1.

Sidh kare 6+√2 ak aprimy sankhaya hai​

Answer»

Step-by-step explanation:

Given,

6+√2 is irrational number,

By the method of contradiction,

let us assume 6+√2 is rational number,

let, 6+√2 = p/q where p & q are co-primes, and b≠0,

6+√2 = p/q

6 = p/q - √2

Squaring on both sides,

(6)² = (p/q - √2)²

[Since (a-b)² = a² +2ab - b²]

(OR)

[(a-b)² = a² - b² + 2ab]

36 = (p/q)² - 2(p/q)(√2) + (√2)²

36 = p²/q² - (2p/q)(√2) + 2

(2p/q)(√2) = p²/q² + 2 - 36

(2p/q)(√2) = p²/q² - 34

√2 = (q/2q)[p²/q² - 34]

Here,

(q/2q)[p²/q² - 34] is rational number,

So,

√2 is also rational number,

But it contradicts the fact that √2 is irrational number,

So our assumption is wrong,

6+√2 is irrational number.



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