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I am currently reading the paper "PRIMES is in p" and have come across some notation that I don't quite understand in this following sentence

Consider a prime $q$ that is a factor of $n$ and let $q^k || n$. Then...

What does the notation $q^k || n$ mean here?

The full paper can be found here and the notation described above is used in the proof on page 2

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  • $\begingroup$ Well I use this as a symbol of representing concatenation, for which $$a\mid\mid b = \left\{10^na + b : b\text{ has $n$ digits}\right\}.$$ but since we have $a = q^k$ and $b = n$ with $q$ being raised to an exponent $k$, then this notation would be different from how I know it to be defined, for all $k > 1$. $\endgroup$
    – Mr Pie
    Commented Feb 19, 2018 at 15:19

3 Answers 3

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Typically, that $q^k$ divides $n$ but no higher power of $q$ does.

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  • $\begingroup$ Another way of writing it is $v_q(n)=k$ $\endgroup$ Commented Feb 19, 2018 at 15:16
  • $\begingroup$ @GTonyJacobs why is $v$ included? How do we even say this? With the two lines $\mid\mid$ then pursuant to one of the answers above, this is read as divides exactly but how would we read the equation, $v_q(n) = k$? $\endgroup$
    – Mr Pie
    Commented Feb 19, 2018 at 15:20
  • $\begingroup$ The '$v$' stands for 'valuation'. For a prime $p$, the $p$-adic valuation of a number is a measure of how many powers of $p$ go into the number. What I wrote above says that the $q$-adic valuation of $n$ is $k$. Example: odd integers have a $2$-adic valuation of $0$. Integers such as $2,6,10$, which are $2$ times an odd integer, have a $2$-adic valuation of $1$. A number such as $96$, which we can divide by $2$ five times before reaching an odd number, would have a $2$-adic valuation of $5$. $\endgroup$ Commented Feb 19, 2018 at 15:24
  • $\begingroup$ It's a nice concept for talking about how the integers interact with a specific prime, and it extends to all rational numbers: $\frac54$ has a $2$-adic valuation of $-2$. $\endgroup$ Commented Feb 19, 2018 at 15:25
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It means "divides exactly" in the sense that $q^{k+1}$ does not divide n.

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It means that $k$ is the higher power that divides $n$. In other words: $$q^k\|n\iff q^k|n\wedge q^{k+1}\nmid n$$

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