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Actually, there is no algebraic problem to show that $-(-x) = x$. This proof can be build upon the concept of the addition of the opposite like this: $- x + x = - x + [- ( - x) ]$, and thus by eliminating $- x$ from both sides we obtain $x = - ( - x )$.

I wonder how we can introduce the concept that $-(-x)=x$ and, the closely related and in some sense more general, $(-a)(-b)=ab$ through an example, without using abstract analysis.

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  • $\begingroup$ Its valid in the real numbers since every real number has an inverse which can be added to the numbers to have the identity of addition which is zero ! $\endgroup$ – Abdallah Abusharekh May 1 '16 at 15:55
  • $\begingroup$ You're proving $x=-(-x)$ by assuming $x=-(-x$). That's not a valid argument at all, and no wonder they can't follow it. $\endgroup$ – Nij Jun 3 '16 at 0:16
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    $\begingroup$ I realize that it has been a while, but giving the OP a closer look I realized that the proof given, is actually correct. I have corrected my answer below, accordingly. For those who don't see it: $$ a+b=a+(-a)\implies b=-a.$$ Now just choose $a=-x$ and $b=x$. $\endgroup$ – gebruiker Jun 9 '16 at 9:48
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    $\begingroup$ @gebruiker: Yeah, I and apparently many others first thought it used circular reasoning and I eventually and finally got the implied punchline in a burst of dazzling wit. I think the remaining, additional, last step, namely that $-x+x \, \mathbf{= 0} = -x+[-(-x)]$ (i.e., that transitivity is being used), would have cleared up things greatly and been trivial to include. $\endgroup$ – Vandermonde Jun 11 '16 at 17:20

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In a comment, the OP has suggested that he actually wants a practical example convincing students that the product of two negative numbers is positive. This is related to, but psychologically distinct from, the question asked. In particular, students often have different mental models for multiplication (a binary operator) and negation (an unary operator). Actually the problem is easy from the negation point of view: negation just "switches the sign", and switching the sign twice takes you back to where you started.

Here is a "real world" example which models multiplication of real numbers of any sign:

If Jill walks $x$ miles per hour north, and travels for $y$ hours, what is her displacement from her starting position?

If $x=3$ and $y=2$, then she ends up $6$ miles north of where she started.

If $x=-3$ and $y=2$, then we have to interpret what walking $-3 \textrm{mph}$ north means. It literally means that every $3$ hours, you will be $3$ miles "less north" than you were before. This is equivalent to walking $3 mph$ south. So she ends up 6 miles south of where she started, aka $-6$ miles north of where she started.

If $x=3$ and $y=-2$, then we have to interpret what walking $-2 \textrm{h}$ means. It means that she has been walking at a constant rate, and we are interested in her position $2$ hours ago. In $2$ hours she walks $6$ miles north, so $2$ hours ago she was $6$ miles south of where she started, aka $-6$ miles north of where she started.

If $x=-3$ and $y=-2$, then she is walking $3 \textrm{mph}$ south, and we are interested in her position $2$ hours ago. She was $6$ miles north of where she is now $2$ hours ago.

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    $\begingroup$ This is a great analogy: Using N/S for one dimension, and future/past as another dimension. $\endgroup$ – Joseph O'Rourke Nov 7 '14 at 23:29
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    $\begingroup$ @JosephO'Rourke I am pretty sure I stole this from Sybilla Beckman's Math for Elementary School teachers book. Not totally sure: it may have just been in the course materials one of the professors at my university wrote up for this course. $\endgroup$ – Steven Gubkin Nov 8 '14 at 4:08
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I would use the (real) number line.

First introduce the concepts of positive numbers (distance to the right of zero) and negative numbers (distance to the left of zero).

Then introduce the concept of negation. Instead of teaching it as flipping with respect to the vertical line at zero, I recommend teaching it as the rotation of $180^\circ$ (counter-clockwise) about the point zero. (I'll explain why later.) This handles the case of why negating a negative number yields a positive number.

Then introduce the concept of addition (of positive numbers) as moving to the right, subtraction (of positive numbers) as moving to the left. Note that the negation of a positive number yields the same result as subtracting a positive number from zero.

Now introduce the concept of addition (of negative numbers) as moving to the left, that is, show that adding a negative number yields the same result as subtracting a positive number.

