Question

prove that 4-√2 is an irrational number

Answer 1

Here your answer goes

Step :-1

Consider ,  4 - $\sqrt{2}$  

Let 4 - $\sqrt{2}$  = ( a/b ) which is rational number

Where a and b are positive co - prime

Step :- 2

$\sqrt{2}$ =  4 -  (a/b)

$\sqrt{2}$  = ( 4b-a )/b

$\sqrt{2}$ is rational

This is a Contradiction

Hence , 4 - $\sqrt{2}$  is rational

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Answer 2

Let 4-√2 is rational number

4-√2 =a/b

 √2   = 4-a/b

√2  =4b-a/b

 √2    =4b-a/b

  LHS is irrational number

RHS is rational number

Contradiction to assumption

therefore,

4-√2 is an irrational number

@skb