Elementary Number Theory – Difference Between Divisors and Proper Divisors

elementary-number-theory

I'm really confused about these two. For example if $n = 6$, then:

Divisors: $2, 3$
Proper Divisors: $1, 2, 3, 6$

Is it right?

Update
From Elementary Number Theory and Its Application by Kenneth H. Rosen 6th edition, page 256:

Because of certain mystical beliefs, the ancient Greeks were interested in those integers that are equal to the sum of all their proper positive divisors. Such integers are called perfect numbers.

Example:
$\sigma(6) = 1 + 2 + 3 + 6 = 12$, we see that $6$ is perfect.

Thanks,

Best Answer

I think most of the time the convention would be: divisors = $\{1,2,3,6\}$ and proper divisors= $ \{1,2,3\} $.

For instance 6 is perfect because $\sigma(n)=2n$ where σ is the sum of its divisors or $s(n)=n$ where $s$ is the sum of its proper divisors . (Note that the notation $s$ may not be completely standard.)

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I believe you can find the value of $(3|p)$ without finding $p\equiv 5\pmod{12}$, by use of the second supplement to quadratic reciprocity. Recall that $$ \left(\frac{2}{p}\right)=(-1)^{(p^2-1)/8}. $$ Since $p\equiv 2\pmod{3}$, you then have $(p|3)=(2|3)=-1$. So altogether, $$ \left(\frac{3}{p}\right)\left(\frac{p}{3}\right)=\left(\frac{3}{p}\right)\left(\frac{2}{3}\right)=\left(\frac{3}{p}\right)(-1)=(-1)^{(p-1)(3-1)/4}=(-1)^{(p-1)/2}=1 $$ This implies $(3|p)=-1$.

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Here is something about period length but nothing about pre-period length.

The first question is whether a fraction is periodic or not. If it's not periodic it's of the form $x/b^r$ for example 36.542 = 36542/1000 = 18271/500. The contrapositive of this tells us that if a fraction is of the form $x/y$ with $b \not | y$ then the expansion is periodic, for example in base 10: $10 \not | 73$ so 15/73 is periodic: $0.2054794520547945\ldots$.

It suffices to consider only fractions between 0 and 1 so let's do that. How do we compute the base $b$ expansion of a fraction? For example how do we compute $0.14285714285714285\ldots$ from $1/7 = p/q$ multiply both sides by 10 and throw away the remainder to get the first digit $d = \text{floor}(\frac{b p}{q})$. To get the rest of the digits we just do the same process on $r/q$, where $r$ is the remainder of the division $\frac{b p}{q}$. Here is an example of this

and if you read the digits column we have the correct digits.

So how can we use this idea of repeatedly applying the division algorithm to find the period length? It seems like a difficult problem so for a warm up lets just find out when the period has length 1, for example $1/3 = 0.333\ldots$ has that property. Why? Well it's obvious from the table

p q d r
1 3 3 1
1 3 3 1
...

It's got period 1 because $p = r$, the remainder of dividing $bp$ by $q$ is $p$, $bp \equiv p \pmod q$. In fact it turns out this solves the problem completely - 1/7 is periodic with period 6 in base 10 because it is periodic with period 1 in base 10^6! So what is the smallest $r$ such that $b^r p \equiv p \pmod q$? It's just that ord thing that was mentioned.

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