Finally, introduce the concept of subtraction (of negative numbers) as moving to the right, that is, show that subtracting a negative number yields the same result as adding a positive number.

Comment: I teach negation as rotation of $180^\circ$ (counter-clockwise) about zero. This helps later on when the complex numbers are introduced. With the imaginary axis (and the complex plane) introduced, I teach multiplication by $i$ as a rotation of $90^\circ$ (counter-clockwise) about zero. This makes it easy to understand why $i^2=-1$.

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    $\begingroup$ Hence $-(-x)=x$, as two rotations of $180^\circ$ about zero lead you back to where you started. $\endgroup$ – J W Nov 9 '14 at 11:11
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This is redundant with several nice answers, but let me try it anyway, as (for me) the intuition is strong.

A gambler (or spendthrift) loses \$10 per day. If they have \$30 today, how much did they have 5 days ago?

So I'm using loses per day as one negative, and past time as another. It is intuitively clear that the further in the past, and the greater the loses per day, the larger the initial pot of money.

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First of all I want to thank a lot the moderator who volunteer to edit my contributions in this nice site.

In fact I have a good practical example to illustrait that $- ( - x ) = x $, but I was a bit greedy to find an other example that meight be easier to accept for students minds.

The example is as follows:

Suppose we have a monybox:

1 - If we add 2 dollars a day, then after 5 days we will have increasing that equal to $2 × 5 = + 10 $ .. That was $+ × + = + $

2 - If we pull 2 dollars a day, then after 5 days we will have decrease equal to $-2 × +5 = - 10 $ .. This was $ - × + = - $

3- If we add 2 dollars a day, then BEFORE 5 days we have a decrease = $ + 2 × -5 = -10 $ .. That was $+ × - = - $

4- Now suppose we PULL 2 dollars a day, then BEFORE 5 days we were had an increase = + 10 That was $ - 2 × -5 = + 10 $

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  • $\begingroup$ I don't think this directly answers the question. As was noted in an answer, the question is about negating a negative number. This is not the same as multiplying two negative numbers. $\endgroup$ – Joel Reyes Noche Nov 9 '14 at 11:01
  • $\begingroup$ the main question was how to explain that multiplying tow negative number is equal to positive number .. also $-(-x)$ means the same thing $\endgroup$ – Abdallah Abusharekh Nov 9 '14 at 15:13
  • $\begingroup$ -(-x) is not the same as multiplying two negative numbers. It's finding the additive inverse of an additive inverse, or subtracting a negative number, which has nothing to do with multiplication. $\endgroup$ – DavidButlerUofA Nov 9 '14 at 18:47
  • $\begingroup$ Well in fact the original question was about multiplying tow negative numbers ( This was the title ). But also $-(-x)$ is multiplying tow negative signes which it is the same as finding the additive inverse of an additive inverse. $\endgroup$ – Abdallah Abusharekh Nov 9 '14 at 19:01
  • $\begingroup$ I now see the problem. Someone (not the OP) edited the main question and changed the title. $\endgroup$ – Joel Reyes Noche Nov 10 '14 at 0:02
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I'm not really understanding the aversion to using abstract analysis. You don't need to ruin it for your students and say "by the way, this is a result of Ring-theoretic axioms, which is waaaaaaay beyond the scope of anything you'll ever learn!" Just show them that (-1)*(1 - 1) = 0 (which they won't dispute), and then distribute the terms, giving you -1 + (-1)^2 = 0, meaning (-1)(-1) = 1. Every other case of a negative multiplied by a negative reduces to this.

If you present it like this, it comes across as a neat trick rather than an abstract algebraic result (like when you show why 0/0 is undefined - it's enough to show that it contradicts our assumptions about real numbers, and you don't need to accompany your explanation with a real-life example). I think trying to make it seem "intuitive" can be distracting, like when people try to reason why an implication being false makes a whole statement true. Students will usually be able to poke holes in your analogies and find destabilizing examples, or they'll just get confused and blame themselves for not understanding what is likely a silly comparison. Sometimes math concepts can't be analogized well with things in real life, and trying to force it to work is bad pedagogy.

