Question

Prove that 3+25 Irrational
​

Answer 1

Question :-

Prove that 3 + 2√5 is an irrational number ?

Answer :-

Required to prove :-

• 3 + 2√5 is an irrational number

Method used :-

$\huge{\dagger}$$\huge{\green{\tt{Contradictory \; method}}}$$\huge{\red{\bigstar}}$

Conditions used :-

• p, q are integers

• q ≠ 0

• p , q are co - primes

• An irrational number is not equal to a rational number

Proof :-

3 + 2√5

Let's assume on the contradictory that 3 + 2√5 is a rational number .

So, equal this value with p/q

( where p and q are integers , q ≠ 0 and p,q are co - primes )

Hence,

$\longrightarrow{\tt{ 3 + 2 \sqrt{5} = \dfrac{ p }{q }}}$

Transpose 3 to the right side

$\longrightarrow{\tt{ 2 \sqrt{5} = \dfrac{p}{q} - 3 }}$

By taking LCM we get,

$\longrightarrow{\tt{ 2 \sqrt{5} = \dfrac{p}{q} - \dfrac{3}{1}}}$

$\longrightarrow{\tt{ 2 \sqrt{5} = \dfrac{p \times 1 }{q \times 1 } - \dfrac{ 3 \times q }{ 1 \times q }}}$

$\longrightarrow{\tt{ 2 \sqrt{5} = \dfrac{ p - 3q }{q}}}$

Now transpose the 2 to the right leaving √5

So,

$\longrightarrow{\tt{ \sqrt{5} = \dfrac{ \dfrac{p - 3q }{q} }{2}}}$

$\longrightarrow{\tt{ \sqrt{5} = \dfrac{ p - 3q }{q } \div \dfrac{2}{1} }}$

$\longrightarrow{\tt{ \sqrt{5} = \dfrac{p - 3q }{q} \times \dfrac{1}{2}}}$

$\longrightarrow{\tt{ \sqrt{5} = \dfrac{p - 3q }{2q}}}$

Here,

$\huge{\dagger}$$\large{\boxed{\rm{ \dfrac{ p - 3q }{2q } \; is \; a \; rational \; number }}}$

Similarly,

We know that √5 is an irrational number

But since it is not mentioned here we can't directly write it as irrational number .

So, let's prove that √5 is an irrational number

Let's assume on the contradictory that √5 is a rational number .

Equal √5 with a/b

( where a,b are integers , b ≠ 0 , p and q are co - primes )

$\tt{ \sqrt{5} = \dfrac{a}{b}}$

By cross multiplication

√5b = a

By squaring on both sides

( √5b )² = ( a )²

5b² = a²

Recall the fundamental theorem of arithmetic ,

According to which

If a divides q²

a divides q ( also )

So,

5 divides a²

5 divides a ( also )

Similarly,

Let's take a = 5k

( where k is any positive integer )

Hence,

√5b = a

√5b = 5k

squaring on both sides

( √5b )² = ( 5k )²

5b² = 25k²

$\rm{ b^2 = \dfrac{ 25 k^2 }{5}}$

b² = 5k²

Interchanging the terms

5k² = b²

So,

5 divides b²

5 divides b ( also )

From the above we can conclude that

a,b have common factor as 5

But according to rational numbers properties a,b should have common factor as 1 because they are co - primes .

This contradicton is due to the wrong assumption that √5 is a rational number

So, our assumption is wrong .

Hence, √5 is an irrational number .

So,

$\huge{\dagger}$$\large{\boxed{\rm{ \sqrt{5} \; is \; an \; irrational \; number }}}$

But we know that

An irrational number is not equal to a rational

number .

This implies,

$\huge{\leadsto{\boxed{\tt{ \sqrt{5} \neq \dfrac{ p - 3q }{2q }}}}}$

And this contradicton is due to the wrong assumption that 3 + 2√5 is a rational number

Our assumption is wrong

Hence, 3 + 2√5 is an irrational number

Hence proved .