Question

prove that 3+2√5 is irrational​

Answer 1

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Let 3+2√5 be a rational no.

Hence,

$= > 3 + 2\sqrt{5} = \frac{a}{b}$

(Here a and b are coprime)

$= > 2 \sqrt{5} = \frac{a}{b} - 3$

$= > 2\sqrt{5} = \frac{a - 3b}{b}$

$= > \sqrt{5} = \frac{a - 3b}{2b}$

• Here,if a-3b/2b is a rational no.
• Then,√5 is also a rational no.
• But this contradicts the fact.
• Our assumption is wrong.
• Hence,3+2√5 is an irrational no.

Hope it helps....

Answer 2

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

Now..

3 + 2√5 = $\dfrac{a}{b}$

Here, a and b are co-prime numbers.

Squaring on both sides..

(3 + 2√5)² = $\dfrac{a}{b}$

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(a + b)² = a² + 2ab + b²

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(3)² + 2 (3) (2√5) + (2√5)² = ${ (\frac{a}{b} )}^{2}$

9 + 12√5 + 20 = ${ (\frac{a}{b} )}^{2}$

29 + 12√5 = ${ (\frac{a}{b} )}^{2}$

12√5 = $\dfrac{ {a}^{2} }{ {b}^{2} }$ - 29

12√5 = $\dfrac{ {a}^{2} \: - \: 29 {b}^{2} }{ {b}^{2} }$

√5 = $(\dfrac{1}{12})$ $(\dfrac{ {a}^{2} \: - \: 29 {b}^{2} }{ {b}^{2} })$

Here;

$(\dfrac{1}{12})$ $(\dfrac{ {a}^{2} \: - \: 29 {b}^{2} }{ {b}^{2} })$ is rational number.

So, √5 is also a rational number.

But we know that√5 is irrational number.

So, our whole assumption is wrong.

6 + 2√5 is irrational

number..