Consider the following theorem: If n is an even integer, then n + 1is odd and provide:
a)direct proof,
b)proof by contraposition, &
c) proof by contradiction
of this theorem.
a) Direct proof:
To prove that if n is an even integer, then n + 1 is odd, we can directly show that the statement holds true.
Let's assume that n is an even integer. By definition, an even integer is any integer that can be divided evenly by 2, meaning n = 2k, where k is an integer.
Now let's evaluate n + 1:
n + 1 = 2k + 1
Since 2k is an even number (multiplying an even number by any integer gives an even number), we can rewrite 2k as 2k + 0 (which doesn't change its parity).
Now, n + 1 becomes:
2k + 0 + 1 = 2k + 1
We can see that 2k + 1 is in the form of an odd number (2k + 1), where k is an integer. Therefore, n + 1 is odd.
b) Proof by contraposition:
To prove the theorem using contraposition, we need to show that if n + 1 is even, then n must be odd.
Assume n + 1 is even. By definition, an even number is any integer that can be divided evenly by 2, so n + 1 = 2k, where k is an integer.
Now, let's subtract 1 from both sides of the equation:
n = 2k - 1
We can rewrite 2k as 2k + 0:
n = 2k + 0 - 1
Simplifying further:
n = 2k - 1
We observe that 2k - 1 is in the form of an odd number (2k - 1), where k is an integer. Therefore, n must be odd.
c) Proof by contradiction:
To prove the theorem using contradiction, we assume the opposite of the desired statement and show that it leads to a contradiction.
Assume that n is an even integer, but n + 1 is not odd.
If n is even, it can be represented as n = 2k, where k is an integer.
If n + 1 is not odd, it must be even as well. So, n + 1 = 2k' where k' is an integer.
Now, let's subtract n from both sides of the equation:
1 = 2k' - 2k
Simplifying further:
1 = 2(k' - k)
Since k' - k is an integer, let's say k' - k = m, where m is an integer.
The equation becomes:
1 = 2m
But this equation is not true since an even number cannot equal an odd number (1 is odd, while 2m is even for any integer m).
The assumption that n + 1 is not odd leads to a contradiction. Therefore, our initial assumption (that if n is an even integer, then n + 1 is odd) must be true.