Question

0.26. Prove that √2+√3 is irrational.​

Answer 1

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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.

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Answer 2

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.

Please Drop Some Thanks.