# How to Approximate Irrational Numbers? (+FREE Worksheet!)

This article teaches you how to Approximate Irrational Numbers in a few simple steps.

## Step by step guide to Approximate Irrational Numbers

Numbers that cannot be written as a fraction are called irrational. An irrational number is a non-repeating, non-terminating decimal and it does not have an exact place on the number line. Square roots of numbers that are not perfect squares are irrational.

We use approximations of irrational numbers to locate them approximately on a number line diagram.

Since the irrational numbers are radical numbers that are not a perfect square, to approximate them, follow the steps below

• Step 1: First, we need to find the two consecutive perfect squares that the number is between. if  is our number, we can do this by writing this inequality: $$a^2< x <b^2$$
• Step 2: Take the square root of each number:$$\sqrt{a^2}< \sqrt{x} <\sqrt{b^2}$$
• Step 3: Simplify the square roots of perfect squares:$$a< \sqrt{x} <b$$, then, $$\sqrt{x}$$ is between $$a$$ and $$b$$.
• Step 4: To find a better estimate, choose some numbers between $$a$$ and $$b$$.

### Approximating Irrational Numbers – Example 1:

Find the approximation of $$\sqrt{22}$$

Solution:

since $$\sqrt{22}$$ is not a perfect square,  is irrational. To approximate $$\sqrt{22}$$ first, we need to find the two consecutive perfect squares that $$22$$ is between. We can do this by writing this inequality: $$16< 22 <25$$. Now take the square root of each number: $$\sqrt{16}< \sqrt{22} <\sqrt{25}$$. Simplify the square roots of perfect squares:

$$4< \sqrt{22} <5$$, then, $$\sqrt{22}$$ is between $$4$$ and $$5$$. To find a better estimate, choose some numbers between $$4$$ and $$5$$ Let’s choose $$4.6$$, $$4.7$$ and $$4.8$$.

$$4.6^2=21.16$$, $$4.7^2=22.09$$, $$4.8^2=23.04$$, $$4.7$$ is closer to $$22$$. Then: $$\sqrt{22}≈4.7$$

### Approximating Irrational Numbers – Example 2:

Find the approximation of $$\sqrt{74}$$

Solution:

since $$\sqrt{74}$$ is not a perfect square,  is irrational. To approximate $$\sqrt{74}$$ first, we need to find the two consecutive perfect squares that $$74$$ is between. We can do this by writing this inequality: $$64< 74 <81$$. Now take the square root of each number: $$\sqrt{64}< \sqrt{74} <\sqrt{81}$$. Simplify the square roots of perfect squares:

$$8< \sqrt{74} <9$$, then, $$\sqrt{74}$$ is between $$8$$ and $$9$$. To find a better estimate, choose some numbers between $$8$$ and $$9$$ Let’s choose $$8.5$$, $$8.6$$ and $$8.7$$.

$$8.5^2=72.25$$, $$8.6^2=73.96$$, $$8.7^2=75.69$$, $$8.6$$ is closer to $$74$$. Then: $$\sqrt{74}≈8.6$$

### Approximating Irrational Numbers – Example 3:

Find the approximation of $$\sqrt{94}$$

Solution:

since $$\sqrt{94}$$ is not a perfect square,  is irrational. To approximate $$\sqrt{94}$$ first, we need to find the two consecutive perfect squares that $$94$$ is between. We can do this by writing this inequality: $$81< 94 <100$$. Now take the square root of each number: $$\sqrt{81}< \sqrt{94} <\sqrt{100}$$. Simplify the square roots of perfect squares:

$$9< \sqrt{94} <10$$, then, $$\sqrt{94}$$ is between $$9$$ and $$10$$. To find a better estimate, choose some numbers between $$9$$ and $$10$$ Let’s choose $$9.6$$, $$9.7$$ and $$9.8$$.

$$9.6^2=92.16$$, $$9.7^2=94.09$$, $$9.8^2=96.04$$, $$9.7$$ is closer to $$94$$. Then: $$\sqrt{94}≈9.7$$

### Approximating Irrational Numbers – Example 4:

Find the approximation of $$\sqrt{26}$$

Solution:

since $$\sqrt{26}$$ is not a perfect square,  is irrational. To approximate $$\sqrt{26}$$ first, we need to find the two consecutive perfect squares that $$26$$ is between. We can do this by writing this inequality: $$25< 26 <36$$. Now take the square root of each number: $$\sqrt{25}< \sqrt{26} <\sqrt{36}$$. Simplify the square roots of perfect squares:

$$5< \sqrt{26} <6$$, then, $$\sqrt{26}$$ is between $$5$$ and $$6$$. To find a better estimate, choose some numbers between $$5$$ and $$6$$ Let’s choose $$5.1$$, $$5.2$$ and $$5.3$$.

$$5.1^2=26.01$$, $$5.2^2=27.04$$, $$5.3^2=28.09$$, $$5.1$$ is closer to $$26$$. Then: $$\sqrt{26}≈5.1$$

## Exercises for Approximating Irrational Numbers

### Find the approximation of each.

1. $$\color{blue}{\sqrt{41}}$$
2. $$\color{blue}{\sqrt{52}}$$
3. $$\color{blue}{\sqrt{59}}$$
4. $$\color{blue}{\sqrt{72}}$$
5. $$\color{blue}{\sqrt{17}}$$
6. $$\color{blue}{\sqrt{10}}$$
1. $$\color{blue}{6.4}$$
2. $$\color{blue}{7.2}$$
3. $$\color{blue}{7.7}$$
4. $$\color{blue}{8.5}$$
5. $$\color{blue}{4.1}$$
6. $$\color{blue}{3.2}$$

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