Comparing Irrational Numbers and Estimating Expressions | Grade 8 Math | LevelKite

A rational number is a number you can write as a fraction $\frac{p}{q}$ , where $p$ and $q$ are integers and $q \neq 0$ . Its decimal form either terminates (stops) or repeats a block of digits forever. An irrational number is a number on the number line that cannot be written that way. Its decimal form never terminates and never repeats. Together, the rational and irrational numbers are the real numbers.

You already know how to find rational numbers close to an irrational number. In this lesson you put those approximations to work. You will decide which of two numbers is greater, which means farther to the right on the number line. You will put a list of numbers in order. And you will estimate an expression that contains an irrational number. An expression is numbers joined by operations, and to estimate it is to find a rational number close to its value.

An irrational number such as $\sqrt{10}$ or $\pi$ is one exact number with one exact place on the number line. Here $\sqrt{n}$ means the principal square root: for $n \ge 0$ , it is the nonnegative number (zero or positive) whose square is $n$ . The number $\pi$ (pi) is the circumference of any circle divided by its diameter. What you cannot do is write either exact place as a terminating or repeating decimal.

A rational number close to an irrational number is an approximation. You mark an approximation with the symbol $\approx$ , read "is approximately equal to". For example, $\sqrt{10} \approx 3.16$ . On a number line, the point for $\sqrt{10}$ is near the mark for $3.16$ , not on it.

A stronger statement uses two rational bounds: a lower bound that is less than the number and an upper bound that is greater than it. The statement $3.1 < \sqrt{10} < 3.2$ says for certain which stretch of the number line holds $\sqrt{10}$ . Next you will see how to check it.

You have already approximated square roots with rational numbers. Squares are the tool that checks those approximations, and here is why you can trust them. Think of two square tiles. The tile with the longer side has the greater area, and the tile with the greater area has the longer side. So comparing areas tells you how the sides compare. A side length is never negative, and that is the condition to keep in mind: the numbers you compare this way must be nonnegative.

The numbers $3.1$ , $\sqrt{10}$ and $3.2$ are all nonnegative, and the square of $\sqrt{10}$ is exactly $10$ . So compare the three squares:

3.1^2 = 9.61 \;<\; 10 \;<\; 10.24 = 3.2^2 \quad\text{so}\quad 3.1 < \sqrt{10} < 3.2

Key concept

Rational bounds decide the question

You can never write every digit of an irrational number, and you do not need to. Trap the number between two rational bounds, then tighten the bounds until they answer your question: which number is greater, or what an expression is close to.

For square roots, squares do the checking. When two numbers are both nonnegative, the greater number has the greater square, and the greater square belongs to the greater number. So for $n \ge 0$ and a rational number $r \ge 0$ , you can compare $\sqrt{n}$ with $r$ by comparing $n$ with $r^2$ .

\text{If } a \ge 0 \text{ and } b \ge 0, \text{ then } a < b \text{ exactly when } a^2 < b^2.

Tip: check the signs before you square. The squares test needs both numbers to be nonnegative. With negative numbers it can give the wrong order: $-3 < -2$ , yet $(-3)^2 = 9$ is greater than $(-2)^2 = 4$ . To compare two negative numbers such as $-\sqrt{10}$ and $-3.2$ , compare their opposites $\sqrt{10}$ and $3.2$ , then reverse the order, because opposites are mirror images across zero: $\sqrt{10} < 3.2$ , so $-\sqrt{10} > -3.2$ .

Cube roots need no sign check. For any real number $n$ , the cube root $\sqrt[3]{n}$ is the one real number whose cube is $n$ , and it has the same sign as $n$ . Cubing keeps the order of any two real numbers, so you can compare cubes directly: $2.7^3 = 7.29 \times 2.7 = 19.683$ , which is less than $20$ , so $2.7 < \sqrt[3]{20}$ .

Worked example

Ordering four numbers

Write $\sqrt{10}$ , $\pi$ , $3.2$ and $\frac{22}{7}$ in order from least to greatest.

