Posted by Anna on .
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.
Domain: positive real numbers
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
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)
≡ ¬(x≥A ∨ y≥A)
≡ x<A ∧ y<A [de Morgan]
⇒ (x+y)/2 < A
⇒ (x+y)/2 ≠ A [since A=(x+y)/2]
Therefore we have proved ¬Q(x,y)⇒¬P(x,y) QED
You can reformulate your answer along the above lines.