Need help in Apostol Calculus proof

zjhok2004
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let b denote a fixed positive integer. Prove the following statement by induction: for every integer n≥0, there exist nonnegative integers q and r such that n= qb+r, 0≤r<b.



Can someone help me on how to solve this question? and how does induction works here?
thank you
 
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Are there any restrictions on q and b? Otherwise the statement is trivially satisfied by q = n, b = 1, r = 0.
 
voko said:
Are there any restrictions on q and b? Otherwise the statement is trivially satisfied by q = n, b = 1, r = 0.

I think you have to prove by using induction
 
voko said:
Are there any restrictions on q and b? Otherwise the statement is trivially satisfied by q = n, b = 1, r = 0.

I don't think you get to choose b.
 
zjhok2004 said:
Can someone help me on how to solve this question? and how does induction works here?
thank you
You can easily check that it's true for n = 0. Now suppose it's true for n, so there exist q and r &lt; b such that n = qb + r. Now consider n + 1. A reasonable first step would be to add 1 to both sides of the equation above:
n + 1 = qb + r + 1
What does this do for you?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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