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.
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
an is the largest integer s. t.
an <= kn * (x - a0 - a1/k - a2/k2 - … - an-1/kn-1)if that makes it clearer?
Given this we know that a1 is the largest integer s.t. k*(x - a0) >= a1
(Why does a1 even exist might be a good question?)
Then show that a1 <= k-1 (just substitute in).
Assume that ai <= k-1 is proven for all integers i=1,…,n (where n might be 1) and show that an+1 <= k-1 is true as well.
Hope this helps? If not please do say so I have more hints but it’s a fine line between hinting and spoonfeeding and I have no feeling for your proficiency (there is nothing wrong with spoonfeeding when necessary so don’t be afraid to ask, but it’s obviously bad when it’s not necessary)
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?
Oh just be clear
Then I assume that the hypothesis holds for some positive integer n-1.
The hypothesis is “0 <= ai <= k-1 for all i <= n-1” right?
Yes, that one.
Yes and yes 😁 I would show both inequalities seperately, i.e. one chain of inequalities for 0 <= ai and another for ai <= k-1 (or < k) (in the base case as well) just for clarity but the argument is solid.👍
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.
About your question of definitions more broadly, generally when an is defined, other subscripts ai, aj, am, axyz, are defined in the same way, even if that isn’t specifically stated.
The a is what carries the properties of the sequence, and any variables in the subscript are just different ways of referencing the index. That can be to communicate a separation between concepts, or to set up relationships like i<=n that will be needed laterThanks, I was thinking the same. I had already wrote a proof that worked if all a_i are defined like that, but wanted to be sure.
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 definedBut that’s just a guess I haven’t checked the validity of.Looking at it again I think you should read the question inductively.
So a_1 is defined as the largest integer such that the inequality for n=1 holds, then given that a_1, a_2 is the largest integer such that the inequality for n=2 holds etc
So if x=0.32 and k=2 then a1=0, a2=1, a3=0 etc
I’ll use this final comment to complain about how lazy some mathematics professors are with formulating their questions that it sometimes takes more time to decipher the question than to come up with the correct answer.
