1. Assume that n is a positive integer. Use the proof by contradiction method to prove:
If 7n + 4 is an even integer then n is an even integer.
2. Prove: n is an even integer iff 7n + 4 is an even integer. (Note this is an if and only if (iff) statement.
To prove both statements, we will use the proof by contradiction method.
1. Proof by contradiction:
Assume that 7n + 4 is an even integer, but n is an odd integer.
Since n is an odd integer, we can express it as n = 2k + 1, where k is an integer.
Substituting n in the equation 7n + 4, we get 7(2k + 1) + 4 = 14k + 7 + 4 = 14k + 11.
Now, let's suppose that 7n + 4 is an even integer, which means that it can be expressed as 7n + 4 = 2m, where m is an integer.
If we substitute the value of 7n + 4 into this equation, we have 14k + 11 = 2m.
Rearranging the equation, we get 2m - 14k = 11.
Now, let's consider the left side of the equation, which is 2m - 14k. Since both 2m and 14k are even integers, their difference should also be even. However, 11 is an odd integer.
This contradiction disproves our assumption that 7n + 4 is an even integer and n is an odd integer. Hence, we can conclude that if 7n + 4 is an even integer, then n must be an even integer.
2. Prove: n is an even integer iff 7n + 4 is an even integer.
To prove the "if and only if" statement, we need to prove two parts:
(a) If n is an even integer, then 7n + 4 is an even integer.
(b) If 7n + 4 is an even integer, then n is an even integer.
(a) If n is an even integer, we can express it as n = 2k, where k is an integer.
Substituting this value into 7n + 4, we get 7(2k) + 4 = 14k + 4 = 2(7k + 2).
Since 7k + 2 is an integer, we can express 2(7k + 2) as the product of an integer (2) and an integer (7k + 2), making it an even integer.
(b) If 7n + 4 is an even integer, we can express it as 7n + 4 = 2m, where m is an integer.
Rearranging the equation, we get 7n = 2m - 4.
Since 2m - 4 is an even integer, its right-hand side will always be even. 7n, therefore, must be an even integer, which implies that n must also be an even integer.
Both parts (a) and (b) have been proven, and therefore we can conclude that n is an even integer if and only if 7n + 4 is an even integer.
Step 1: Assume that n is an odd integer.
Step 2: Since n is odd, we can write it as n = 2k + 1, where k is an integer.
Step 3: Substitute the value of n into the expression 7n + 4:
7(2k + 1) + 4 = 14k + 7 + 4 = 14k + 11
Step 4: The expression 14k + 11 is an odd integer, as it can be written as 2(7k + 5) + 1.
Step 5: This contradicts the assumption that 7n + 4 is an even integer.
Step 6: Therefore, our initial assumption that n is an odd integer must be false.
Step 7: Hence, n must be an even integer.
To prove the if and only if statement "n is an even integer iff 7n + 4 is an even integer," we need to prove both directions:
Forward direction:
Assume that n is an even integer.
Since n is even, we can write it as n = 2k, where k is an integer.
Substitute the value of n into the expression 7n + 4:
7(2k) + 4 = 14k + 4 = 2(7k + 2).
The expression 2(7k + 2) is an even integer, since it can be written in the form 2m, where m = 7k + 2 is an integer.
Therefore, if n is an even integer, then 7n + 4 is an even integer.
Backward direction:
Assume that 7n + 4 is an even integer.
We need to prove that n is an even integer.
Suppose n is an odd integer.
Using the proof by contradiction method:
1. Assume n is an odd integer.
2. Write n as n = 2k + 1, where k is an integer.
3. Substitute the value of n into the expression 7n + 4:
7(2k + 1) + 4 = 14k + 7 + 4 = 14k + 11.
4. The expression 14k + 11 is an odd integer, as it can be written as 2(7k + 5) + 1.
5. This contradicts the assumption that 7n + 4 is an even integer.
6. Therefore, our initial assumption that n is an odd integer must be false.
7. Hence, n must be an even integer.
Since we have proved both directions, we can conclude that n is an even integer iff 7n + 4 is an even integer.