Let x be a positive real number. Prove that if x -? 2/x > 1, then x > 2 by (a) a

Question

Let x be a positive real number. Prove that if x -? 2/x > 1, then x > 2 by (a) a direct proof, (b) a proof by contrapositive, and (c) a proof by contradiction.

Explanation / Answer

A) Assume that x is a positive real number and that x - 2/x >1. First, we form a common denominator to combine like terms: (x^2 - 2) / x > 1 Now we subtract 1 from both sides: (x^2 - 2) - 1/ x > 0 Again combine terms with common denominator: (x^2 - x -2) / x > 0 Factor the numerator: (x - 2)(x+1) / x > 0 Since x is positive (x+1)/x is always positive, we can divide both sides of the inequality by (x+1)/x. This gives us: (x-2) > 0 Adding 2 to both sides shows that: x>2 Thus if x is a positive real number and x-2/x > 1, then x>2. B) Assume that x is a positive real number and x< =2. Starting with x 0 Factor the numerator: (x - 2)(x+1) / x > 0 Since x is positive (x+1)/x is always positive, we can divide both sides of the inequality by (x+1)/x. This gives us: (x-2) > 0 Adding 2 to both sides shows that: x>2 But our assumption was that x1, then x>2

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