Compare $\sqrt{10}$ with $3.2$ . Both are nonnegative, so the greater square belongs to the greater number. The square of $\sqrt{10}$ is $10$ , and $3.2^2 = 3.2 \times 3.2 = 10.24$ , which is greater than $10$ .
$\left(\sqrt{10}\right)^2 = 10 < 10.24 = 3.2^2 \quad\text{so}\quad \sqrt{10} < 3.2$
Compare $\sqrt{10}$ with $\frac{22}{7}$ the same way, because both are nonnegative. Square the fraction: $22^2 = 484$ and $7^2 = 49$ . Then write $10$ with the denominator $49$ , as $\frac{490}{49}$ . With the same positive denominator, the smaller numerator gives the smaller fraction.
$\left(\frac{22}{7}\right)^2 = \frac{484}{49} < \frac{490}{49} = 10 \quad\text{so}\quad \frac{22}{7} < \sqrt{10}$
Compare $\pi$ with $\frac{22}{7}$ using decimal forms, because squaring $\pi$ does not give a number you know exactly. The decimal form of $\pi$ begins $3.14159$ , so $\pi < 3.142$ . Dividing $22$ by $7$ gives $3.\overline{142857}$ , where the bar marks the digits that repeat, so $\frac{22}{7} > 3.142$ . The rational number $3.142$ separates the two.
$\pi < 3.142 < \frac{22}{7}$
Join the three results into one chain. Each number is less than the next, so this is also their left-to-right order on a number line. The rational number $\frac{22}{7}$ lies between the two irrational numbers, and that is what tells $\pi$ and $\sqrt{10}$ apart.
$\pi < \frac{22}{7} < \sqrt{10} < 3.2$

Answer

From least to greatest, the order is $\pi$ , $\frac{22}{7}$ , $\sqrt{10}$ , $3.2$ .

A tempting mistake. Round each number to two decimal places, see $\pi \approx 3.14$ and $\frac{22}{7} \approx 3.14$ , and decide that $\pi = \frac{22}{7}$ . This fails because rounding throws away the very digits where the numbers differ: the decimal form of $\pi$ begins $3.1415$ , while $\frac{22}{7} = 3.\overline{142857}$ begins $3.1428$ . There is a deeper reason too. A rational number has a decimal form that terminates or repeats, and an irrational number has one that does neither, so a rational number and an irrational number are never equal. When rounded values match, the comparison is not finished. Tighten the bounds until a rational number separates the two, as $3.142$ does here.

Now turn to expressions, such as $\sqrt{10} + \pi$ or $\pi^2$ . When an expression contains an irrational number, its value often has no terminating or repeating decimal either. So you report an estimate, marked with $\approx$ , and it helps to know how far off the estimate can be.

Bounds tell you that. Replace each irrational number with its lower and upper bounds, then carry both bounds through the operation. An operation that keeps order turns lower bounds into a lower bound for the result. Some operations reverse order, so check each one.

The table uses $3.1 < \sqrt{10} < 3.2$ and $3.14 < \pi < 3.15$ . The bounds for $\pi$ come from its decimal form, which begins $3.14159$ . When both bounds of an expression round to the same number, every value between them rounds to that number too.

Carrying bounds through an operation

Expression	Carry the bounds through	Why the order holds	Nearest whole number
$\sqrt{10} + \pi$	$3.1 + 3.14 < \sqrt{10} + \pi < 3.2 + 3.15$ , so $6.24 < \sqrt{10} + \pi < 6.35$	Adding smaller numbers gives a smaller sum. This holds for all real numbers.	$6$
$2\sqrt{10}$ , which means $2 \times \sqrt{10}$	$2 \times 3.1 < 2\sqrt{10} < 2 \times 3.2$ , so $6.2 < 2\sqrt{10} < 6.4$	Multiplying by a positive number keeps the order. A negative multiplier would reverse it.	$6$
$10 - \sqrt{10}$	$10 - 3.2 < 10 - \sqrt{10} < 10 - 3.1$ , so $6.8 < 10 - \sqrt{10} < 6.9$	Taking away more leaves less, so the upper bound $3.2$ gives the lower bound $6.8$ .	$7$

Worked example

Estimating the square of pi

Estimate $\pi^2$ to the nearest tenth.

