Is it true that $a \bmod n\equiv (a \bmod n)\bmod n$?
Is is possible to show intuitively why ?
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The answer is yes. One explanation is simply that for any $x$ between $0$ and $n-1$, $x \bmod n = x$. – Ben Grossmann Oct 24 '19 at 10:04
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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.
Bernard
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