Question

Prove that 3+ √5 is an irrational number​

Answer 1

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.

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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

Answer 2

Step-by-step explanation:

let us assume, to the contrary, that 3+√5 is rational.

that is,we can find co-prime a and b. (b is not equal to 0) such that 3+√5 = a .

b

therefore, 3 - a = -√5

b

rearranging this equation, we get

-√5 = 3- a = 3b - a

b b

since a and b are integers, we get 3 - a

b is rational, and so -√5 is rational.

But this contradicts the fact that -√5 is irrational.

this contradiction has arisen because of our incorrect assumption that 3+√5 is rational.

so,

we conclude that 3+√5 is irrational.