Question

prove that (√2 + √3) is irrational.​

Answer 1

Step-by-step explanation:

Given √2 + √3

To prove: √2 + √3 is an irrational number.

Proof:

Letus assume that √2 + √3 is a rational number.

So it can be written in the form a/b

√2 + √3 = a/b

Here a and b are coprime numbers and b ≠ 0

Solving

√2 + √3 = a/b

√2 = a/b – √3

On squaring both the sides we get,

=> (√2)2 = (a/b – √3)2

We know that

(a – b)2 = a2 + b2 – 2ab

So the equation (a/b – √3)2 can be written as

(a/b – √3)2 = a2/b2 + 3 – 2 (a/b)√3

Substitute in the equation we get

2 = a2/b2 + 3 – 2 × √3 (a/b)

Rearrange the equation we get

a2/b2 + 3 – 2 = 2 × √3 (a/b)

a2/b2 + 1 = 2 × √3 (a/b)

(a2 + b2)/b2 × b/2a = √3

(a2 + b2)/2ab = √3

Since, a, b are integers, (a2 + b2)/2ab is a rational number.

√3 is a rational number.

It contradicts to our assumption that √3 is irrational.

∴ our assumption is wrong

Thus √2 + √3 is irrational.

Thank you .

Answer 2

Answer:

here is your answer

Step-by-step explanation:

Answer:

Given √2 + √3

To prove: √2 + √3 is an irrational number.

Proof:

Letus assume that √2 + √3 is a rational number.

So it can be written in the form a/b

√2 + √3 = a/b

Here a and b are coprime numbers and b ≠ 0

Solving

√2 + √3 = a/b

√2 = a/b – √3

On squaring both the sides we get,

=> (√2)2 = (a/b – √3)2

We know that

(a – b)2 = a2 + b2 – 2ab

So the equation (a/b – √3)2 can be written as

(a/b – √3)2 = a2/b2 + 3 – 2 (a/b)√3

Substitute in the equation we get

2 = a2/b2 + 3 – 2 × √3 (a/b)

Rearrange the equation we get

a2/b2 + 3 – 2 = 2 × √3 (a/b)

a2/b2 + 1 = 2 × √3 (a/b)

(a2 + b2)/b2 × b/2a = √3

(a2 + b2)/2ab = √3

Since, a, b are integers, (a2 + b2)/2ab is a rational number.

√3 is a rational number.

It contradicts to our assumption that √3 is irrational.

∴ our assumption is wrong

Thus √2 + √3 is irrational.