Question

Show that there is no positive integer 'n' for which Vn-1+ Vn+1 is rational

Answer 1

By contradiction we can prove that there is no positive integer 'n' for which √(n-1) + √(n+1) is rational.

Given: To show that there is no positive integer 'n' for which √(n-1) + √(n+1) rational.

Let us assume that √(n-1) + √(n+1) is a rational number.

So we can describe by some p / q such that

√(n-1) + √(n+1) = p / q , where p and q are some number and q ≠ 0.

                         

Let us rationalize √(n-1) + √(n+1)

Multiplying √(n-1) - √(n+1) in both numerator and denominator in the LHS we get

{√(n-1) + √(n+1)} × {{√(n-1) - √(n+1)} / {√(n-1) - √(n+1)}} = p / q

=> {√(n-1) + √(n+1)}{√(n-1) - √(n+1)} / {√(n-1) - √(n+1)} = p / q

=> {(√(n-1))² - (√(n+1))²} / {√(n-1) - √(n+1)} = p / q

=> {n - 1 - (n + 1)] / {√(n-1) - √(n+1)} = p / q

=> {n - 1 - n - 1} / {√(n-1) - √(n+1)} = p / q

=> -2 / {√(n-1) - √(n+1)} = p / q

Multiplying {√(n-1) - √(n+1)} × q / p on both sides we get:

{-2 / {√(n-1) - √(n+1)}} × {√(n-1) - √(n+1)} × q / p = p / q × {√(n-1) - √(n+1)} × q / p

-2q / p = {√(n-1) - √(n+1)}

So {√(n-1) - √(n+1)} = -2q / p

Therefore, √(n-1) + √(n+1) = p / q                  [equation 1]

√(n-1) - √(n+1) = -2q / p                                 [equation 2]

Adding equation 1 and equation 2, we get:

{√(n-1) + √(n+1)} + {√(n-1) - √(n+1)} = p / q -2q / p

=> 2√(n-1) = (p² - 2q²) / pq

squaring both sides

{2√(n-1)}² = {(p² - 2q²) / pq}²

4(n - 1)  = (p² - 2q²)² / p²q²

Multiplying 1 / 4 on both sides

1 / 4 × 4(n - 1)  = (p² - 2q²)² / p²q² × 1 / 4

(n - 1) =  (p² - 2q²)² / 4p²q²

Adding 1 on both sides:

(n - 1) + 1 =  (p² - 2q²)² / 4p²q² + 1

n = (p² - 2q²)² / 4p²q² + 1

= ((p⁴ - 4p²q² + 4q⁴) + 4p²q²) / 4p²q²

= (p⁴ + 4q⁴) / 4p²q²

n = (p⁴ + 4q⁴) / 4p²q², which is rational  

Subtracting equation 1 and equation 2, we get:

{√(n-1) + √(n+1)} - {√(n-1) - √(n+1)} = p / q - (-2q / p)

=>√(n-1) + √(n+1) - √(n-1) + √(n+1) = p / q - (-2q / p)

=>2√(n+1) = (p² + 2q²) / pq

squaring both sides, we get:

{2√(n+1)}² = {(p² + 2q²) / pq}²

4(n + 1) = (p² + 2q²)² / p²q²

Multiplying 1 / 4 on both sides

1 / 4 × 4(n + 1)  = (p² + 2q²)² / p²q² × 1 / 4

(n + 1) =  (p² + 2q²)² / 4p²q²

Adding (-1) on both sides

(n + 1) - 1 =  (p² + 2q²)² / 4p²q² - 1

n = (p² + 2q²)² / 4p²q² - 1

= (p⁴ + 4p²q² + 4q⁴ - 4p²q²) / 4p²q²

= (p⁴ + 4q⁴) / 4p²q²

n =  (p⁴ + 4q⁴) / 4p²q², which is rational.

But n is rational when we assume √(n-1) + √(n+1) is rational.

So, if √(n-1) + √(n+1) is not rational, n is also not rational. This contradicts the fact that n is rational.

Therefore, our assumption √(n-1) + √(n+1) is rational is wrong and there exists no positive n for which √(n-1) + √(n+1) is rational.

Hence by contradiction we can prove that there is no positive integer 'n' for which √(n-1) + √(n+1) is rational.

Know more about "irrational numbers" here: brainly.com/question/17450097

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