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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.

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    $\begingroup$ Your definition of "factor" is wrong, what is written there is the definition of "multiple" (of $x$). $\endgroup$
    – Alex M.
    Sep 1, 2015 at 9:13
  • $\begingroup$ @AlexM. Ooh! Thanks for pointing out, I need to think more about it. $\endgroup$
    – Solace
    Sep 1, 2015 at 9:26

1 Answer 1

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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 $2\times 2 = 0$.

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