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Disquisitiones Arithmeticae

I was reading the beginning of the wonderful Disquisitiones Arithmeticae by Carl Gauss.

Unfortunately i got stuck at the very beginning only!!(at the first theorem)

It goes like this: $$\textit{Let m successive integers a,a+1,a+2.....,a+m-1 and another integer,A, be given,then one and only one of these m integers}$$ $$\textit{will be congruent to A modulo m.}$$

Gauss gave his proof not very clear(maybe it was the mistake of the translator).

Does anyone have a good proof? $$\textbf{[For if you want to refer to Gauss's proof(unclear), please tell me]}$$

Note by Anandmay Patel
5 months, 4 weeks ago

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" Any number $$A$$ will be $$0$$ or $$1$$ or $$2$$................or $$m-1$$ modulo $$m$$; which account to a total of $$m$$ different possibilities.

Also; in a series of $$m$$ consecutive integers; two of them cannot have the same result modulo $$m$$ since that would make the difference between them a multiple of $$m$$ and hence total numbers in the series will become at least $$m+1$$. · 5 months, 4 weeks ago

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(Continued)

But, since no two share same result modulo $$m$$ and there are $$m$$ numbers and only $$m$$ possibilities hence; each possibility must occur exactly once and hence; all $$A$$ will have one and exactly one number in the series which is exactly the same modulo $$m$$." · 5 months, 4 weeks ago

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I got it.Nice explanation;however,i am not getting what do you mean by the same result mod m and exactly same mod m. We are supposed to say with respect to A.Do you mean with respect to A while saying it? · 5 months, 4 weeks ago

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I havent yet started modular arithmetic formally so I dont know the exact terns for that;
Same result modulo $$m$$ means leaves the same remainder when divided by $$m$$
Exactly same mod $$m$$ means leaves the same remainder as A leaves when divided by $$m$$ · 5 months, 4 weeks ago

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No problem.Thanks · 5 months, 4 weeks ago

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