Thursday

October 2, 2014

October 2, 2014

Posted by **Anna** on Thursday, October 7, 2010 at 9:56pm.

If A is the average of two positive real numbers then one of the two numbers is greater than or equal to A.

Proof.

Domain: positive real numbers

P: A=(x+y)/2

Q: x is greater than or equal to A or y is greater than or equal to A

Method: Assume P is true and Q is false and find a contradiction.

- Discret Mathematical Structures -
**MathMate**, Thursday, October 7, 2010 at 11:12pmNOTE: Contrapositive is not the same as contradiction.

Given: p → q

If we would like to prove by*contrapositive*, we need to prove:

¬q → ¬p

since

p → q ≡ ¬q → ¬p

In the given case,

P(x,y) : "A is the average of two real numbers x and y"

Q(x,y) : (x≥A)∨(y≥A)

We will attempt to prove that:*∀x,y∈ℝ ¬Q(x,y) → ¬P(x,y)*

¬Q(x,y)

≡ ¬(x≥A ∨ y≥A)

≡ x<A ∧ y<A [de Morgan]

So

¬Q(x,y)

⇒ (x+y)/2 < A

⇒ (x+y)/2 ≠ A [since A=(x+y)/2]

⇒ ¬P(x,y)

Therefore we have proved ¬Q(x,y)⇒¬P(x,y) QED

You can reformulate your answer along the above lines.

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