Use proof by contraposition to prove the following statement:

Question

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.

Answer 1

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.