Start with rational bounds for $\pi$ . Its decimal form begins $3.14159$ , so $\pi$ is greater than $3.14$ and less than $3.15$ .
$3.14 < \pi < 3.15$
Square all three numbers. They are all nonnegative, so their squares stay in the same order.
$3.14^2 < \pi^2 < 3.15^2$
Work out each square by splitting one factor into two parts. Since $3.14 = 3 + 0.14$ , add $3.14 \times 3 = 9.42$ and $3.14 \times 0.14 = 0.4396$ . Since $3.15 = 3 + 0.15$ , add $3.15 \times 3 = 9.45$ and $3.15 \times 0.15 = 0.4725$ .
$3.14^2 = 9.42 + 0.4396 = 9.8596 \qquad 3.15^2 = 9.45 + 0.4725 = 9.9225$
Round at the end. A number rounds to $9.9$ when it is at least $9.85$ and less than $9.95$ . Both bounds are in that stretch, so every number between them rounds to $9.9$ , and $\pi^2$ is one of those numbers.
$9.8596 < \pi^2 < 9.9225 \quad\text{so}\quad \pi^2 \approx 9.9$

Answer

To the nearest tenth, $\pi^2 \approx 9.9$ . Its exact value lies between $9.8596$ and $9.9225$ .

Worked example

Will two square rugs fit along a wall?

Two square rugs are laid side by side, each with one edge against a wall that is $8$ feet long. One rug has an area of $30$ square feet, and the other has an area of $6$ square feet. Do the rugs fit along the wall? About how many feet of wall do they cover?

Write the expression. Let $s$ be the side length, in feet, of the larger rug. Its area in square feet is $s^2$ , so $s^2 = 30$ . Two numbers have the square $30$ , namely $\sqrt{30}$ and $-\sqrt{30}$ , but a length is not negative, so $s = \sqrt{30}$ . In the same way, the smaller rug has a side of $\sqrt{6}$ feet. Side by side, the rugs cover the sum of the two sides.
$\left(\sqrt{30} + \sqrt{6}\right) \text{ feet}$
Trap each side between two numbers that are one tenth apart by comparing squares, which is valid because all the numbers are nonnegative. For $\sqrt{30}$ , test $5.4$ and $5.5$ : $5.4^2 = 29.16$ and $5.5^2 = 30.25$ , and $30$ lies between these squares.
$29.16 < 30 < 30.25 \quad\text{so}\quad 5.4 < \sqrt{30} < 5.5$
Do the same for $\sqrt{6}$ with $2.4$ and $2.5$ : $2.4^2 = 5.76$ and $2.5^2 = 6.25$ , and $6$ lies between these squares.
$5.76 < 6 < 6.25 \quad\text{so}\quad 2.4 < \sqrt{6} < 2.5$
Add the two lower bounds and add the two upper bounds. Adding smaller numbers gives a smaller sum, so the total lies between the two results.
$5.4 + 2.4 < \sqrt{30} + \sqrt{6} < 5.5 + 2.5 \quad\text{so}\quad 7.8 < \sqrt{30} + \sqrt{6} < 8$
Answer both questions from the bounds. The total is less than $8$ , so the rugs fit. The total is also more than $7.8$ , so it lies past the halfway point $7.5$ between $7$ and $8$ , and it rounds to $8$ .
$\sqrt{30} + \sqrt{6} < 8 \quad\text{and}\quad \sqrt{30} + \sqrt{6} \approx 8$

Answer

Yes, the rugs fit. Together they cover between $7.8$ and $8$ feet of wall, which is about $8$ feet but a little less.

Takeaway. An irrational number is exact, but any decimal you write out for it is only an approximation, so mark it with $\approx$ . To compare or order numbers, trap each irrational number between rational bounds and tighten the bounds until they separate the numbers you are comparing. For nonnegative numbers, comparing squares does this quickly. To estimate an expression, carry the lower and upper bounds through each operation, check what the operation does to their order, and round only at the end.