1. ## Proof by induction

Hi Folks,

I'm having difficulty making progress on the following proof, and I'm hoping for some hints to help me along.

Let $\displaystyle b$ denote a fixed positive integer. Prove the following statement by induction: For every integer $\displaystyle n \geq 0$, there exist non-negative integers $\displaystyle q$ and $\displaystyle r$ such that

$\displaystyle n = qb + r, 0 \leq r < b$.

So I can state the assertion and show the initial case is true:
$\displaystyle A(n): n = qb + r, 0 \leq r < b$

$\displaystyle A(0): 0 = 0 \cdot b + 0, n = q = r = 0$

I've tried picking a value for b and creating a table of related values for q and r as n counts up from 0. But I can't figure out how to express the patterns I see to form a usable general case and inductive step.

Again, just looking for hints right now to help get me out of the rut I'm in.

Thanks,
Scott

2. Originally Posted by ScottO
Hi Folks,

I'm having difficulty making progress on the following proof, and I'm hoping for some hints to help me along.

Let $\displaystyle b$ denote a fixed positive integer. Prove the following statement by induction: For every integer $\displaystyle n \geq 0$, there exist non-negative integers $\displaystyle q$ and $\displaystyle r$ such that

$\displaystyle n = qb + r, 0 \leq r < b$.

So I can state the assertion and show the initial case is true:
$\displaystyle A(n): n = qb + r, 0 \leq r < b$

$\displaystyle A(0): 0 = 0 \cdot b + 0, n = q = r = 0$

I've tried picking a value for b and creating a table of related values for q and r as n counts up from 0. But I can't figure out how to express the patterns I see to form a usable general case and inductive step.

Again, just looking for hints right now to help get me out of the rut I'm in.

Thanks,
Scott
this is a wrong subforum for this question. you should've posted it in pre-algebra or discrete math subforum. anyway, suppose $\displaystyle n=qb+ r,$ where $\displaystyle q \geq 0, \ 0 \leq r < b.$

thus $\displaystyle n+1=qb + r + 1.$ we have $\displaystyle r + 1 \leq b$ because $\displaystyle r < b.$ now if $\displaystyle r + 1 < b,$ then we're done and if $\displaystyle r+1=b,$ then $\displaystyle n+1=qb+b=(q+1)b,$ done again!