Am I meant to assume a_i is defined the same way as a_n for each of 1<= i <= n-1 ?
If so, I think I see the proof by induction on n, but the question just says a_i is defined for each 1<=i<=n-1, not that it is defined in that way. Is the question just overly vague or am I missing something obvious?
If only a_n is defined as the greatest integer such that the sum from 0 to n of each a_i/k^i is <=x, then I think there are counterexamples to the hypothesis, right?
Like if x=0.32, k=2, n=2, then a_0=0, and the inequality is satisfied by a_1 = 2 > k-1 = 1 and a_2 = -3 < 0.
Yes, another way to look at it is to reform the inequality as
if that makes it clearer?
Given this we know that a~1~ is the largest integer s.t. k*(x - a~0~) >= a~1~
(Why does a~1~ even exist might be a good question?)
Then show that a~1~ <= k-1 (just substitute in).
Assume that a~i~ <= k-1 is proven for all integers i=1,...,n (where n might be 1) and show that a~n+1~ <= k-1 is true as well.
Hope this helps? If not please do say so
Yeah, that's helpful. So I show that 0 <= a_1 <= k-1 from a_1 <= k(x - a_0) < a_1 + 1 by showing 0 <= (x - a_0) < 1 which implies 0 <= k(x - a_0) < k. That's the base case. Then I assume that the hypothesis holds for some positive integer n-1. From there I manipulate the inequalities to show that they imply 0<= a_n <=k-1.
Then having already assumed 0 <= a_i <= k-1 for all 1 <= i <= n-1, that means it is the case for all 1<= i <= n. Since I have shown it to be true for n=1, it is true for n=2. Since it is true for n=1 and n=2, it is true for n=3, and so on for all positive integers n.
Is this the way? Also, I think this technique is an example of strong induction and not regular induction, is that right?
Yes and yes 😁 I would show both inequalities seperately, i.e. one chain of inequalities for 0 <= a~i~ and another for a~i~ <= k-1 (or < k) (in the base case as well) just for clarity but the argument is solid.👍
Oh just be clear
The hypothesis is "0 <= a~i~ <= k-1 for all i <= n-1" right?
Yes, that one.