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
To satisfy the above, set a8 = 1 and b8 = 1, therefore 3(1) + 5(1) = 3 + 5 = 8
If n = 9, then 3a9 + 5b9 = 9
Set a9 = 3 and b9 = 0, therefore 3(3) + 5(0) = 9 + 0 = 9
If n = 10, then 3a10 + 5b10 = 10
Set a10 = 0 and b10 = 2, therefore 3(0) + 5(2) = 0 + 10 = 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
Since for every integer n, 8 ≤ n ≤ k, there are ak-2 and bk-2, then it follows that:
3ak-2 + 5bk-2 = k - 2 (Starting Point)
3ak+1 + 5bk+1 = k+1 (Goal)
Now I know where to begin and the goal I need to reach for my inductive step but I am unsure as to how to reach my goal.
Any guidance would be much appreciated!