Example of a Proof by Contradiction
Suppose a ∈ Z. If a2 is even, then a is even.
Proof. For the sake of contradiction, suppose a2 is even and a is not even.
Then a2 is even, and a is odd.
Since a is odd, there is an integer c for which a = 2c +1.
Then a2 = (2c +1)2
= 4c2 +4c +1
= 2(2c2 +2c)+1, so a2 is odd.
Thus a2 is even and a2 is not even, a contradiction.
Explanation of the proof:
The given proof uses the method of proof by contradiction to show that if a2 is even, then a is even. Let's break down the proof step by step:
1. Suppose for the sake of contradiction that a2 is even and a is not even. This means we assume the opposite of what we want to prove, which is that a is odd.
2. We know that if a is odd, there exists an integer c such that a = 2c + 1. This is because odd numbers can be expressed in the form 2c + 1, where c is an integer.
2
3. Substitute the value of a = 2c + 1 into a^2 to obtain a2 = (2c + 1)2.
4. Expand the expression (2c + 1)2 to get a2 = 4c2 + 4c + 1.
5. Rearrange the terms to get a2 = 2(2c2 + 2c) + 1.
6. From step 5, we can see that a2 can be written in the form 2k + 1, where k = 2c2+ 2c is an integer.
7. Therefore, we have shown that a2 is odd because it takes the form of 2k + 1, where k is an integer.
8. However, this contradicts our initial assumption that a2 is even.
9. Since we have arrived at a contradiction, our initial assumption must be false.
10. Hence, we can conclude that if a is even, then a must be even.
By using contradiction, we have established the desired result, which is that if a2 is even, then a is even.