Question

prove that
$\sqrt{?} 7$
is an irrational number ​

Answer 1

Question :-

Prove that √7 is irrational number

Answer :-

Required to prove :-

• √7 is an irrational number

Method used :-

• Contradictory Method

Concept used :-

➜ p , q are integers

➜ q ≠ 0

➜ p and q are co - primes

Proof :-

Let's assume on the contradictory that √7 is an irrational number

Equal √7 with p/q

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

This implies ;

$\leadsto{\rm{ \sqrt{7} = \dfrac{p}{q} }}$

Cross multiplication

$\leadsto{\rm{ \sqrt{7}q = p }}$

Squaring on both sides

$\leadsto{\rm{ ( \sqrt{7} q {)}^{2} = ( p {)}^{2} }}$

$\leadsto{\rm{ 7{q}^{2} = {p}^{2} }}$

Recall the Fundamental theorem of arithmetic

According to which ;

>> If a divides q²

>> a divides q also

Here,

➜ 7 divides p²

➜ 7 divides p also

Now,

Let's consider the value of p as 7k

( where k is any positive integer )

So,

$\leadsto{\tt{ \sqrt{7}q = 7k }}$

By squaring on both sides

$\leadsto{\tt{ ( \sqrt{7}q {)}^{2} = ( 7k {)}^{2} }}$

$\leadsto{\tt{ 7{q}^{2} = 49{k}^{2} }}$

$\leadsto{\tt{ {q}^{2} = \dfrac{ 49 {k}^{2} }{ 7 } }}$

$\leadsto{\tt{ {q}^{2 } = 7 {k}^{2} }}$

$\implies{\tt{ 7 {k}^{2} = {q}^{2} }}$

Here,

➜ 7 divides q²

➜ 7 divides q also

From the above we can conclude that ;

p and q have common factor as 7

But,

According to the condition;

p , q should have common factor as 1 . Since, p and q are co - primes .

So,

This contradicton is due to the wrong assumption that ;

√7 is an irrational number

So, Our assumption is wrong

Therefore,

$\large{\underline{\underline{\rm{ \sqrt{7} \ is \ an \ irrational \ number }}}}$

$\large{\text{\pink{Hence proved}}}$ $\huge\checkmark$