Proving Equivalence of Decimals in Number Theory

[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^∞ a_n/10ⁿ Σ₁=^∞b_n/10ⁿ then a_n and b_n must be equivalent, for if not then let a_N and b_N be the first pair that differ then a_N-b_N≥1, where a_n and b_n are decimals. It follows that

Σ₁^∞ a_n/10ⁿ-Σ₁^∞ b_n/10ⁿ≥1/10^N -Σ_N+1^∞ (a-b)/10ⁿ≥1/10^N -∑_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.

 

[Mark44]

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^∞ a_n/10ⁿ Σ₁=^∞b_n/10ⁿ then a_n

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/.

and b_n must be equivalent, for if not then let a_N and b_N be the first pair that differ then a_N-b_N≥1, where a_n and b_n are decimals. It follows that

Σ₁^∞ a_n/10ⁿ-Σ₁^∞ b_n/10ⁿ≥1/10^N -Σ_N+1^∞ (a-b)/10ⁿ≥1/10^N -∑_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...

By assumption here $\sum_{n = 1}^{N - 1} a_n = \sum_{n = 1}^{N - 1} b_n$. The series' terms are different starting at a_N and b_N. 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.