Question

Prove that 3 + √5 is an irrational number​

Answer 1

Let us assume that 3+√5 is rational.

3+√5 =a/b, where a and b are integers and co primes. b is not equal to 0.

√5 = a/b - 3

√5= (a-3b)/b

√5 = (a-3b) / 2b

(a-3b) / 2b is a rational number, since a and b are integers.

Therefore√5 is also a rational number.

It contradicts the fact that√5 is irrational.

Therefore, Our assumption is wrong.

Therefore, 3+√5 is irrational.

Answer 2

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.

__________________________________

How √5 is a irrational number.?

→ √5 = a ÷ b [a and b are co-prime numbers]

b√5 = a

Now, squaring on both side we get,

5b² = a² ........(1)

b² = a² ÷ 5 

Here 5 divide a²

and 5 divide a also

Now,

a = 5c [Here c is any integer]

Squaring on both side

a² = 25c²

5b² = 25c² [From (1)]

b² = 5c²

c² = b² ÷ 5

Here 5 divide b²

and 5 divide b also

→ a and b both are co-prime numbers and 5 divide both of them.

So, √5 is a irrational number.

Hence, proved