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To quote from Wikipedia's article on negative numbers

Negative numbers represent opposites. If positive represents movement to the right, negative represents movement to the left. If positive represents above sea level, then negative represents below level. If positive represents a deposit, negative represents a withdrawal.

This leads one to think that anything one does to a negative number will result in opposite (or just opposite signed) results to that of its positive counterpart.

This is obviously not true though because, for example $(-3)^2=9$ and $3^2=9$.


With that in mind, I ask, are negative numbers really the opposite of positive numbers? What makes them fundamentally different from positives in a way that leads to this discrepancy?

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    $\begingroup$ Positives are closed under multiplication. $\endgroup$ – alphacapture May 26 '16 at 1:46
  • $\begingroup$ Some functions (such as the squaring function $x \mapsto x^2$ or the absolute value function $x \mapsto |x|$) simply don't care if you give them a number $x$ or its opposite $-x$. $\endgroup$ – pjs36 May 26 '16 at 2:05
  • $\begingroup$ Also $(-3)0=0$ and $(3)0=0$. $\endgroup$ – vadim123 May 26 '16 at 2:15
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    $\begingroup$ $(-3)^2 \ne 6$... $\endgroup$ – user21820 Jul 3 '16 at 10:02
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    $\begingroup$ So has my answer addressed your question? In a nutshell, you can look at the number line from the other side so that every number on the line is effectively negated, and since vector addition is independent of how we look at it, the addition of numbers is unaffected as well; if $a+b=c$ then $(-a)+(-b)=(-c)$ too. However, the scaling that does not move the points on the line is still the $1$-scaling, where this "$1$" is a scale factor. In my answer the $r$-scaling is defined as the one bringing $1$ on the line to $r$; agreeing with this. Thus multiplication is not preserved by negation. $\endgroup$ – user21820 Jul 4 '16 at 23:57
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None of the answers so far has addressed the underlying question, which is why it seems that multiplication is asymmetric and satisfies "positive times positive is positive" while "negative times negative is positive". The reason is that multiplication is not as simple as it seems. $\def\nn{\mathbb{N}}$ $\def\zz{\mathbb{Z}}$ $\def\rr{\mathbb{R}}$

This answer is necessarily long to explain in reasonable detail how one builds real arithmetic intuitively from scratch, rather than purely by axiomatization. This is unavoidable since the field axioms can show that $-1 \times -1 = 1$ but give no insight whatsoever into why it should be like this.

Viewpoint 1 is sort of the more fundamental and relies on less intuition, but you will need to follow slowly and carefully in order to completely grasp and apprciate it, so if you want a quick answer skip to Viewpoint 2 and stop once I begin to tie it to Viewpoint 1.

Viewpoint 1

First we can start with the natural numbers $\nn$, which you intuitively understand, whether using blocks or using strings of symbols, or treating each natural number as a way to represent the number of times of repetition of some procedure.

Real numbers $\rr$ represent displacement, as you mentioned in your question. But what does $a \times b$ represent? You cannot multiply displacement by another displacement! So something else is meant here. Let's see what we can define intuitively. Define $k \times x$ to be "$k$ times as much as $x$" for any $k \in \nn$ and $x \in \rr$. Of course $0_\nn \times x = 0_\rr$ and $1_\nn \times x = x$, for any $x \in \rr$. Note that I explicitly distinguish between $0_\nn$ ($0$ times) and $0_\rr$ (zero amount). Also note that $x \times k$ is not assigned any meaning (yet), since it makes no sense to talk about "$x$ as much as $k$ times".

Note that $\nn$ embeds into $\rr$ in the following manner. You can treat $k_\rr$ as $k \times 1_\rr$, which by our definition means "$k$ times of $1_\rr$". In the physical world this corresponds to the amount "$k$ units".

Now you can verify intuitively that two crucial properties hold:

$( a +_{_\nn} b ) \times x = a \times x + b \times x$ for any $a,b \in \nn$ and $x \in \rr$.

$( a \times_{_\nn} b ) \times x = a \times ( b \times x )$.

