To prove the statement 3n > n for all integers n ≥ 3 using mathematical induction, we will follow two main steps: the base case and the inductive step.
Step 1: Base Case
First, we check the base case where n = 3.
Substituting into the inequality, we have:
- Left side: 3n = 3(3) = 9
- Right side: n = 3
Since 9 > 3 holds true, the base case is verified.
Step 2: Inductive Step
Now we assume the statement is true for some integer k ≥ 3. This is our inductive hypothesis:
3k > k.
Next, we need to prove that the statement holds for k + 1: that is, we need to show:
3(k + 1) > k + 1.
Expanding the left side:
- Left side: 3(k + 1) = 3k + 3
Now, we can substitute our inductive hypothesis into this inequality:
We know that:
- Since 3k > k, adding 3 to both sides gives us:
- 3k + 3 > k + 3
So we need to show:
k + 3 > k + 1.
Now, simplifying this gives us:
- 3 > 1
This statement is true. Thus, we have shown that if 3k > k holds, then 3(k + 1) > k + 1 also holds.
Having established both the base case and the inductive step, we can conclude by the principle of mathematical induction that:
3n > n is true for all integers n ≥ 3.