What I understand is that the only difference between plus and minus is direction.

I've never understood this: That +1 + +1 is +2, why is -1 - -1 not -2?

(In the first, we are moving from right to the right to the right, so we get right. In the second, we are moving from left to the left to the left, so why don't we get left?)

What's a logically/ philosophically logical/ true explanation? What are the base axioms that everything is defined upon?

Indeed, why does a negative(left) number times a positive(right) number equal negative(left)? Shouldn't it have an undefined direction with respect to left–right?

  • $\begingroup$ Minus times positive is simpler so I answer here. If you are dealing with integers, multiplying by positive integers actually means "repeating addition for $N$ times" so you are doing $(-A)+(-A)+(-A)+(-A)+\dots$ and it will give you a negative number since it persistently go to negative direction. $\endgroup$ – ElfHog Feb 2 '18 at 4:08

The $\,\color{blue}{+}\,$ or $\,\color{blue}{-}\,$ sign attached to the number is part of the number itself. The $\,\color{red}{+}\,$ and $\,\color{red}{-}\,$ signs between the numbers are operations, even though (confusingly enough) we use the same symbols for both.

I've never understood this: That +1 + +1 is +2,

In light of the above, this is the same as $\,(\color{blue}{+1}) \color{red}{+} (\color{blue}{+1}) = (\color{blue}{+2}) = 2 \cdot (\color{blue}{+1})\,$.

why is -1 - -1 not -2?

Because replacing the number $\,(\color{blue}{+1}) \to (\color{blue}{-1})\,$ in the above gives $\,(\color{blue}{-1}) \color{red}{+} (\color{blue}{-1}) = 2 \cdot (\color{blue}{-1})=(\color{blue}{-2})\,$.

In fact, both are particular cases of the general identity $\,a + a = 2 \cdot a\,$, the first one for $\,a=\color{blue}{+1}\,$, and the second one for $\,a=\color{blue}{-1}\,$. In the same way, the identity $\,a-a=0\,$ gives $\,(\color{blue}{+1}) \color{red}{-} (\color{blue}{+1}) = 0\,$ and $\,(\color{blue}{-1}) \color{red}{-} (\color{blue}{-1}) = 0\,$ for $\,a=\color{blue}{+1}\,$ and $\,a=\color{blue}{-1}\,$, respectively.

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    $\begingroup$ Do you mean that just as <bank> has two distinct meanings, the <+> in <(+1)> is distinct in meaning from the middle <+> in <(+1) + (+1)>? $\endgroup$ – Pacerier Feb 2 '18 at 4:35
  • $\begingroup$ @Pacerier That's precisely what I meant to say. The convention for addition is that $\,+a = a\,$ and $\,-a\,$ is the unique number such that $\,a + (-a) = 0\,$. We could just as well use $\,a^{\%\%}\,$ and $\,a^{\#\#}\,$ as notations, instead, and all algebra would still work out the same, though it would arguably look worse ;-) $\endgroup$ – dxiv Feb 2 '18 at 4:40
  • $\begingroup$ Hm, that I see that 3 can mean either directionless 3 or +3; and +3 can mean only +3, I understand that <(-2)*(3)> equals <(-6)>, but why does <(-2)*(+3)> equal <(-6)>? Shouldn't it be undefined? $\endgroup$ – Pacerier Feb 2 '18 at 5:30
  • $\begingroup$ Along the same line as above, I could argue that $\,6+(-2)\cdot(+3)$ $=(+2)\cdot(+3)+(-2)\cdot(+3)$ $=(+3)\cdot\big((+2)+(-2)\big)=(+3) \cdot 0 = 0\,$, and therefore $\,(-2)\cdot(+3) =(-6)\,$ by the definition of the additive inverse. For (many) more ways to look at it, see Why is negative times negative = positive? $\endgroup$ – dxiv Feb 2 '18 at 5:33
  • $\begingroup$ I see it leads to the question of that it's obvious that <(+2)*(3)> equals <(+6)>, but why does <(+2)*(+3)> equal <(+6)> instead of undefined? That (+2)*(+3) is not an axiom, what is its relation to what axiom(s)? $\endgroup$ – Pacerier Feb 2 '18 at 5:51

The result of the subtraction $x - y$ of integers $x$ and $y$ is defined as the unique integer $z$ which satisfies $x = y + z$.

The $-$ sign in $-1$ is a syntactic convention to denote the inverse of 1 for addition. Using this notation in the definition above gives $0 - 1 = (-1)$, which a posteriori justifies the choice of notation.

When the sign of an integer is important, another convention is to write $+x$ for positive integers $x$. This other syntactic convention is justified by (part of) the definition of addition: $0 + x = x$.

It is tempting to interpret every $+$ or $-$ sign as denoting an addition or subtraction. To make this work, it is necessary to add a "sign rule" to deal with the (purely syntactic cases) when two sign symbols appear next to each other. That way, we can forget about the previous syntactic conventions and compute only at the syntactic level to reach the correct result.

  • $\begingroup$ Hi nicolas, what are your thoughts on math.stackexchange.com/questions/2632224/… ? $\endgroup$ – Pacerier Feb 2 '18 at 7:41
  • $\begingroup$ @Pacerier The sign rule I mention in the last part of my answer is only meant to rewrite expressions with $+$ and $-$ symbols next to each other. It is not about integer multiplication. $\endgroup$ – Nicolas Pelletier Feb 2 '18 at 12:53

I will try to answer philosophically (with math terms).

"Minus" means "retrieving", or "to add opposite things" (adding additive inverse). So if you are doing $A - A$, it means you want to add "inverse of $A$" to $A$ [written as $A+(-A)$], and philosophically it is zero (nothingness) since you are adding oneself (A) to its inverse.

Think it in another way, if you have 2 apples, and you eat both, it is $2-2$ and it becomes none. So if you loan a person 1 dollar ($-1$), and the person "cancel" (minus) the loan [by returning you money], then the mathematical equation is indeed $-1 - (-1) = 0$ since the person has cleared all loans by returning money to you.

  • $\begingroup$ That + is defined as "add in direction1", why does + + not equal "add the act of adding in direction1"? why is + + not undefined? Indeed, why does 2 + +2 = 2 + 2 instead of undefined? $\endgroup$ – Pacerier Feb 2 '18 at 4:33
  • $\begingroup$ @Pacerier Good question, hmm let me see. If you mean $2 (++) 2$ then I guess it is undefined since it really means "adding the act of adding". However, normally when we write $2++2$ we do mean $2+(+2)$ and this means that, there is originally a "$+2$", and we adding it to $2$ (which is in fact also $+2$, just with the sign hidden as in convention), so it generates $4$. $\endgroup$ – ElfHog Feb 2 '18 at 4:36

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