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    $\begingroup$ When operations on negative numbers are first being introduced, I think it is important that they are grounded in reality. Rather than being just a formal abstract system, we expect students to use negative numbers to model reality. Essentially sign models directional quantities, like north/south, future/past, etc. Thus it is important to have models of multiplication which show that the everything actually makes sense. $\endgroup$ – Steven Gubkin Sep 2 '15 at 17:43
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    $\begingroup$ I agree that showing that other axioms force the (-1)(-1) = 1 is a good adjunct to this, but might actually be a bit too abstract for children at the level where these concepts are introduced. It is also good to look at patterns: 3(2) = 6, 3(1) = 3, 3(0) =0. We are subtracting three each time, so reasonable that 3(-1) = -3, 3(-2) = -6 to fit pattern. Then decrease other factor. 2(-2) = -4, 1(-2) = -2, 0(-2) = 0 which is going up by two each time. So reasonable that (-1)(-2) = 2, (-2)(-2) =4, ..., again to fit the pattern. Children at this level should be exploring in all these ways. $\endgroup$ – Steven Gubkin Sep 2 '15 at 17:49
  • $\begingroup$ OP indicated he already know the abstract analysis approach, so I think looking for additional approaches is very reasonable. That does not mean to only use one approach or the other. It is best to attack from many angles. $\endgroup$ – Steven Gubkin Sep 2 '15 at 17:50
  • $\begingroup$ So basically, I think that the question (and existing answers) do not display an aversion to abstract analysis. Just seeking more than just abstract analysis. $\endgroup$ – Steven Gubkin Sep 2 '15 at 17:51
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This might be contained in some other answer in one form or another, but I think it might be useful to simply appeal to the definition of $-x$.

Definition: whatever the number $x$ is, we denote by $-x$ the number which yields $0$ when added to $x$.

This is both mathematically sound in a very general setting, simple enough for pupils (I guess), and easy to relate to all interpretations that where given (North/South, debt/credit, etc.).

Now, this definition makes it easy to deduce $-(-x)=x$ ; I'll make it on an example since this is how I would proceed myself. By its very definition, $-(-3)$ is the number which yields $0$ when added to $-3$. We know that $-3+3=0$ (this is the definition of $-3$) so that the number we seek is $3$. I.e. $-(-3)=3$.

This is a bit formal in some way, but I feel that getting back to the basic definition and relate it to the meaning of a mathematical object of notation should never be avoided unless necessary.

Of course, there is something hidden behind this: the definition only makes sense if we know for sure that there exist a unique number with the given property. But this can be kept behind the scene for quite a few years I guess.

I will not get at multiplying negative numbers, as I have little to add to other answers.

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You could have also proven this algebraically, by doing this:

To show that $-(-x)=x$ is the same as showing $-(-x)-x=0$. We then simply say $$-(-x)-x=-1(-x+x)=0 \tag{QED}$$

Or without using the distributive law, it could go something like this:

If the students already believe that $-1\times-1=1$, then $$--a=-1\times-1\times a=1\times a=a\tag{QED}$$

${ }$


As for an example: I suppose you can appriciate something intuitive. How I was taught, was with the example of a big bowl of water. It goes like this:

We have a man (/women/machine/etc...) standing by a big bowl of water. With him the man has two piles of blocks. Red blocks and blue blocks.

  • Now the red blocks are hot. They have a positive temperature.
  • The blue blocks are cold. They have a negative temperature.

At the begining, the temperature can be whatever you like. It will now probably speak to the students' imagination that

  • adding a red block will increase the temperature of the water. So adding something positive leads to an increase.
  • adding a blue block will cause a decrease in temperature. So adding something negative leads to a decrease.
  • removing a red block will make the water colder. So subtracting something positive leads to a decrease.

And now for your case:

  • removing a blue block will make the water warmer. So substracting something negative will lead to an increase.

The magnitudes of the decreases and increases depend on the sizes of the blocks of course. I'll take it you won't find that very hard to explain.

So there we have it. An algebraic example and an intuitive example.

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    $\begingroup$ The idea is nice. But removing something hot from water will in fact not make it colder (it will stop heating it but this is not the same), except you look into the future. I am howver worried this makes things a bit complicated to follow. $\endgroup$ – quid Nov 7 '14 at 14:22
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    $\begingroup$ @quid It's a visualization. Visualizations often don't formally mach with the real world to the fullest extend. I'm not going to say I know for sure this is a good method. I'm just saying that this is how I was taught the concept and it seemed to work for the majority of my classmates and me. $\endgroup$ – gebruiker Nov 7 '14 at 14:31
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    $\begingroup$ @quid While money is less fun, it does work here. Cash and debts. Forgiving a debt is effects your net worth in the same way as gaining an equal amount of cash. $\endgroup$ – Steven Gubkin Nov 9 '14 at 15:31
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I wrote a great blog post on why a negative times a negative equals a positive. I liken it to pulling weeds. I'll preview the post here and then provide a link.

