askchapo

Am I meant to assume a_i is defined the same way as a_n for each of 1<= i <= n-1 ?

Yes, another way to look at it is to reform the inequality as

a~n~ is the largest integer s. t.
a~n~ <= k^n^ * (x - a~0~ - a~1~/k - a~2~/k^2^ - ... - a~n-1~/k^n-1^)

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

a_i and a_n are defined the same way, they're just examining two different numbers in the sequence, but each can only be defined after the previous numbers in the sequence have been defined. In your example, after determining a_0=0, you next have to evaluate n=1. This shows a_1 can't be 2, it has to be the largest integer such that 0+a_1/2 <= 0.32, so a_1 has to be 0. Next you evaluate a_2, which given a_0 and a_1 has to be 1.

This process is essentially expressing a number in a different base, but focusing on the non-integer part. What it's asking you to prove is similar to proving that a number in binary will only have digits 0 or 1, and a number in octal will only have digits 0-7.

Edit: read my follow up comments

~~There is information missing, so you should ask your professor or move on to a different question.~~

~~It might be that it means to say a_0<=a_1<=…<=a_n integers have been defined~~

~~But that’s just a guess I haven’t checked the validity of.~~