What are Irrational Numbers in Math? (Definition & Examples)

About Irrational Numbers

Definition of Irrational Numbers

The set of real numbers that cannot be written in the form of $\frac{p}{q}$, where p and q are integers, is known as irrational numbers. The decimal expansion of an irrational number is neither terminating nor repeating.

Classify Real Numbers as Rational and Irrational Numbers

A rational number is any number that can be expressed as a fraction $\left(\frac{p}{q}\right)$ or as a ratio. The numerator (p) and denominator (q), when q is not zero, are included to form a rational number. A whole number or an integer are also part of the set of rational numbers.

Rational numbers	Irrational numbers
It can be stated as a fraction or ratio, such as $\frac{p}{q}$, where q $\neq$ 0, and p and q are integers.	It cannot be stated as a fraction or as a ratio.
The decimal expansion is either terminating or non-terminating recurring (repeating)	At any point in time, the decimal expansion is non-terminating and non-recurring.
Example: 0.33333, 0.656565.., 1.75, $\frac{9}{5}$	Example: π, $\sqrt{13}$ , e

$irrational1$

Irrational numbers include the square roots of numbers that are not perfect squares. They can’t be represented as the product of two numbers because of this. The decimal form of an irrational number will neither terminate or repeat.

How do we determine the Value of Square roots or Cube roots?

To determine the value of a number within a square root or cube root, knowing the squared and cubed values of numbers from one to ten is necessary. For example, if we need to find the value of $\sqrt{13}$, We know that,

$\sqrt{16}$ = 4

$\sqrt{9}$ = 3

So, $\sqrt{13}$ is between 3 and 4. The value is closer to four than to three. The same logic can be applied to find the cube roots as well.

$root13_num_line$

Solved Irrational Numbers Examples

Example 1:

Classify the given real numbers as irrational, rational, or integer.

Solution:

$\pi$, $-\overline{0.29},-\sqrt[3]{17},\sqrt{16},\sqrt{82}$

$\pi$ : It is an irrational number as the decimal form of the number never ends or repeats.

$-\overline{0.29}$ : It is a rational number as it is a recurring non-terminating decimal.

$\sqrt[3]{17}$ : This is an irrational number as 17 is not a perfect cube.

$\sqrt{16}$ : This is a rational number as 16 = 4, which is also an integer.

$\sqrt{82}$ : It is not a perfect square, hence it is an irrational number.

Example 2:

Find the length $“x”$ of the diagonal of the square whose side is 4 units. In between which two whole numbers would the value of $“x”$ lie?

Solution:

$irrational3$

Using Pythagoras theorem we can say that:

$x=\sqrt{(4)^2+(4)^2}$

$x=(16)+(16)$

$x=32$

Now, we know that:

5 = $\sqrt{25}$ and 6 = $\sqrt{36}$

Hence, $\sqrt{32}$ would lie between 5 and 6.

Example 3:

Find the length $“x”$ of the diagonal of the rectangle whose length is 9 units, and width is 3 units. In between which two whole numbers would the value of $“x”$ lie?

Solution:

$irrational4$

Using Pythagoras theorem we can say that:

$x =\sqrt{(9)^2+(3)^2}$

$x=81+9$

$x=90$

Now, we know that:

10 = $\sqrt{100}$

9 = $\sqrt{81}$

Hence, $\sqrt{90}$ is between 9 and 10.

Example 4:

Which among the two is greater in value,$\sqrt{40}$ or $\sqrt[3]{80}$?

Solution:

Now, we know that,

$\sqrt{36}$ = 6 and $\sqrt{49}$ = 7

This means that 40 lies between 6 and 7.

Now we also know that,

$\sqrt[3]{64}$ = 4 and $\sqrt[3]{125}$ = 5

This means that $\sqrt[3]{80}$ lies between 4 and 5.

So very clearly $\sqrt{40}$ > $\sqrt[3]{80}$

$cube_root_80$

Example 5:

Jessica just learnt about irrational numbers. She wrote down an equation a² = 0.81 b and wanted to check whether irrational numbers could result from this equation. She wants to substitute various values for b and note down the corresponding values for a. She starts off by substituting b = 9, is the corresponding value of a an irrational number?

Solution:

a² = 0.81 b

Since,

b = 9

a² = 0.81 $\times$ 9

= 7.29

Now to find out the value of a we need to take the square root of both sides.

So, if a² = 7.29

Then,

a = $\sqrt{7.29}$

= $\sqrt{\frac{7.29}{100}}$

= $\frac{27}{10}$

a = $\frac{27}{10}$, from the definition of irrational numbers we know that, a number is irrational if it cannot be written in the form of $\frac{p}{q}$, where p and q are integers.

Here $\frac{27}{10}$ is of the form $\frac{p}{q}$where both 27 and 10 are integers. Hence, $\frac{27}{10}$ is a rational number. This proves that Jennifer needs to use other numbers as values for b to find an irrational value for a!