Here I've used "$+_{_\nn}$" and "$\times_{_\nn}$" for addition and multiplication in $\nn$ to distinguish it from "$+$" and "$\times$" in $\rr$. Based on these two properties, we say that $\nn$ acts on $\rr$ via $\times$. This notion of action is ubiquitous one, even if you don't realize it. For example, $\nn$ acts on physical entities via $\def\of{\text{ times }}$$\of$ as follows: "$k \of X = \text{$k$ instances of $X$}$". We say "$2$ apples" by which we mean "$2 \of apples$" = "2 instances of apples". We intuitively understand that we can handle abstract numbers to perform counting of physical entities, which is why we have:

$( 2 + 3 ) \of apples = 2 \of apples + 3 \of apples$.

$( 2 \cdot 3 ) \of apples = 2 \of ( 3 \of apples )$.

[likewise for all other physical entities, which is the key insight; it works not only for apples!]

Similarly, $\nn$ acts on (real-world) actions via $\of$ as follows: "$k \of A = \text{$A$ repeated $k$ times}$". Again we can check that both properties hold.

All these are examples of the action of $\nn$ on some collection. Some of these can be extended meaningfully to action of $\zz$ on the same collection, but not others. What we would want is that $(-_{_\zz}k) \times x = -( k \times x )$ for any $k \in \zz$ and $x \in \rr$. For displacement this would mean that "$(-1_\zz) \times x$" is simply "the opposite displacement to $x$".

Already here we can see the asymmetry, because $(-1_\zz) \times (-1_\rr) = 1_\rr$ and $1_\zz \times 1_\rr = 1_\rr$. The former means "$-1$ times the opposite to unit displacement", while the latter just means "unit displacement", and both are the same. So the answer to your question under this viewpoint is that the asymmetry comes from the concept of iteration that is intrinsic to the action of $\zz$ on $\rr$.

Just for completeness, $\zz$ does not act on physical entities, because there are no direct opposites to physical entities that when combined with the original yields nothing. Using the concept of "lack" or "debt" simply changes the collection; originally it is the collection of physically existing entities, subsequently it is the collection of records of relative quantities of physical entities. It is essentially the same way we construct $\zz$ from $\nn$ mathematically. Also $\zz$ only acts on reversible actions.

Viewpoint 2

Real numbers are scale factors, or better still the scalings themselves with centre at the origin. $2$ is the $2$-times zoom on the origin. $1$ is the identity scaling that does nothing at all. $0$ is the scaling that collapses everything into the origin. $-1$ is the scaling that takes every point to the point opposite the origin. Multiplication of real numbers is the same as composing the two scalings together (doing one after another). Note that intuitively the order of scalings does not matter, giving us commutativity of multiplication in $\rr$.

Obviously $-1 \times -1 = 1$, answering the question of asymmetry. The underlying reason is actually the same as in the other viewpoint, because here positive is special simply because $1$ is the identity scaling (keep it as it is), while there positive is special because $1$ is the unit amount (exactly as it is). Notice the core similarity!

Now the above only explains the multiplicative structure of the reals, so where does the additive structure come in? We can in fact combine it with Viewpoint 1 by considering the real numbers themselves as positions on a line through the origin (equivalently displacements from the origin), and then define that a scaling $r$ is precisely the scaling centred at $0$ on the line that brings the point $1$ on the line to the point $r$. Here we have used the intuition that there is such a scaling and that it is unique, so there is a one-to-one correspondence between the scalings and points on the line. To keep compatible with the above paragraph, we still define a composed scaling $r \times s$ as the composition scaling by $r$ then by $s$.

Since the scaling $( r \times s )$ brings the point $x$ to the same point if we did first scaling $s$ and then $r$, we do have the second property for action of scalings $\rr$ on points $\rr$. This is clearer in symbols, where we write $\def\on{\text{ on }}$"$r \on x$" to denote "the result of scaling $r$ applied on $x$":

$( r \times s ) \on x = r \on ( s \on x )$.

From the definition of scaling and commutativity of scaling, we get:

$r \times x = ( r \times x ) \on 1 = r \on ( x \on 1 ) = r \on x$.

