Question

prove that is irrational
$3 + 2 \sqrt{5}$
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Answer 1

Answer:

Step-by-step explanation:

→ Let take that 3 + 2√5 is a rational number.

→ So we can write this number as

→ 3 + 2√5 = a/b

→ Here a and b are two co prime number and b is not equal to 0

→ Subtract 3 both sides we get

→ 2√5 = a/b – 3

→ 2√5 = (a-3b)/b

→ Now divide by 2 we get

→ √5 = (a-3b)/2b

Here a and b are integer so (a-3b)/2b is a rational number so √5 should be a rational number But √5 is a irrational number so it contradict the fact

→ Hence result is 3 + 2√5 is a irrational number

Answer 2

Answer:

provre 3+2√5 is an irrational number.

Step-by-step explanation:

Let 3+2√5 be rational number.

3+2√5=p/q

p/q=Z(integers) q is not equals to 0 and

HCF(p,q)=1

2√5=p/q-3=p-3q/q

√5=p-3q/2q

p-3q/2q is rational number.

so, √5 is also rational as it is equal to that .

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

:3+2√5 is an irrational number.