say that the point lies between the ticks at $1.41$ and $1.42$ .

Worked example

Locate a negative irrational number

A number line runs from $-3$ to $3$ with a step of $0.5$ between ticks. Between which two neighboring ticks does the irrational number $-\sqrt{6}$ lie, and which of those two ticks is it nearer to?

Say what the number means. Two numbers have a square of 6: the square root of 6, which is positive, and its opposite, which is negative. The opposite of a number is the same distance from 0 on the other side. Find the bounds of the positive one first, because the squaring rule needs nonnegative ticks.
$(\sqrt{6})^2 = 6 \quad \text{and} \quad (-\sqrt{6})^2 = 6$
Bound the square root of 6. The ticks 2 and 2.5 are nonnegative. The square of 2 is 4 and the square of 2.5 is 6.25, and 6 is between those squares. So the square root of 6 is between 2 and 2.5.
$2^2 = 4 < 6 < 6.25 = 2.5^2 \;\Rightarrow\; 2 < \sqrt{6} < 2.5$
Use the midpoint test. Add the two ticks and divide by 2 to get the midpoint, 2.25. Its square is 5.0625, which is less than 6. Both 2.25 and the square root of 6 are nonnegative, so the larger square belongs to the larger number: the square root of 6 is greater than 2.25 and nearer to 2.5.
$\frac{2 + 2.5}{2} = \frac{4.5}{2} = 2.25, \qquad 2.25^2 = 5.0625 < 6 \;\Rightarrow\; \sqrt{6} > 2.25$
Take opposites to cross to the negative side. Taking the opposite of every number flips each point to the other side of 0 at the same distance. Left and right swap, so each inequality sign reverses. Distances between points stay the same, so the nearer tick is the mirror image of 2.5, which is -2.5. The lower bound is now the tick farther from 0.
$2 < \sqrt{6} < 2.5 \;\Rightarrow\; -2.5 < -\sqrt{6} < -2$
Read the number line below. The markers at 2 and 2.5 are the bounds of the square root of 6. The markers at -2.5 and -2 are their mirror images across 0, and they are the bounds of its opposite. The ticks at -2 and 2 are the exact points of those integers, but for the irrational numbers they are only bounds.

Answer

$-\sqrt{6}$ lies between the ticks at $-2.5$ and $-2$ and is nearer to $-2.5$ . This location is approximate: no tick on the line is exactly $-\sqrt{6}$ .

The markers at

2

and

2.5

bound

\sqrt{6}

. Their mirror images at

-2.5

and

-2

bound

-\sqrt{6}

, which is nearer to

-2.5

.

The big idea: an irrational number has one exact point on the number line, but on a line with rational ticks that point always falls in a gap, never on a tick. To locate it, trap it between two neighboring ticks. Square nonnegative tick values to find a lower bound and an upper bound, then use the midpoint test to see which tick is nearer. For a negative number such as $-\sqrt{6}$ , find the bounds of its positive opposite first, then reflect them across $0$ . Finer ticks make the gap narrower, but a tick still only approximates the number, so write $\approx$ and name the bounds.

Next, you will use these locations to compare and order rational and irrational numbers and to estimate the value of expressions with irrational numbers.