$r \on s = r \times s = s \times r = s \on r$.

We now define addition of points $x,y$ on the line to be vector addition (add the displacements from the origin), and then define addition of scalings $r,s$ to be exactly the same according to the correspondence between scalings and points! Note that by intuition (or 'definition' of scaling/zoom) we can see that the result of vector addition is the same no matter what scale we are observing it from, and so scaling before adding should give the same as adding before scaling. Symbolically:

$r \on x + r \on y = r \on ( x + y )$.

From the above results we can finally get the first property for action of $\rr$ on $\rr$:

$( r + s ) \on x = x \on ( r + s ) = x \on r + x \on s = r \on x + s \on x$.

These properties correspond to the properties of multiplication of real numbers (commutativity, associativity and distributivity over addition), which are the most difficult to justify intuitively. The rest are left as an exercise.

Therefore we have established that arithmetic of $\rr$ captures facts about scaling (or zooming) in the real world, at least according to our intuition. Even if it turns out that it is not exactly true, it certainly serves as a good approximation, and hence investigating theorems about $\rr$ give us approximate facts about the real world.

Footnotes

Of course the above viewpoints cannot be used to prove that real numbers exist, but at least give a reasonable justification for why we believe $\rr$ is useful in describing the real-world. Nevertheless, the explanation can be converted into a rigorous construction of the real numbers in any formal system that is powerful enough to permit construction of simple collections, functions from $\nn$ to $\nn$, equivalence relations and equivalence classes.

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  • $\begingroup$ How did you manage to explain so well and in so great depth? It never occurred to me that the equation $-1\times - 1 = 1$ had so much "behind the scenes". +1 for the excellent "viewpoints". $\endgroup$ – Paramanand Singh Feb 2 '17 at 11:11
  • $\begingroup$ @ParamanandSingh: It took me quite a while to convince myself of the meaning of the real numbers, and these two viewpoints were the results. Thanks for your kind encouragement! =) $\endgroup$ – user21820 Feb 2 '17 at 11:14
  • $\begingroup$ @ParamanandSingh: Oh and by the way, since Viewpoint 2 involves some transformations of a line, it also naturally inspires the question of whether there is a similar kind of structure that captures some of the transformations of a plane. In particular, multiplying by $-1$ can be understood as a reflection or a rotation of the line 180 degrees about $0$, and that cries out for other rotations. If we limit to rotation plus scaling, we get complex multiplication and can intuitively see that it is commutative! $\endgroup$ – user21820 Feb 2 '17 at 11:35
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This leads one to think that anything one does to a negative number will result in opposite (or just opposite signed) results to that of its positive counterpart.

Well don't think that, because this is overgeneralizing.

It is "good" to pour a cup of water into the mouth of a thirsty man. It is "evil" to punch him in the nose. Good and evil are opposites. However if you change both activities in the same way, they need not remain opposites. If you punch him in the nose 1000 times, that is still evil. If you pour 1000 cups of water into his mouth, that is evil as well, as he will drown.

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One way to define the set of all integers, from the set of "natural numbers" (positive integers) is to define an equivalence relation on the set of positive integers by "(a, b) is equivalent to (c, d) if and only if a+ d= b+ c". We then define the integers as the collection of all "equivalence classes". It is fairly easy then to show that if a>b for some pair (a, b) then c> d for all (c, d) equivalent to (a, b) and, conversely, that if a< b for some pair (a, b) then c< d for all (c, d) equivalent to (a, b). Finally, there exist a unique equivalence class consisting of all pairs (a, a).

The positive integers, as a subset of all integers, is the collection of all equivalence classes containing (a, b) with a> b. The negative integers is the collection of all equivalence classes containing (a, b) with a< b. Finally, "0" is the set of all pairs (a, a).

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When you study more advance mathematics you will understand that there are two square classes for Real numbers namely positive one and negative one. So basically you can divide all real numbers into piece of two one having square roots and one doesn't have. But for complex number you cannot differentiate between positive and negative since all numbers are square.

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