Question

prove that each of the following numbers is irrational:
$\sqrt{3 + \sqrt{2} }$
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Answer 1

Answer :

$\large\fbox {\red\bf To prove :}$

• √3 + √2 is an irrational number

Step-by-step explanation :

$\large\fbox { \green\bf Proof :}$

Let us assume that √3 + √2 as  rational

So it can be written as a/b

→ √3 + 2 = a/b

Where a and b are co-primes and b ≠ 0

→ √2 = (a/b) - √3

Now , squaring on both sides

→ (√2)² = { (a/b) - √3 }²

We know that

• (a - b)² = a² + b² - 2ab

Substitute above formula in above equation

→ 2 = { (a/b)² + 3 - 2 × √4 (a/b) }

→ (a/b)² + 3 - 2 = 2 × √3 × (a/b)

→ (a² + b² / b² ) × (b/2a) = √3

→ (a² + b² / 2ab) = √3

Where ,

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

a³ is a rational number

It is contradiction to our assumption that √3 is irrational

∴ , Our assumption is wrong

Thus , √2 + √3 is irrational

Hence , proved

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