Sum question

1. Apr 19, 2015

moriheru

I have a question, in the field of number theory (Hardy and Wright chapter 9 representation of numbers by decimals) concerning the prove by contradiction of the statement:
If Σ1 an/10n Σ1=bn/10n then an and bn must be equivalent, for if not then let aN and bN be the first pair that differ then aN-bN≥1, where an and bn are decimals. It follows that

Σ1 an/10n1 bn/10n≥1/10NN+1 (a-b)/10n≥1/10N -∑N+1 9/10n=0

the proof goes on but I only need to know why the sums are greater or smaller than each other...

Thanks for any clarifications, if this is to vague I shall try and add information.

2. Apr 19, 2015

Staff: Mentor

You have a typo in the line above, with '=' between the summation sign and the general term in the sum.
As an aside, what you have is very tedious to type, with all the SUB and SUP tags. LaTeX is much easier to type. See https://www.physicsforums.com/help/latexhelp/.
By assumption here $\sum_{n = 1}^{N - 1} a_n = \sum_{n = 1}^{N - 1} b_n$. The series' terms are different starting at aN and bN. Since all of the terms in both series are decimal digits (i.e., 0, 1, 2, 3, ..., 8, 9), if $a_N \neq b_N$, then $a_N - b_N$ has to be at least 1. There is a tacit assumption here that $a_N > b_N$. If this isn't so, you can write $b_N > a_N$ and carry on from there.