Need help in Apostol Calculus proof

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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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  • #2
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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.
 
  • #3
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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.

I think you have to prove by using induction
 
  • #4
jbunniii
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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.

I don't think you get to choose b.
 
  • #5
jbunniii
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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 [itex]n = 0[/itex]. Now suppose it's true for [itex]n[/itex], so there exist [itex]q[/itex] and [itex]r < b[/itex] such that [itex]n = qb + r[/itex]. Now consider [itex]n + 1[/itex]. A reasonable first step would be to add 1 to both sides of the equation above:
[tex]n + 1 = qb + r + 1[/tex]
What does this do for you?
 

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