Question

prove 2-√2 / 2+√2 is an irrational number...


(don't use rationalise the denominator method)​

Answer 1

let root 2 be rational no

=>root 2=p/q (where p nd q r integers q is not = 0 and q and p are co primes or having no common factor other than 1)

squaring both sides

2=(p/q)^2

2=p^2/q^2

2×q^2=p^2

therefore 2 is a factor of p^2

=>2 is also a factor if p

let p=2a

2q^2=(2a)^2

2q^2=4a^2

q^2=4a^2/2

q^2=2a^2

=>2 is also a factor of q^2

2 is also a factor of q

so , q and p are not co primes

=>our assumption is wrong that root 2 is rational no

=>root 2 is irratonal

similar substitute ur values like I did

Step-by-step explanation:

like in pic change the values

Answer 2

Answer:

Step-by-step explanation:

Given :

To prove that, $\frac{2 - \sqrt{2} }{2 +\sqrt{2} }$ is irrational,.

Solution :

We know that ,

Sum or difference of a rational number is always irrational,.

We know,

2 is rational number , $\sqrt{2}$ is irrational,

Hence, the sum or difference of the numbers is irrational,

⇒ 2 + √2 & 2 - √2 are irrationals,.

We know that,

Division of two irrationals is irrational if there isn't any factors existing to remove the irrationality,.

E.g, $\frac{2\sqrt{2}}{\sqrt{2}} = \frac{2 \ times \sqrt{2}}{\sqrt{2}}$ (factor exist) hence, it is rational.

Since, there is  no factors exist except √2 ,

⇒ $\frac{2-\sqrt{2}}{2 + \sqrt{2} } = \frac{\sqrt{2}(\sqrt{2} - 1) }{\sqrt{2}(\sqrt{2} + 1) } = \frac{\sqrt{2} - 1}{\sqrt{2} + 1}$

As we can't remove the irrationality (irrational number $\sqrt{2}$

hence it is irrational,.