To provide full context of the practice question I'm attempting, it is as follows:
For every integer n such that n ≥ 8, there exist nonnegative integers an and bn such that 3an + 5bn = n.
Write a proof of the claim using the strong form of mathematical induction with the integer 10 as breakpoint — such that that n = 8, n = 9 and n = 10 would all be considered in the basis.
This is my attempt at the proof, but I can't seem to get past applying the I.H. in order to reach my goal in the inductive step.
If n = 8, then 3a8 + 5b8 = 8
If n = 9, then 3a9 + 5b9 = 9
If n = 10, then 3a10 + 5b10 = 10
Let k be an integer such that k ≥ 10. It is necessary and sufficient to use the following:
- Inductive Hypothesis: 3an + 5bn = n for every integer n such that 8 ≤ n ≤ k.
3ak+1 + 5bk+1 = k+1
Now I have to complete the remainder of the proof but I'm not even sure if I was proving it correctly so far or how to complete it from where I'm at.
Any guidance would be much appreciated!