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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