This proof works for any prime number: 2, 3, 5, 7, 11, etc. Let’s prove for 5.

First, we will assume that the square root of 5 is a rational number. Next, we will show that our assumption leads to a contradiction.

Let us assume √5 is a rational number.

**Statement A**

$5 =yx y=0$

such that $yx $ is in its lowest form (there is no common factor of x and y).

$5 =yx $

$5=y_{2}x_{2} $ (Squaring both sides)

5y^{2} = x^{2} or

x^{2} = 5y^{2} **Equation 1**

**Statement B**

x^{2} is a multiple of 5. Therefore x is a multiple of 5 (as 5 is a prime number).

Assume Statement B to be true for now. We will prove it later. Using Statement B, we can write:

x = 5a (for some integer a)

Replacing x by 5a in Equation 1:

5^{2}a^{2} = 5y^{2} or

5y^{2} = 5^{2}a^{2}

y^{2} = 5a^{2}

y^{2} is a multiple of 5. So y is a multiple of 5.

We can write y = 5b, where b is some integer.

Using x = 5a and y = 5b in **Statement A:**

**Statement C**

$5 =yx =5b5a $

5 is a common factor of x and y.

In statement A we assumed there is no common factor of x and y. Statement C contradicts it. This means we cannot find integers x, y such that $yx $ is in the lowest form or such x, y do not exist. Therefore √5 cannot be represented as a rational number.

You can use the same argument for √2. We can prove this for any prime number.

## ✩ Known Irrationals

- Square Root of Prime ($Prime $): √2, √3, √5, √7, √11, √13, √17, √19 …
- Special Numbers: Pi ( π ) , Euler’s number ( e ), Golden Ratio
- Logarithms of primes with prime base: log
_{2}3, log_{3}5…

## x^{2} is Multiple of Prime p ⇒ x is Multiple of Prime p

Why does this proof work only for prime numbers? Where did we use the fact that p is a prime number? We used it in the Statement B.

In the proof above, we used the fact that if x^{2} is a multiple of a prime number p, then x is also a multiple of the same prime number. Let us see why this holds.

Let us say x^{2} is multiple of prime number p. We do a prime factorization of x:

**Statement D**

x = p_{1}.p_{2}.p_{3}…p_{N}

p_{1}, p_{2}, …p_{N} are prime numbers.

Squaring x we get:

x^{2} = (p_{1}.p_{2}.p_{3}…p_{N}).(p_{1}.p_{2}.p_{3}…p_{N})

x^{2} = p_{1}^{2}.p_{2}^{2}.p_{3}^{2}…p_{N}^{2}

We know that x^{2} is multiple of p, so p must be among p_{1}, p_{2}…p_{N}

So from Statement D above, x is a multiple of p.

## ✩ Irrational Result of Operations

Following operations between **rational and irrational** numbers result in an **irrational number**. Whatever the order of operations, the outcome is always an irrational number.

- Rational + Irrational: [ 3 + √2 ], [ 4 + √7 ], …
- Rational − Irrational: [ 5 – √2 ], [ √3 – 6 ], …
- Rational × Irrational: [ 4 × π = 4π ], [ 6 × √3 = 6√3 ], …
- Rational ÷ Irrational: [ 2 ÷ √2 ], [ π ÷ 2 ], …

## Practice Questions

### 1 Question

Prove that √3 is irrational.

#### Answer

Can you assume √3 is rational and represent it accordingly?

### 2 Question

$12 $ is rational or irrational?

#### Answer

Can you express $12 $ as a product of a rational and an irrational number?

### 3 Question

Prove that √7 is irrational.

#### Answer

Can you assume √7 is rational and represent it accordingly?

### 4 Question

Prove that the length of the hypotenuse of a right triangle with sides 2 units and 3 units is irrational.

#### Answer

Can you use the Pythagoras Theorem (c^{2} = a^{2} + b^{2}) to find the hypotenuse?

### 5 Question

$7 63 $ is rational or irrational?

#### Answer

Can you express 63 = 9 × 7 and simplify?

## Answers

### 1 Answer

Let us assume √3 is a rational number.

$3 =yx y=0$

where $yx $ is in its lowest form.

$3 =yx $

$3=y_{2}x_{2} $ (Squaring both sides)

3y^{2} = x^{2} or

x^{2} = 3y^{2} Equation 1

From Equation 1, we can say that x^{2} is a multiple of 3.

Therefore x is a multiple of 3.

We can now write:

x = 3a (for some integer a)

Replace x by 3a in Equation 1:

3^{2}a^{2} = 3y^{2} or

3y^{2} = 3^{2}a^{2}

y^{2} = 3a^{2}

y^{2} is a multiple of 3. So y is a multiple of 3.

We can write y = 3b, where b is some integer.

Using x = 3a and y = 3b in our original assumption we get:

$3 =yx =3b3a $

3 is a common factor of x and y.

This contradicts our assumption that there is no common factor of x and y. So we cannot find integers x, y such that $yx $ is in its lowest form (such x, y do not exist). Therefore √3 is not rational. It is an irrational number.

### 2 Answer

$12 =4×3 =23 $

$12 $ is product of a rational number (2) and an irrational number (√3), so it is irrational.

### 3 Answer

Let us assume √7 is a rational number.

$7 =yx y=0$

where $yx $ is in its lowest form.

$7 =yx $

$7=y_{2}x_{2} $ (Squaring both sides)

7y^{2} = x^{2} or

x^{2} = 7y^{2} Equation 1

From Equation 1, we can say that x^{2} is a multiple of 7.

Therefore x is a multiple of 7.

We can now write:

x = 7a (for some integer a)

Replace x by 7a in Equation 1:

7^{2}a^{2} = 7y^{2} or

7y^{2} = 7^{2}a^{2}

y^{2} = 7a^{2}

y^{2} is a multiple of 7. So y is a multiple of 7.

We can write y = 7b, where b is some integer.

Using x = 7a and y = 7b in our original assumption we get:

$7 =yx =7b7a $

7 is a common factor of x and y.

This contradicts our assumption that there is no common factor of x and y. Therefore no x, y exist such that $yx $ is in its lowest form. So √7 is an irrational number.

### 4 Answer

Using the Pythagoras Theorem, we get:

hypotenuse^{2} = 2^{2} + 3^{2}

hypotenuse^{2} = 4 + 9 = 13

$hypotenuse=13 $ units.

As 13 is a prime number, its square root is irrational.

∴ The length of the hypotenuse is irrational.

### 5 Answer

$7 63 =7 9×7 $

$=7 37 =3$, a rational number

## Related

**Examples of Irrational Numbers ➤**