# Square root of a Prime (5) is Irrational (Proof + Questions)

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

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

(Squaring both sides)

5y2 = x2 or

x2 = 5y2 Equation 1

Statement B

x2 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:

52a2 = 5y2 or

5y2 = 52a2

y2 = 5a2

y2 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 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 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

1. Square Root of Prime (): √2, √3, √5, √7, √11, √13, √17, √19 …
2. Special Numbers: Pi ( π ) , Euler’s number ( e ), Golden Ratio
3. Logarithms of primes with prime base: log23, log35

## x2 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 x2 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 x2 is multiple of prime number p. We do a prime factorization of x:

Statement D

x = p1.p2.p3…pN

p1, p2, …pN are prime numbers.

Squaring x we get:

x2 = (p1.p2.p3…pN).(p1.p2.p3…pN)

x2 = p12.p22.p32…pN2

We know that x2 is multiple of p, so p must be among p1, p2…pN

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.

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

## Practice Questions

### 1 Question

Prove that √3 is irrational.

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

### 2 Question

is rational or irrational?

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

### 3 Question

Prove that √7 is irrational.

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.

Can you use the Pythagoras Theorem (c2 = a2 + b2) to find the hypotenuse?

### 5 Question

is rational or irrational?

Can you express 63 = 9 × 7 and simplify?

Let us assume √3 is a rational number.

where is in its lowest form.

(Squaring both sides)

3y2 = x2 or

x2 = 3y2 Equation 1

From Equation 1, we can say that x2 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:

32a2 = 3y2 or

3y2 = 32a2

y2 = 3a2

y2 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 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 is in its lowest form (such x, y do not exist). Therefore √3 is not rational. It is an irrational number.

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

Let us assume √7 is a rational number.

where is in its lowest form.

(Squaring both sides)

7y2 = x2 or

x2 = 7y2 Equation 1

From Equation 1, we can say that x2 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:

72a2 = 7y2 or

7y2 = 72a2

y2 = 7a2

y2 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 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 is in its lowest form. So √7 is an irrational number.

Using the Pythagoras Theorem, we get:

hypotenuse2 = 22 + 32

hypotenuse2 = 4 + 9 = 13

units.

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

The length of the hypotenuse is irrational.