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Is it true that $a \bmod n\equiv (a \bmod n)\bmod n$?

Is is possible to show intuitively why ?

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  • $\begingroup$ The answer is yes. One explanation is simply that for any $x$ between $0$ and $n-1$, $x \bmod n = x$. $\endgroup$ Oct 24, 2019 at 10:04

1 Answer 1

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$a\bmod n$ is by definition the congruence class of the remainder in the division of $a$ by $n$ .

$(a \bmod n)\bmod n$ is the congruence class of the remainder in the division of this remainder by $n$. But, as this remainder is less than $n$, the latter remainder is the first remainder itself.

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  • $\begingroup$ How in the world do you have a reputation of 1. Is SE joking right now? $\endgroup$
    – Joker
    May 25, 2022 at 19:20

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