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Can we define subtraction in $\mathbb{N}$ like this:

$\forall a,c\in \mathbb{N} (0\notin \mathbb{N})$, $$\exists b\in \mathbb{N}:a+b=c\iff b=c-a$$

I am acquainted with the fact that for $a=2$ and $c=1$ there's no $b\in \mathbb{N}$ such that $2+b=1$ can mean $b=1-2$, would this be a reason for this definition to not work?

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    Well if $b=1-2$, then $b\notin \Bbb{N}$, which contradicts your hypothesis that such a $b$ exists. – tmaj Aug 04 '21 at 17:14
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    You can define it, as you've noted. It's just that $a-b$ isn't always defined. We say that the naturals are not closed under subtraction. – Rushabh Mehta Aug 04 '21 at 17:15
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    You are defining subtraction of $a$ from $c$, so $c>a$ must be one of your hypotheses since you are carrying out subtraction on a set of natural numbers. – Aman Kushwaha Aug 04 '21 at 17:16
  • @tmaj haven't picked on that. Thanks! –  Aug 04 '21 at 17:21
  • @tmaj I'm confused because of what Don Thousand said so, can't I define it that way? –  Aug 04 '21 at 17:24
  • @AmanKushwaha So what you're telling me is that I can't define it that way since $c>a$ is not part of my hypothesis and without it $b-a$ is not natural? Meaning that whenever defining an operation I must make sure my definition ensures closure? –  Aug 04 '21 at 17:26
  • @HannyBoy Ensuring closure is not my idea. It's you who wrote there always exists such a $b \in \mathbb N$ – Aman Kushwaha Aug 04 '21 at 17:28
  • @HannyBoy if you want to change the quantifiers, you can. Adding a condition of c>a is another thing you can do. There's nothing wrong in defining "subtraction" on the set of natural numbers such that the set of natural numbers is closed under "subtraction". – Aman Kushwaha Aug 04 '21 at 17:42
  • Another approach I've seen is to define some way to extend the subtraction function to a function on all of $\mathbb{N} \times \mathbb{N}$ - for example, define $a - c$ to be the unique $b$ such that $a = c + b$ if $a \ge c$, whereas you define $a - c$ to be 0 if $a < c$. That has the advantage of being able to freely write expressions without worrying about whether some "preconditions" are satisfied. On the other hand, you do still need to have appropriate hypotheses for theorems - for example, with this definition, $a - (b - c) = (a + c) - b$ does not necessarily hold for all $a,b,c$. – Daniel Schepler Aug 04 '21 at 18:06

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So first of all one can prove that $$ \forall a \,b \in \mathbb{N} : b \leq a \implies ~\exists c \in \mathbb{N} : b + c= a $$ and that this number $c \in \mathbb{N}$ is unique. Therefore in the case $b \leq a$ we can set $a - b := c$. The subtraction $-$ defined in this way is not defined on the full domain $\mathbb{N} \times \mathbb{N}$, but only on $\{(a,b) \mid b \leq a\} \subseteq \mathbb{N} \times \mathbb{N}$. If you want a function that is defined on the whole domain $\mathbb{N} \times \mathbb{N}$, you can define the so called truncated subtraction by $$ a \ominus b := \begin{cases} 0 & \text{if }~ b > a \\ a - b & \text{if }~ b \leq a \end{cases} $$

Léreau
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  • May you walk me through the meaning of "not defined" in an intuitive way, please? :) –  Aug 04 '21 at 18:20
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    @HannyBoy There exists no $c \in \mathbb{N}$ such that $2 + c = 1$. So subtraction can not be defined / there is no value that makes sense for the tuple $(1,2) \in \mathbb{N} \times \mathbb{N}$. The only tuples it is defined on are in the subset I indicated. – Léreau Aug 04 '21 at 19:16