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Discreet Mathematical Structures

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Use proof by contraposition to prove the following statement:

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 -

    NOTE: 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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