Question

Prove that $2-3{\sqrt{5}}$ is an irrational number

Answer 1

SOLUTION :  

Let us assume , to the contrary ,that 2 - 3√5 is rational. Then,it will be of the form a/b where a, b are co primes integers and b ≠0.

2 - 3√5 = a/b

2 - a/b = 3√5

(2b - a)/b = 3√5

(2b - a)/3b = √5

since, a & b is an integer so,(2b - a)/3b   is a rational number.  

∴ √5 is rational  

But this contradicts the fact that √5 is an irrational number .

Hence, 2 - 3√5 is an irrational .

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Answer 2

Answer:

Step-by-step explanation:

Given :-

Let x be 2 - 3√5  be a rational number.

2 - 3√5 = x

2 - x = 3√5

(2 - x)/3 = √5

x is rational, 2-x is rational and then (2 - x)/3 is also rational number.

√5 is a rational numbers, which is a contradiction.

Hence, 2 - 3√5 is an irrational number.

Hence, Proved.

Extra Information :-

Irrational Numbers :-

A number is an irrational if and only if, its decimal representation is non-terminating (non-repeating).