Question

Show that 2+ /2 is not a rational number

Answer 1

Answer:

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,

Now,

Transpose 2 to the right side

Taking LCM we get ,

Here,

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,

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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,

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,

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

Step-by-step explanation:

Answer 2

$prove \: that \: \sqrt{2 \:} \: is \: irrational \\ \\ \\ assume \: that \: \sqrt{2 \: } \: is \: rational$

$\sqrt{2} = \frac{a}{b} \\ \: \: where \: a \: and \: b \: are \: integers \: \: \: \\ b≠0 \: \: a \: and \: b \:are \: coprimes$

$b \sqrt{2} = a \: \: \: \: - (1) \\ squaring \: both \\ 2 {b}^{2} = {a}^{2} \\ 2 \: divides \: { a}^{2} \\ 2 \: divides \: a \: \: \: \: (A)$

$a = 2c \: \: \: \: \: for \: some \: integer \: c \\ substituting \: for \: a \: \: \: \: \: 2 {b}^{2} = 4 {c}^{2} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {b}^{2} = 2 {c}^{2}$

$2 \: divides \: {b}^{2} \\ 2 \: divides \: b \: - (B)$

3 divides both a and b .

This tells us that a and b have more than 1 factor. So we prove that

$\sqrt{2 \: } is \: irrional$

$prove \: that \: \:2 + \sqrt{2} \: is \: irrational \\ assume \: that \: \: 2 + \sqrt{2} \: is \: \ \: rational \\ 2 + \sqrt{2} = \frac{p}{q} \: \: \: \: where \: p \: and \: q \: are\: \\ integers \: \: q≠0 \: \: p \: and \: q \: are \: \\ coprimes$

$\sqrt{2 } = \frac{p}{q} - 2 \\ \\ \sqrt{2 } = \frac{p - 2q}{q}$

Here RHS

$\frac{p - 6q}{q} \: is \: \: rational$

but LHS √ 2 is irrational

This contradiction has arisen due to our wrong assumption.