Use mathematical induction to prove that the statement is true for every positive integer n. Show your work.
2 is a factor of n2 - n + 2
To prove that 2 is a factor of n^2 - n + 2 for every positive integer n using mathematical induction, we will follow these steps:
Step 1: Base Case
First, we need to prove that the statement is true for the smallest possible value of n, which is usually n=1.
When n=1, we have:
n^2 - n + 2 = 1^2 - 1 + 2 = 2, which is divisible by 2.
So, the statement is true for n=1.
Step 2: Inductive Hypothesis
Assume that the statement is true for some positive integer k, where k >= 1. That is, assume that 2 is a factor of k^2 - k + 2.
Step 3: Inductive Step
We need to prove that, based on the assumption in step 2, the statement is true for the next positive integer, k+1.
That is, we need to prove that 2 is a factor of (k+1)^2 - (k+1) + 2.
Expanding the expression:
(k+1)^2 - (k+1) + 2 = k^2 + 2k + 1 - k - 1 + 2
= k^2 + k + 2k + 2
= k^2 + 3k + 2
Using the assumption that 2 is a factor of k^2 - k + 2 (from step 2):
k^2 - k + 2 = 2m, where m is an integer.
Now, let's substitute this into our expression for (k+1)^2 - (k+1) + 2:
k^2 + 3k + 2 = 2m + 3k + 2
= 2m + 3k + 2 + 6k - 6k
= 2m + 3k + 6k + 2 - 6k
= 2(m + 3k) - 6k + 2
Since (m + 3k) is an integer, let's call it p:
2(m + 3k) - 6k + 2 = 2p - 6k + 2
Now, we need to prove that 2p - 6k + 2 is divisible by 2.
2p - 6k + 2 = 2(p - 3k) + 2
The first term, 2(p - 3k), is divisible by 2 (because p - 3k is an integer).
Adding 2 to a number divisible by 2 will also result in a number divisible by 2.
So, 2p - 6k + 2 is divisible by 2.
This completes the inductive step.
Step 4: Conclusion
Since we have proven that the statement is true for n=1 (base case) and that if it is true for k, then it is also true for k+1 (inductive step), we can conclude that the statement is true for every positive integer n by the principle of mathematical induction.
Therefore, 2 is a factor of n^2 - n + 2 for every positive integer n.
To prove that the statement "2 is a factor of n² - n + 2 for every positive integer n" is true using mathematical induction, we will follow these steps:
Step 1: Base Case
First, we need to test the statement for the smallest possible value of n, which is usually 1 in this case. Let's substitute n = 1 into the expression n² - n + 2 and check if the statement holds:
1² - 1 + 2 = 1 - 1 + 2 = 2
As we can see, 2 is indeed a factor of 2. So the statement is true for n = 1, and the base case is satisfied.
Step 2: Inductive Hypothesis
Assume that the statement is true for some positive integer k. In other words, assume that 2 is a factor of k² - k + 2.
Step 3: Inductive Step
We need to prove that the statement is also true for (k + 1), assuming it's true for k. So we need to show that 2 is a factor of (k + 1)² - (k + 1) + 2.
(k + 1)² - (k + 1) + 2 = k² + 2k + 1 - k - 1 + 2
= k² + k + 2k + 1 - k - 1 + 2
= k² + 2k,
= k(k + 2).
Since we have assumed that 2 is a factor of k² - k + 2, we can write k² - k + 2 as 2m (where m is an integer), and thus:
k(k + 2) = 2m(k + 2),
= 2(m(k + 2)).
This implies that k(k + 2) is divisible by 2, which means 2 is a factor of (k + 1)² - (k + 1) + 2.
Step 4: Conclusion
By proving the base case and the inductive step, we can conclude that 2 is a factor of n² - n + 2 for every positive integer n, using mathematical induction.