(1) Would you say that getting good stuff is a good thing or a bad thing?
(2) How about getting bad stuff?
(3) Losing good stuff?
(4) Losing bad stuff?

http://lifesavertutoring.com/blog.php

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  • $\begingroup$ Thank you Mr Frank to your nice criative questions that gives a meaning to signs multiplcation $\endgroup$ – Abdallah Abusharekh May 6 '15 at 19:29
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    $\begingroup$ 404 error at that link. $\endgroup$ – JoeTaxpayer Jun 21 '16 at 15:04
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Forgive me if this is already above. I only see one reference to it in an answer

Instead of teaching it as flipping with respect to the vertical line at zero, I recommend teaching it as the rotation of 180∘ about the point zero.

I would argue that (the well-known analogy of) flipping/reflecting about the vertical line at zero is a very good way, actually, and I do this often. Here are three reasons.

  • Kinesthetic - simply take your arms and (say) flip your elbow to your other elbow. That's $-1$; do it again for $-(-1)=1$. Do it with the ends of your arms for $-(-2)=2$.
  • Doesn't require wondering about why we are rotating when there is no "meaning" (in the eyes of students likely to ask this) of things "above" the number line.
  • It's easy.

I agree that if one is going to be thinking of negation in a broader context, with students already very comfortable with $\mathbb{R}^2$ or even $\mathbb{C}$ as such (rather than as the place where graphs of functions live) then rotation is the way to go, since $-(i)=-i$ would just be wrong under this analogy. But I find it very useful for a quick rationale. (And if they bring up $-(1,1)=(-1,-1)$ doesn't work, you have a live one!)

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None of the other answers get at the basic psychological problem: People begin with an understanding of numbers as magnitudes (i.e amounts of things). When introduced to zero and negative numbers, they can usually fit zero in as the magnitude of 'nothing left', but they try to grasp negatives as somehow pushing magnitude beyond nothing, which they find utterly baffling. What you have to do is get them to understand that the sign of a number has nothing to do with magnitude, it is a completely new and different feature of a number.

Since this is about time the number line gets introduced, the most natural approach is to say that numbers can have direction as well as magnitude, and sign indicates direction. So, for example: The difference between 1, +1, and -1 is exactly analogous to the difference between 'a mile', 'a mile east', and 'a mile west'. [if they ask 'what about a mile north?', tell them they'll get to that a few years down the road :-) ]

You can then talk about numbers using the number line via the metaphor of walking: The number 3, for example, means starting at the origin and walking three paces; if you walk east, that's +3, if west, that's -3 (zero means don't walk). Addition means you start at one number and walk the other, using sign to know in which direction to walk.

When you get to '-(-x) == x', you can explain that applying 'minus' in multiplication means flipping the entire number line over about the origin - east becomes west, west becomes east. So if 'x' is 'right of origin, facing east', '-x' flips it over so now it is 'left of origin, facing west'. So naturally, '-(-x)' flips it over twice, leaving you back with what you started.

[note: if you can get kids to use this model early, they'll probably have a much better time if and when they come to complex numbers]

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I won't try to give a specific explanation here because I think that depends highly on the level of knowledge of the student. Rather, I'd like to emphasize some general points that I think are helpful for teachers to keep in mind when explaining and motivating these and related matters about the extension of "number systems" (here the extension from positive to negative integers).

First, it is often helpful to look at the history of how these concepts evolved. For example, we need to have available negative integers, rationals, and (imaginary) quadratic surds in order to give a single (uniform) formula for the solution of a quadratic equation (prior ancient solutions of the quadratic bifurcated into motley cases because they had to avoid working with not yet understood negative or "imaginary" or "irrational" numbers).

Yet, once we have such numbers available, the solution of the quadratic is childs-play, using simple algebra by completing the square. Deserving explicit emphasis about such algebra is that the arithmetical laws used (commutative, associative and distributive laws) hold true for all the types of numbers involved (i.e. they are all commutative rings). Because of the persistence of these laws, even though we may make detours through fractions, "irrational" and "imaginary" numbers, any results we deduce about "natural" integers still hold true. This is one of the primary reasons that we desire to preserve these laws when we make extensions to our number systems. It allows us to give "universal" proofs that work in any of these rings. Further, problems may simplify when passing to extension rings, e.g. nonlinear problems may reduce to linear problems in algebraic extension rings.

