Question

Prove that
$\sqrt{2 } + \sqrt{3}$
is irrational.



Please answer immideately
It is urgent.​

Answer 1

Step-by-step explanation:

Given:-

√2+√3

Required To Prove:-

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

Proof:-

Let us assume that √2+√3 is a rational number.

It is in the form of p/q

√2+√3 = a/b

Where a and b are co - primes

=>√3 = (a/b) + √2

On squaring both sides, then

=>(√3)^2 = [(a/b/+√2]^2

This is in the form of (a+b)^2

Where a = (a/b) and b = √2

We know that (a+b)^2 = a^2+2ab+b^2

=>3 = (a/b)^2 + 2(a/b)(√2) + (√2)^2

=>3 = (a/b)^2 + 2√2a/b + 2

=>3-2 = (a^2/b^2 )+(2√2a/b)

=>1 = (a^2+2√2ab)/b^2

=>1×b^2 = a^2 + 2√2 ab

=>b^2 = a^2 + 2√2 ab

=> b^2-a^2=2√2ab

=>(b^2-a^2)/2ab = √2

=>√2 = (b^2-a^2)/(2ab)

=>√2 is in the form of p/q

=>√2 is not a raational number

This is a contradiction to our assumption.

=>√2 is an irrational number

Therefore,√2+√3 is an irrational number.

Hence,Proved.

Used Method:-

Method of Contradiction or Indirect method is used for proving the given number is an irrational number.

Note:-

The sum of two irrational numbers is also an irrational number.

Answer 2

$\tt \: Let \: assume \: that \: \sqrt{2} + \sqrt{3} \: is \: not \: irrational.$

So,

$\rm :\implies\: \sqrt{2} + \sqrt{3} \: is \: rational.$

$\tt \: Let \: \sqrt{2} + \sqrt{3} = \dfrac{x}{y} \: where \: x \: and \: y \: are \: integers$

$\tt \: such \: that \: y \ne 0 \: and \: x \: and \: y \: are \: co - prime.$

$\rm :\longmapsto\: \tt \: \sqrt{3} = \dfrac{x}{y} - \sqrt{2}$

On squaring both sides, we get

$\rm :\longmapsto\:\tt \: 3 = 2 + \dfrac{ {x}^{2} }{ {y}^{2} } - 2 \times \sqrt{2} \times \dfrac{x}{y}$

$\rm :\implies\:\dfrac{2 \sqrt{2}x }{y} = \dfrac{ {x}^{2} }{ {y}^{2} } - 1$

$\rm :\implies\:\dfrac{2 \sqrt{2}x }{y} = \dfrac{ {x}^{2} - {y}^{2} }{ {y}^{2} }$

$\rm :\implies\:\sqrt{2} = \dfrac{ {x}^{2} - {y}^{2} }{2xy}$

Now,

• Since x and y are integers,

So,

$\rm :\longmapsto\:\dfrac{ {x}^{2} - {y}^{2} }{2xy} \: is \: rational \: number.$

$\rm :\implies\: \sqrt{2} \: is \: rational.$

$\tt \: which \: is \: contradiction \: as \: \sqrt{2} \: is \: irrarional.$

$\rm :\implies\:our \: assumption \: is \: wrong \: that \: \sqrt{2} + \sqrt{3} \: is \: not \: irrational.$

$\tt \: Hence \: \sqrt{2} + \sqrt{3} \: is \: irrational.$