I have a homework problem, the solution has been published and I understand it and am not looking for that. Rather I've hit an impassé that contradicts the solution and I would be very grateful if you can point out where I'm making the mistake.
The problem: Decimal representation of real numbers For $ b \in \Bbb{N} $ and $ b >= 2$ and $ x \in [0, 1) $ We define $x_1, x_2, .... $ recursively as: $x_1 = max ({k \in \Bbb{Z}\space | \space k <= bx} )$
and $$ x_n = [b^n(x - \sum_{k=1}^{n-1} \frac {x_k}{b^k})]$$
So there are several parts of the problem and in one we have to prove an inequality by induction
Here is where I failed because I ended up with $x_{n+1} = 0 $ somehow:
$$x_{n+1} = [b^{n+1}(x - \sum_{k=1}^{n} \frac {x_k}{b^k})] = \space [b^{n+1}(x - \sum_{k=1}^{n-1} \frac {x_k}{b^k} \space - \frac{x_n}{b^n})]$$
$$ = [b^{n+1}(x - \sum_{k=1}^{n-1} \frac {x_k}{b^k}) \space - \space b^{n+1}\frac{x_n}{b^n}]$$ $$ = [b^{n+1}(x - \sum_{k=1}^{n-1} \frac {x_k}{b^k}) \space - \space b{x_n}]$$ $$ = [bb^{n}(x - \sum_{k=1}^{n-1} \frac {x_k}{b^k}) \space - \space b{x_n}]$$ $$ = [b(b^{n}(x - \sum_{k=1}^{n-1} \frac {x_k}{b^k})) \space - \space b{x_n}]$$ grouping the elements $$ = [b(x_n) \space - \space b{x_n}]$$ $$ = 0 $$
which is obviously wrong? Thank you very much for your help. Usually I have an idea where something could be wrong, but this time I'm completely stumped.
The problem is that you substitute $x_n$ for $(x - \sum_{k=1}^{n-1}\frac{x_k}{b^k})b^n$. Notice that you are neglecting the integer part.