Prove by strong induction that every natural number n≥6 can be written in the form n=3k+4t for some k,t∈N.
I'm not entirely comfortable with the concept of strong induction. I believe you assume that P(1),P(2),P(3),...P(n) are true to prove P(n+1) is. Now, how does that fit into this problem? Would the base case be where n=6?
Answer
P(1) through P(5) are vacuously true because 1 through 5 are not greater than or equal to 6. Your base cases are 6,7,8. Show by example that you can do each of them. Intuitively, you can just add enough 3s to get to any number. So assume it is true up to n, then to do n+1 you say that you can do n−2 and add a 3
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