Math, asked by rohit7145, 7 months ago

prove that 2-3√5 is an irrational number​

Answers

Answered by pradnya250604
4

Answer:

Let us assume that 2 - 3√5 is a rational number.

So it can be written in the form a/b

2 - 3√5 = a/b

Here a and b are co prime numbers and b ≠ 0

Solving 2 - 3√5 = a/b we get,

⇒-3√5= a-2b/b

⇒√5=a-2b/-3b= 2b-a/3b

This shows (2b-a/3b) is a rational number. But we know that But √5 is an irrational number.

So it contradicts our assumption.

Our assumption of 3 + 2√5 is a rational number is incorrect.

3 + 2√5 is an irrational number

Hence proved

Answered by viny10
80

\huge\star\:\:{\orange{\underline{\purple{\mathcal{Proof :-}}}}}

Let,

x = 2 -  3 \sqrt{5}

be a rational number.

3 \sqrt{5}  = 2 - x

 \sqrt{5}  =  \frac{2 - x}{3}

Since x is rational , 2 - x is rational and hence

 \frac{2 - x}{3}

is also rational number

 =  >  \sqrt{5}

is a Rational Number , which is a contradiction.

{\purple{\boxed{\large{\bold{Hence}}}}}

2 - 3 \sqrt{5}

must be an irrational number.

 \boxed {Hence Proved}

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