Question

prove that 3+√5 is an irrational number​

Answer 1

Answer:

____✨Here ia the answer ✨____

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

Now,

3 + √5 = (a ÷ b)

[Here a and b are co-prime numbers]

√5 = [(a ÷ b) - 3]

√5 = [(a - 3b) ÷ b]

Here, {(a - 3b) ÷ b} is a rational number.

But we know that √5 is a irrational number.

So, {(a - 3b) ÷ b} is also a irrational number.

So, our assumption is wrong.

3 + √5 is a irrational number.

Hence, proved.

Answer 2

Step-by-step explanation:

let , 3+√5 is a rational number

since 3+√5=p/q

√5=p/q-3

√5=p-3qwhole/q

hence p/q is a rational number

since p-3qwhole/q is also a rational number

=√5 is also a rational number

but we know that√5 is an irrational number

so,

this is contradiction to our assumption

hence 3+√5 is an irrational number

hope this helps you