0
$\begingroup$

Basic question about Factors and Multiples:

When we define factors as

For an integer $x$, any integer which is completely divisible by $x$ is a factor of $x$.

and multiples as

For an integer $y$, any integer $z$ whose product with $y$ is an integer (or is it whole number? I am confused :s) is a multiple of $y$.

are we talking about negative integers and $0$ as well? or are we just talking about the positive integers or just whole numbers or just natural numbers? I am confused here.

I mean are there negative factors and multiples as well? and what about $0$?

PS: Please tweak my definitions if you find a mistake in them, but please keep the jargon simple if you do so. Thank you.

$\endgroup$
  • 2
    $\begingroup$ Your definition of "factor" is wrong, what is written there is the definition of "multiple" (of $x$). $\endgroup$ – Alex M. Sep 1 '15 at 9:13
  • $\begingroup$ @AlexM. Ooh! Thanks for pointing out, I need to think more about it. $\endgroup$ – Solace Sep 1 '15 at 9:26
2
$\begingroup$

We say that two numbers are associate if you can get one multiplying the other for a unit (that is to say, an invertible element).

If you are working on $\mathbb N$, then the only invertible element is $1$, so you only consider the positive multiples of a number.

If you are working on $\mathbb Z$, we have that $1$ and $-1$ are the only units (invertible elements), so we could consider both positive and negative multiples of a number. Usually, we say that the factors of a number are unique up to associates (for example, $6=2\times 3$ and $6=(-2)\times(-3)$, but $2$ and $-2$ (respectively $3$ and $-3$) are associates).

If you are working on a field (that is, every nonzero element is invertible), for example $\mathbb Q$, then every number is associate, factor and multiple of another one. For example, $6=\frac 6 5 \cdot 5$, so $6$ is a multiple of $5$ in $\mathbb Q$. As there are no prime numbers in a field, there is not a unique factorization.

$0$ is a multiple of any number in any ring. Also, there can exist zero divisors in some rings, such in $\mathbb Z _4$, where $2\times 2 = 0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.