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nooby question.

  1. I heard many times that 0 is a pair number.
  2. I'm fairly sure that the definition of pair is multiple of 2.
  3. Yet I heard too that multiples of a prime number p are only 1 and p, therefore excluding 0.

So in the above contradiction, where is/are the false argument/s? I feel like despite what I heard, 0 is not a pair number or maybe only in informatics (that I started studying) which is still weird (the divergence I mean)

Thanks for your time.

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    $\begingroup$ I think by "pair" you mean even. $0$ is in fact an even number. The definition of even (in $\Bbb Z$) is a multiple of two. (More generally I think we can define it through parity maps, homomorphisms from groups of units into $\Bbb Z/2$). There are many more multiples of a prime $p$ than just $1$ and $p$. All of $$\cdots,~-3p,~-2p,~-p,~0,~+p,~+2p,~+3p,~\cdots$$ are integer multiples of $p$ (note $1$ isn't among this list unless $p=\pm1$, not prime). Rather, the only (positive) divisors of a prime $p$ are $1$ and $p$. $a$ is a divisor of $b$ if and only if $b$ is a multiple of $a$. $\endgroup$
    – anon
    Nov 29, 2012 at 19:37
  • $\begingroup$ Finally, every integer is a divisor of $0$, and $0$ is a multiple of every integer, because for any $x\in\Bbb Z$ we have $0=0\cdot x$ is an integer times $x$ i.e. an integer multiple of $x$. $\endgroup$
    – anon
    Nov 29, 2012 at 19:39
  • $\begingroup$ Yeah even, sorry I'm not English my vocabulary is messed up. And in that regard I think I also mixed up multiple with "divisor" (if that's a word). $\endgroup$
    – sinekonata
    Nov 29, 2012 at 23:59

2 Answers 2

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Multiples of a prime number (or of any other number) are $\ldots,-2p,-p,0,p,2p,\ldots$ while the divisors of a prime number are only $\pm1,\pm p$ and the positive divisors are just $1,p$. So there is no contradiction.

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  • $\begingroup$ So yes there was a contradiction among my statements and the argument that was incorrect was the 3rd one. I mixed up multiple and divisor as you pointed out. Thanks, I really am a noob... ^___^ $\endgroup$
    – sinekonata
    Nov 30, 2012 at 0:03
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There is no contradiction.

  • We say that $k$ is even if there is some $n$ such that $k=2n$. Zero is even because $0=2\cdot 0$.

  • We say that $p$ is a prime number if whenever $p=m\cdot n$ and $m\neq p$ then $m=1$. Zero is not a prime number because $0=3\cdot 0$ and $3\neq 0$ but $3\neq 1$.

While zero itself is not a divisor of any number, in fact we have that any number divides zero. Therefore to the question in the title, yes: zero is a multiple of any number.

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