Prove that (√2 + √3)2 is an irrational number

Q) Prove that (√2 + √3)² is an irrational number, given that √6 is an irrational number.

Ans:

Let’s start by considering (√2 + √3)² is a rational number (by the method of contradiction)

If (√2 + √3)² is a rational number, then it can be expressed in the form of $\frac{p}{q}$ , where p and q are integers and q ≠ 0.

∴ (√2 + √3)² = $\frac{p}{q}$

Since (a + b)² = a² + b² + 2 a b

∴ [(√2)² + (√3)² + 2 x √2 x √3] = $\frac{p}{q}$

∴ (2 + 3 + 2√6) = $\frac{p}{q}$

∴ 5 + 2√6 = $\frac{p}{q}$

∴ 2√6 = $\frac{p}{q}$ – 5 = $\frac{p - 5 q}{q}$

∴ √6 = $\frac{p - 5 q}{2 q}$ ……….(i)

Since p and q are integers, so $\frac{p - 5 q}{2 q}$ is also a rational number.

Since, in equation (i), LHS = RHS.

Therefore, if RHS is rational, then LHS is also rational.

Therefore √6 is a rational number.

But it contradicts the given condition (∵ given that √6 is an irrational number).

It means that our assumption that “(√2 + √3)² is a rational number” is wrong.

Therefore, it is confirmed that (√2 + √3)² is an irrational number.

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