**This is the first note in the series Number Theory: Divisibility. All numbers involved in this note are integers, and letters used in this note stand for integers without further specification.**

Numbers involved in this note are integers, and letters used in this book stand for integers without further specification.

Given numbers \(a\) and \(b\), with \(b \neq 0\), if there is an integer \(c\), such that \(a=bc\), then we say \(b\) divides \(a\), and write \(b \mid a\). In this case we also say \(b\) is a *factor* of \(a\), or \(a\) is a *multiple* of \(b\). We use the notation \(b \nmid a\) when \(b\) does not divide \(a\) (i.e., no such \(c\) exists).

Several simple properties of divisibility could be obtained by the definition of divisibility (proofs of the properties are left to readers).

\((1)\) If \(b \mid c,\) and \(c \mid a,\) then \(b \mid a,\) that is, divisibility is transitive.

\((2)\) If \(b \mid a,\) and \(b \mid c,\) then \(b \mid (a \pm c),\) that is, the set of multiples of an integer is closed under addition and subtraction operations.

By using this property repeatedly, we have, if \(b \mid a\) and \(b \mid c\), then \(b \mid (au+cv)\), for any integers \(u\) and \(v\). In general, if \(a_1,a_2,\cdots,a_n\) are multiples of \(b\), then \[b \mid (a_1+a_2+\cdots+a_n).\]

\((3)\) If \(b \mid a\), then \(a=0\) or \(\lvert a \rvert \geq \lvert b \rvert\). Thus, if \(b \mid a\) and \(a \mid b\), then \(\lvert a \rvert = \lvert b \rvert\).

Clearly, for any two integers \(a\) and \(b\), \(a\) is not always divisible by \(b\). But we have the following result, which is called the division algorithm. It is the most important result in elementary number theory.

\((4)\) (The division algorithm) Let \(a\) and \(b\) be integers, and \(b > 0\). Then there is a unique pair of integers \(q\) and \(r\), such that \[a=bq+r \hspace{2mm} \text{and} \hspace{2mm} 0 \leq r < b.\]

The integer \(q\) is called the (incomplete)

quotientwhen \(a\) is divided by \(b\), \(r\) called theremainder. Note that the values of \(r\) has \(b\) kinds of possibilities, \(0,1,\cdots,b-1\). If \(r=0\), then \(a\) is divisible by \(b\).It is easy to see that the quotient \(q\) in the division algorithm is in fact \(\lfloor\frac{a}{b}\rfloor\) (the greatest integer not exceeding \(\frac{a}{b}\)), and the heart of the division algorithm is the inequality about the remainder \(r\): \(0 \leq r < b\). We will go back to this point later on.

The basic method of proving \(b \mid a\) is to factorize \(a\) into the product of \(b\) and another integer. Usually, in some basic problems this kind of factorization can be obtained by taking some special value in algebraic factorization equations. The following two factorization formulae are very useful in proving this kind of problems.

\((5)\) If \(n\) is a positive integer, then \[x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\cdots+xy^{n-2}+y^{n-1}).\]

\((6)\) If \(n\) is a positive odd number, then \[x^n+y^n=(x+y)(x^{n-1}-x^{n-2}y+\cdots-xy^{n-2}+y^{n-1}).\]

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestThanks much for posting this. I would love if you could name the book you had referred to in the beginning of this note!

Log in to reply

I'm not sure if I want to reveal it because I'm scared I'm taking a little too much information :D

Log in to reply

I like the fact that although it's meant to be a comment with a negative connotation, there's still a smiley icon at the back which doesn't really fit the sentence (lol). Also, I think I have an idea which book he may have

referredto...Log in to reply

Log in to reply

Log in to reply

Log in to reply

Uh-Oh!

Log in to reply