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Is this true? 2 does not divide 1? Does a proof exist for this? Is it because the outcome will be a fraction and that is not allowed in discrete mathematics? I am new to this.

Anymee
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  • We define "divides" in number theory precisely for the integers. So if $a,b$ are integers, we say $a\mid b$ if there is an integer $c$ such that $ac=b$. – Thomas Andrews Sep 05 '17 at 15:43
  • $1 \div 2 = 0.5$, which is not an integer. Look at your definition of what "$m$ divides $n$" means to see whether this matters – Henry Sep 05 '17 at 15:44
  • (We define divisibility in other contexts, but it isn't an interesting property on the rationals and the reals, precisely because it is trivial.) – Thomas Andrews Sep 05 '17 at 15:44
  • If 2 divides 1, you can write $1=2k$ where $k$ is an integer. Can you do this? – Eleven-Eleven Sep 05 '17 at 15:44
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    It's not a matter of what's "allowed in discrete mathematics." Instead it's a matter of the definition of the concept of a divisor. We say an integer $a$ divides an integer $b$ if there exists an integer $k$ such that $ak=b$. – symplectomorphic Sep 05 '17 at 15:44
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    And you can actually prove it, from Peano axioms, if necessary. Basically, you want to for $\forall n:2n\neq 1$. This can be done inductively for $n$ a non-negative integer using the Peano axioms, but it is tedious. – Thomas Andrews Sep 05 '17 at 16:04
  • @ThomasAndrews: Supposing that you use PA− plus induction, I think an easier way would be to first prove $\forall n ( n = 0 \lor 2n > 1 )$. – user21820 Sep 09 '17 at 17:26

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If you carefully write out the definition, to say that $2$ divides $1$ means that there is an integer $m$ for which

$$1 = 2m.$$

However, there is no solution to this equation in the integers (e.g. division by $2$ requires that $m = 1/2$, which is non-integral). The key point in the above definition is the bolded part.

  • If the downvoter would care to leave a comment so that I can improve this post, I would greatly appreciate it. –  Sep 05 '17 at 17:31
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    I can't read the downvoter's mind, since I'm not the one, but I disagree with "division by 2 requires that m = 1/2, which is non-integral". Division by 2 cannot even be defined as an operator on natural numbers, much less requires anything. Unless one starts with an axiomatization of say a bigger structure as in fleablood's answer, where one can do division, one would have to prove via induction that your statement is true, and cannot simply invoke the idea that m must be 1/2 or whatever. – user21820 Sep 09 '17 at 17:23
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How deep do you want to go? First we need to define integers. And then we need to define $1$ and $2$ and what "$a$ divides $b$" mean.

"$a$ divides $b$" means that there is a number $k$ so that $a*k = b$. This is usually taken in context and it is usually assumed that $k$ is an integer. (It's trivial if $a$ and $b$ are elements of a field, such as the rational, real, or complex numbers and $a \ne 0$ and $k$ can be any element of the field-- the $k = \frac ba$ will always be the case that $a*k = b.)

So "$2$ divides $1$" means that there is an integer $k$ so that $2k = 1$. There is not.

Why not? And is there a proof? Well, that really depends an the axioms and definitions for your system.

If it were up to me.... well, I'd do something of the nature of taking it as axiomatic that our numbers form an Ordered Field of which the Integers are a subset. This means among other things that $0 < 1$ and that $2 = 1+ 1 > 1$, and whenever $k > 0; a<b$ we have an axiom that $ak > bk$. This means that if $2k = 1$ then $0 < k < 1$. (if $k < 0$ then $2>0$ would mean $2k < 2*0 = 0 < 1$. If $k > 1$ then $2k > 2*1 =2 > 1$). Part of the definitions of Integers would be an axiom that $1$ is the smallest integer that is greater than $0$. So $k$ being between $0$ and $1$ can not be an integer.

That's a little breezy but I hope it gives the idea.

fleablood
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For $a$ to divide $b$ the following must hold (by definition of "divides"): $$b = a \cdot m, m \in \mathbb{Z}$$ When $b=1$ and $a=2$ there is no such m.

Zonko
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