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I came across this notation in a solution manual. It appears in the solution to Exercise 1.2.2. as $3|2^p$. Any idea what it could mean?

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    $\begingroup$ Generally it means $3$ divides $2^p$ $\endgroup$ Commented Sep 6, 2021 at 17:14
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    $\begingroup$ I think you should put that as the answer $\endgroup$
    – LL 3.14
    Commented Sep 6, 2021 at 17:16
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    $\begingroup$ It can also be seen as "such that" in some places too though. But in the context they said, it definitely means "divides" as said. $\endgroup$
    – Acyex
    Commented Sep 6, 2021 at 17:23

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Generally $a \mid b$ means $a$ divides $b$ where $a,b\ne 0$ are integers. Since $\frac{b}{a}$ is a rational number it is sometime cumbersome to say that $\frac{b}{a}$ is an integer even though it means the same thing. Even we can write in words that " $b$ is divisible by $a$" but $a \mid b$ convention is lot more easy to follow.

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It means "divides", i.e. "3 divides $2^p$. Formally:

There exists an integer $k\in\Bbb Z$ such that $3\cdot k = 2^p$.

However, depending over which set or ring you are acting, divisibility is defined the same (in a broader context) but might spell out differently in the details, for example when you are over a ring of polynomials like $\Bbb Z[x]$ or over the ring of integers of some algebraic number field like $\Bbb Z[\sqrt{-1}]$.

By the way, the symbol for "does not divide" is $\nmid$.

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    $\begingroup$ You meant to say $k \cdot 3 = 2^p$. $\endgroup$
    – Umberto P.
    Commented Sep 6, 2021 at 17:25
  • $\begingroup$ Ooops, yes of course. It's fixed. $\endgroup$ Commented Sep 6, 2021 at 17:39

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