Let $p, N$ be positive integers with $p$ divides $N$.
Why for every integer $X$, $[[X \bmod N] \bmod p ] = [X \bmod p ]$?
And how do I show that $[[X \bmod p ] \bmod N]$ need not equal $[X \bmod N]$?
Modulo arithmetic (modulo of modulo)
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number-theory
elementary-number-theory
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0For the first question, that is by definition. Think about what $a + bN$ is modulo $N$ and $p$. For the second, just find a counterexample. With small enough $p$ and $N$ you can simply look at all the possibilities and find one, e.g., try finding a counterexample with $p=2$ and $N=4$. – 2017-01-20
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1@Alex I suggest you turn your comment into an answer. I'm getting a mixed message from you saying "by definition" and also saying "think about it" so you might want to revise that in the process. – 2017-01-20
1 Answers
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For the first question, that is by definition. What is $a + bN$ modulo $N$, and modulo $p$?
For the second, just find a counterexample. (If it's not true for a particular choice of $p$ and $N$, it's not true.) With small enough $p$ and $N$ you can simply look at all the possibilities and find one, e.g., try finding a counterexample with $p=2$ and $N=4$.