Question

Prove that 3 + 2√5 is irrational​

Answer 1

Step-by-step explanation:

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Answer 2

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⭐Solution :-⭐

☑ Method 1.

Assume that 3+2√5 is rational. Then,

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

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

is rational but √5 is an irrational number, therefore, the assumption is wrong.

Hence, 3+2 √5 is irrational.

☑ Method 2.

Let us assume 3+2√5 is rational.

So we can write this number as

$= > 3 + 2 \sqrt{5} = \frac{a}{b} ....(1)$

Here a and b are two co-prime number and b is not equal to zero.

Simplify the equation (1) subtract 3 both sides, we get

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

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

Now divide by 2 we get,

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

Here a and b are integer so ,

$\frac{a - 3b}{2b}$

is a rational number, so √5 should be a rational number.

But √5 is a irrational number, so it is contradict.

Therefore, 3+2 √5 is irrational number.

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