In the old days (pre-axiomatic method) the constraint to preserve the laws of arithmetic when we extend our number systems was sometimes called the Permanence Principle, attributed to Hankel or Peacock. Nowadays, with the axiomatic method available, we simply express it in that language, e.g we desire rational, quadratic and complex numbers to be commutative ring extensions of the ring of integers.

This notion of the great utility of the preservation of arithmetical laws can be explained at an elementary level. Doing so yields motivated explanations that are a bit more faithful to the underlying mathematics (and its history) than do alternative "real world" answers to questions like the OP.

For example, below I copy one way I explain the Law of Signs, emphasizing the crucial role played by the distributive law. With a little effort one could simplify the exposition to make it much more elementary, yet still be faithful to the "law preservation" viewpoint.


Law of Signs proof: $\rm\,\ (-x)(-y) = (-x)(-y) + x(\overbrace{-y + y}^{\large =\,0}) = (\overbrace{-x+x}^{\large =\,0})(-y) + xy = xy$

Equivalently, evaluate $\rm\:\overline{(-x)(-y) +} \overline{ \underline {x(-y)}} \underline{ +\,xy}\, $ in 2 ways, noting each over/under term $ = 0$

Said more conceptually, $\rm\:(-x)(-y)\ $ and $\rm\:xy\:$ are both inverses of $\rm\ x(-y)\ $ so they are equal by uniqueness of inverses: if $\,a\,$ has two additive inverses $\,\color{#c00}{-a}\,$ and $\,\color{#0a0}{-a},\,$ then

$$\color{#c00}{-a}\, =\, \color{#c00}{-a}+\overbrace{(a+\color{#0a0}{-a})}^{\large =\,0}\, =\, \overbrace{(\color{#c00}{-a}+a)}^{\large =\,0}+\color{#0a0}{-a}\, =\, \color{#0a0}{-a} $$

This proof of the Law of Signs uses well-known laws of positive integers (esp. the distributive law), so if we require that these laws persist in the larger system of positive and negative integers, then the Law of Signs is a logical consequence of these basic laws of positive integers.

These fundamental laws of "numbers" are axiomatized by the algebraic structure known as a ring, and various specializations thereof. Since the above proof uses only ring laws (most notably the distributive law), the Law of Signs holds true in every ring, e.g. rings of polynomials, power series, matrices, differential operators, etc. In fact every nontrivial ring theorem (i.e. one that does not degenerate to a theorem about the underlying additive group or multiplicative monoid), must employ the distributive law, since that is the only law that connects the additive and multiplicative structures that combine to form the ring structure. Without the distributive law a ring degenerates to a set with two completely unrelated additive and multiplicative structures. So, in a sense, the distributive law is a keystone of the ring structure.

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Here's what my math teacher once told me: if a good thing happens to a good person it's a good thing. If a bad thing happens to a bad person it's a good thing. If a good thing happens to a bad person or a bad thing happens to a good person it's a bad thing.

This doesn't help in any way shape or form of you're trying to explain why these things happen; however, if the student is not naturally mathematically inclined then this is the easiest analogy to help them remember the rules.

If the student in question does not struggle in math then this analogy will just mess them up.

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  • $\begingroup$ For the math to work, a bad thing happening to a bad person should be good. $\endgroup$ – JoeTaxpayer May 28 '16 at 18:16
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    $\begingroup$ I've edited the second sentence to match the spirit and intent of the third. The bad/bad seemed a typo in light of the third sentence. $\endgroup$ – JoeTaxpayer Jun 21 '16 at 16:44
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I reformat the 2nd most upvoted answer with MathJax beneath, from Reddit.

sjets3 14.1k points 7 months ago

Imagine you are watching a movie. The first number is how the person in the movie is moving. The second number is how you are watching the film (normal or in reverse).

$1 \times 1$ is a person walking forward, you watch it normal. Answer is you see a person walking forward, which is 1.

