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.
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.
To prove the statement using proof by contraposition, we assume that Q is false, meaning neither x nor y is greater than or equal to A. Our goal is to find a contradiction to show that if Q is false, then P must also be false.
Assume that A is the average of two positive real numbers x and y, as given in the statement: A = (x + y)/2.
Now, we assume that Q is false. This means that neither x nor y is greater than or equal to A. Therefore, we can say that x < A and y < A.
Next, let's consider what happens when we add these two inequalities: x < A and y < A. Adding these together gives us x + y < 2A.
Now, let's divide both sides of the inequality by 2: (x + y)/2 < A.
But wait, this is exactly the opposite of what P states! P states that A = (x + y)/2. Therefore, if (x + y)/2 < A, it contradicts the assumption that A = (x + y)/2.
Since assuming Q is false leads to a contradiction, we can conclude that Q must be true. Therefore, one of the two numbers x or y is greater than or equal to A. This completes the proof by contraposition.
In summary, we assumed that Q is false, and then showed that it leads to a contradiction. This implies that if Q is false, then its negation, which is the original statement P, must also be false. Therefore, the statement "If A is the average of two positive real numbers, then one of the two numbers is greater than or equal to A" is proven using proof by contraposition.