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

$5 = \frac{x}{y} y \neq = 0$

such that $\frac{x}{y}$ is in its lowest form (there is no common factor of x and y).

$5 = \frac{x}{y}$

$5 = \frac{x ^{2}}{y ^{2}}$ (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

$5 = \frac{x}{y} = \frac{5 a}{5 b}$

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 $\frac{x}{y}$ 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 ( $P r im e$ ): √2, √3, √5, √7, √11, √13, √17, √19 …
Special Numbers: Pi ( π ) , Euler’s number ( e ), Golden Ratio
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…p_N

p₁, p₂, …p_N are prime numbers.

Squaring x we get:

x² = (p₁.p₂.p₃…p_N).(p₁.p₂.p₃…p_N)

x² = p₁².p₂².p₃²…p_N²

We know that x² is multiple of p, so p must be among p₁, p₂…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 (c² = a² + b²) to find the hypotenuse?

5 Question

$\frac{63}{7}$ is rational or irrational?

Answer

Can you express 63 = 9 × 7 and simplify?

Answers

1 Answer

Let us assume √3 is a rational number.

$3 = \frac{x}{y} y \neq = 0$

where $\frac{x}{y}$ is in its lowest form.

$3 = \frac{x}{y}$

$3 = \frac{x ^{2}}{y ^{2}}$ (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:

$3 = \frac{x}{y} = \frac{3 a}{3 b}$

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 $\frac{x}{y}$ is in its lowest form (such x, y do not exist). Therefore √3 is not rational. It is an irrational number.

2 Answer

$12 = 4 \times 3 = 23$

$12$ 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.

$7 = \frac{x}{y} y \neq = 0$

where $\frac{x}{y}$ is in its lowest form.

$7 = \frac{x}{y}$

$7 = \frac{x ^{2}}{y ^{2}}$ (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:

$7 = \frac{x}{y} = \frac{7 a}{7 b}$

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 $\frac{x}{y}$ is in its lowest form. So √7 is an irrational number.

4 Answer

Using the Pythagoras Theorem, we get:

hypotenuse² = 2² + 3²

hypotenuse² = 4 + 9 = 13

$h y p o t e n u se = 13$ units.

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

∴ The length of the hypotenuse is irrational.

5 Answer

$\frac{63}{7} = \frac{9 \times 7}{7}$

$= \frac{3 7}{7} = 3$ , a rational number

Examples of Irrational Numbers ➤

Product of rational & irrational numbers is irrational ➤

Sum of rational & irrational number is irrational ➤

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

✩ Known Irrationals

x2 is Multiple of Prime p ⇒ x is Multiple of Prime p

✩ Irrational Result of Operations

Practice Questions

1 Question

Answer

2 Question

Answer

3 Question

Answer

4 Question

Answer

5 Question

Answer

Answers

1 Answer

2 Answer

3 Answer

4 Answer

5 Answer

Related

x² is Multiple of Prime p ⇒ x is Multiple of Prime p