$1 \times -1$ is a person walking forward, you watch it in reverse. You see a person walking backwards. -1

$-1 \times 1$ is a person walking backward, you watch it normal. You see a person walking backwards. -1

$-1 \times -1$ is a person walking backwards, but you watch it in reverse. What you will see is a person that looks like they are walking forward. 1

As a funny example of the last para. overhead, compare https://gfycat.com/PopularFrighteningCormorant (original video) with https://i.imgur.com/ZCw2C81.gifv (film person walking backward then play backward).

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Not sure this is what you are looking for, but multiplying a positive by a negative is very straightforward, i.e. $+8\cdot-1$. If we take eight negative ones then we get: $-1 + -1+ -1+ -1+ -1+ -1+ -1+ -1 = -8$. However what is not straightforward is what does it mean to have $-1\cdot 8$. Here we can argue that $-1\cdot8 = 8\cdot-1$ (because we assume multiplication is commutative) and thus make sense of it. However, this doesn't really help when we multiply a negative by a negative.

We really need a new definition when we multiply by a negative number. I think first you need to define what a negative number is: $-8 = 0 - 8$. When we multiply by a negative, we aren't adding, rather we are subtracting! So when we say $-1\cdot8$ we are subtracting one "8" value: $0 - 8 = -8$. If we do $-2\cdot3$ we are subtracting two $3$'s: $0 - 3 - 3 = -6$.

So in that regard, if we do $-1\cdot-1$ then we should subtract one negative one: $-1\cdot-1 = 0 - -1 = +1$. Likewise if we do $-3\cdot-4$ then we should subtract three negative fours: $0 - -4 - -4 - -4 = 0 - -12 = +12$.

That's how I would make sense of multiplying by a negative.

Can we prove that $0--x = x$?

Well obviously from a mathematical standpoint it's easy to prove but it also easy to understand from an intuitive standpoint (explained later). Mathematically, subtraction is the inverse to addition just as division is the inverse of multiplication.

So we have: $x + y = z$ therefore the inverse is $x = z - y$. Thus if we have $0 - -x = a$, then $0 = a + -x$. If we recognize that adding a negative is the same as subtracting the positive (which is basically the definition of negative numbers) then we realize that $0 = a - x$. Now we recognize that the inverse of this is that $a = 0 + x$ which equals $x$ because $0$ is the additive identity (i.e. $0 + z = z$), so we arrive at $0 - -x = a = x$.

Intuitive Explanation

What can negative numbers mean intuitively? My example would be debt. If you loan me $x$ dollars then I am in debt to you by $x$ dollars--my balance can be said to be negative, that is $-x$ represents that I owe you $x$ dollars. On the other hand, if I loan you $y$ dollars, then my balance can be said to be positive that is you owe me $y$ dollars ($y$ dollars is the amount I "hold"--assuming you will eventually pay me back).

If you loaned me $x$ dollars, then my current balance is $-x$. If you decide that you are going to forgive that loan, i.e. you are going to subtract $x$ dollars of debt from me, then my new balance would be $-x - -x$. Why subtraction? Because you are subtracting debt from me--you are taking away debt (which is a negative entity). By forgiving my debt, you would make my new balance $0$.

Now let's say that you give me $x$ dollars (not loan me $x$ dollars). How can you do this if all there is is loans and loan forgiveness? It's simple: instead of loaning me $x$ dollars you forgive me $x$ dollars! This means you subtract, from my initial balance of $0$, $-x$ dollars (since a loan means I go negative). By simply giving me $x$ dollars (with no expectation of repayment) you subtract the debt of $-x$ dollars from me: $0 - -x = x \leftarrow$ you gave me $x$ dollars (with no expectation of repayment).

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  • $\begingroup$ I assume you mean $--12=+12$ or, more precisely, that $0 - -12 = +12$. I think subtraction/addition is easier to prove than multiplication (I certainly took liberties with the definition of multiplication which relies on the definition of addition). $\endgroup$ – Jared Nov 7 '14 at 9:17
  • $\begingroup$ well in fact I have proved in the question that - ( - x ) = + x .. I'm seeking a practical example to convince the students that multiplying 2 negative numbers is a positive number ! $\endgroup$ – Abdallah Abusharekh Nov 7 '14 at 9:21
  • $\begingroup$ Thanks a lot my friend .. sorry for I couln't be on line most the time .. We have no electricity in Gaza $\endgroup$ – Abdallah Abusharekh May 1 '16 at 15:57
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Here is another way to show/prove this by treating multiplication as repeated addition:

