Question

prove that 1+1√2 is irrational with contradictory method​

Answer 1

Question :-

prove that 1 + 1√2 is irrational number with contradictory method .

Answer :-

Given :-

1 + 1√2

Required to prove :-

1 + 1√2 is an irrational number ?

Condition mentioned :-

By contradictory method

Conditions used :-

Here conditions refer to the properties ;

• p, q are integers
• q ≠ 0 ( q is not equal to zero )
• p, q are co - primes

Similarly ,

If a divides q²

a divides q also .

At Last,

An irrational number is not equal to a rational number

number Q' ≠ Q

Here, Q' ( Q dash ) represents irrational numbers .

Here, Q' ( Q dash ) represents irrational numbers .Q represents rational numbers .

Proof :-

Given :-

1 + 1√2

Let's assume on the contradictory that 1 + 1√2 is a rational number .

So, equal the number with p/q .

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

$\longrightarrow{\tt{ 1 + 1 \sqrt{2} = \dfrac{p}{q}}}$

Transpose +1 to the right side

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

By taking LCM on the right side

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

So, here we can take 1√2 as √2

Since, any number divided or multiplied with 1 will give back the same number .

Hence,

$\boxed{\tt{ \dfrac{p-q}{q} \; is \; a\; rational \; number}}$

$\Rightarrow{\tt{ \sqrt{2} = \dfrac{p - q}{q}}}$

We know that

√2 is an irrational number .

But in the question it is not mentioned so we have to prove that √2 is an irrational number .

$\rule{250}{6}$

So,

Lets assume on the contradictory that √2 is a rational number

Which is equal to a by b .

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

So,

$\mathrm{ \sqrt{2} = \dfrac{a}{b}}$

By cross multiplication we get,

√2b = a

Squaring on both sides

( √2b )² = ( a )²

2b² = a²

Now recall the fundamental theorem of arithmetic

According to which ,

If a divides q²

Then, a divides q also .

So,

Here,

2b² = a²

2 divides a²

So, 2 divides a also .

However,

Let take the value of a as 2k

Substitute this in the above one .

So,

√2b = a

√2b = 2k

Squaring on both sides

( √2b)² = ( 2k )²

2b² = 4k²

$\rm{ b^2 = \dfrac{4k^2}{2}}$

$\implies{\rm{ b^2 = 2k^2 }}$

$\implies{\rm{ 2k^2 = b^2 }}$

Here,

2 divides b²

So, 2 divides b also .

From the above we can conclude that ,

a, b have common factor 2 .

But according to the condition ,

a, b are co-primes which means a, b should have 1 as the common factor .

So,

This contradiction is due to the wrong assumption that √2 is a rational number .

So, our assumption is wrong .

Hence,

√2 is an irrational number

$\rule{250}{6}$

From the above we came to know that √2 is an irrational number .

Hence,

$\huge{\boxed{\sf{ \sqrt{2} \neq \dfrac{ p - q}{q}}}}$

This is because ,

An irrational number is not equal to a rational number

so,

This contradiction is due to the wrong assumption that 1 + 1√2 is a rational number .

Our assumption is wrong .

Hence

1 + 1√2 is an irrational

number .

Hence proved .