Prove that √2+√3 is irrational.

On rewriting the above equation:

√2 = a/b – √3

On squaring the above equation on both 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 the:

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, and b are integers, (a2 + b2)/2ab is a rational number. √3 is a rational number. It contradicts our assumption that √3 is irrational.

∴ our assumption is wrong

Thus √2 + √3 is irrational.

Summary:

Prove that √2+√3 is irrational.

√2+√3 is irrational. An irrational number is a real number that cannot be expressed as the ratio of integers. The symbol used to indicate the irrational symbol is “P”. As irrational numbers are explained negatively, the set of real numbers (R) which are not the rational number (Q) is known as an irrational number.