Consider

$3 \cdot 5 = 5+5+5 = 15$

as "adding $5$ three times". Now

$3 \cdot (-5) = (-5) + (-5) + (-5) = (-15)$

is clear because "adding $-5$ three times is $-15$". Then we want the commutative law to hold for negative numbers as well, so

$(-5) \cdot 3 = (-15)$

but we have to leave the "repeated addition" model of multiplication here, because adding something $-5$ times does not make sense. If we now flip the sign of the second factor again, we want the product to flip its sign as well, just as it happened in the second step. $(-5) \cdot 3$ shouldn't be the same as $(-5) \cdot (-3)$. We get

$(-5) \cdot (-3) = 15 \qquad \text{"qed"}$

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Proof that $-(-x)=x$:

By definition, for all $a\in R$, we have $-a\in R$ and $a+(-a)=0$.

Apply this definition for $a=x$ where $x\in R$ to obtain $x+(-x)=0$.

Now, apply this definition for $a=-x$ to obtain $-x + (-(-x)) =0 $.

By substitution, $-x + (-(-x))=x+(-x)$.

Cancelling $-x$ from both sides, we obtain $-(-x) = x$.

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  • $\begingroup$ The question was not about proving the statement, but about making it more intuitive and concrete. I don't see how your answer addresses that issue. $\endgroup$ – Xander Henderson Jan 14 at 17:35
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    $\begingroup$ @XanderHenderson Use 2 instead of x. $\endgroup$ – Dan Christensen Jan 15 at 15:26
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I reformat the most upvoted answer (also my favorite) with MathJax beneath, from Reddit.

Zerotan 42.2k points 7 months ago

Repost from 2 years back:

I give you three \$20 notes: $+3 × +20 =$ you gain $60

I give you three \$20 debts: $+3 × -20 =$ you lose $60

I take three \$20 notes from you: $-3 × +20 =$ you lose $60

I take three \$20 debts from you: $-3 × -20 =$ you gain $60

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The idea of using the concept of direction and flipping directions in order to understand the concept of negative numbers and their multiplication is helpful but maybe misleading.

Math is an expression/product of Logic instead, and therefore logical operators and analogies are better suited for this.

So,

Think of the positive sign as a symbol indicating something that is true.

In which case you can think of a positive number n as meaning it is true that something exists n times

Similarly, think of the negative sign as a symbol indicating something that is NOT true.

In which case you can think of a negative number n as meaning it is NOT true that something exists n times

When you multiply a positive X by a positive Y you are saying that it is true Y times in the world that it is true that something exists X times.

So 2 x 3 means it's true 3 times that (it is true that 2 unicorns exist)

In other words, you are saying that:

it is true that 2 unicorns and it is true that 2 unicorns and it is true that 2 unicorns

which can be written as:

it is true that (2 unicorns and 2 unicorns and 2 unicorns)

which can be written as:

it is true that 6 unicorns, also written as +6 unicorns.

When you multiply a positive X by a negative Y you are saying that it is NOT true Y times in the world that something is true X times.

So 2 x -3 means it is NOT true 3 times that (it is true that 2 unicorns)

In other words, you are saying that:

it is not true that 2 unicorns and it is not true that 2 unicorns and it is not true that 2 unicorns

which is logically equivalent to saying:

it is not true that (2 unicorns and 2 unicorns and 2 unicorns)

which can be written as:

it is not true that 6 unicorns, also written as -6 unicorns.

Now, when you multiply a negative X by a negative Y you are saying that it is NOT true Y times in the world that it is NOT true that something exists X times.

So -2 x -3 means it is NOT true 3 times that (it is NOT true that 2 unicorns)

In other words, you are saying that:

not true that (not true that 2 unicorns) and not true that (not true that 2 unicorns) and not true that (not true that 2 unicorns)

But if it's not true that something is not true then that something is actually true, so the above is logically equivalent to:

(true that 2 unicorns) and (true that 2 unicorns) and (true that 2 unicorns) which can be written as:

true that (2 unicorns and 2 unicorns and 2 unicorns)

meaning:

true that 6 unicorns, also written as +6 unicorns

And that is why -2 x -3 = +6

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  • $\begingroup$ Question: What kind of students are you showing this exposition to? $\endgroup$ – Chris Cunningham Feb 25 at 17:50

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