Question

prove that 3_√5 is an irrrational numver​

Answer 1

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.

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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

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Answer 2

First of all, we have to know that √5 is an irrational number. Since we already know that √5 is an irrational number, we are proceeding towards the solution with only simple solution.

Let 3 - √5 be a rational number which we are denoting with "r".

Accordingly, 3 - √5 = r (say)

=> - √5 = r - 3

=> - (√5) = - (- r + 3)

=> √5 = (3 - r)

This contradicts our initial assumption.

Since, (3-r) is equal to √5, so, (3-r) should be a irrational number because √5 is a irrational number.

So, the equation stands somewhat like this :

3 - √5 = r

As we concluded that, r is not rational number, it has to be irrational, so, 3-√5 is also irrational (because 3-√5 is equal to r)

Hence, we can say that 3 - √5 is an irrational number. [PROVED]