Question

show that 2+√2 is not a rational number.
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Answer 1

Required to prove :-

• 2 + √2 is not a rational number

Method used :-

• Contradictory method

Conditions used :-

• p , q are integers

• q ≠ 0

• p and q are co - primes

• An irrational number is not equal to a rational number

Solution :-

We need to prove that 2 + √2 is not a rational number .

So,

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

Equal 2 + √2 with p/q

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

So,

$\rm 2 + \sqrt{2} = \dfrac{p}{q}$

Now,

Transpose 2 to the right side

$\rm \sqrt{2} = \dfrac{p}{q} - 2$

Taking LCM we get ,

$\rm \sqrt{2} = \dfrac{p}{q} - \dfrac{2}{1} \times \dfrac{q}{q}$

$\rm \sqrt{2} = \dfrac{p - 2q}{q}$

Here,

$\large \boxed{\text{ \dfrac{p - 2q}{q} \: is \: a \: rational \: number}}}$

But,

we know that √2 is an irrational number but since, it is not mentioned we need to prove that √2 is an irrational number .

So,

__________________________________________________

Lets assume that √2 is a rational number

So,

equal √2 with a/b

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

Hence,

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

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

a divides q ( also )

So,

This implies ;

2 divides a²

2 divides a ( also )

Similarly ,

Let the value of a = 2k

where k is any positive integer

So,

√2b = 2k

Squaring on both sides

( √2b )² = ( 2k )²

2b² = 4k²

b² = 4k²/2

b² = 2k²

This implies ,

2k² = b²

Hence,

2 divides b²

2 divides b ( also )

From the above we can conclude that ;

2 is the common factor of both a & b

But,

According to the properties of rational numbers ;

where p , q are co - primes which means they should have common factor as 1

Hence,

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

Our assumption is wrong

So,

√2 is an irrational number

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From the above it is clear that ;

√2 is an irrational number

However,

We know that ;

Irrational number is not equal to a rational number

Hence,

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

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

So, our assumption is wrong .

Hence,

2 + √2 is not a rational number

Answer 2

Answer:

2 + √2 is an irrational no.

Step-by-step explanation:

Let 2 + √2 be a rational no. r

then,

2 + √2 = r

=> √2 = r - 2

Since,

√2 is an irrational no.

So,

r - 2 is also an irrational no .

=> r is an irrational no.

Hence,

our assumption r is a rational number is wrong.

So,

2 + √2 is an irrational no.