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Someone recently asked me why a negative * a negative is positive, and why a negative * a positive is negative, etc.

I went ahead and gave them a proof by contradiction like so:

Assume $(-x) * (-y) = -xy$

Then divide both sides by $(-x)$ and you get $(-y) = y$

Since we have a contradiction, then our first assumption must be incorrect.

I'm guessing I did something wrong here. Since the conclusion of $(-x) * (-y) = (xy)$ is hard to derive from what I wrote.

Is there a better way to explain this? Is my proof incorrect? Also, what would be an intuitive way to explain the negation concept, if there is one?

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You haven't proven that -xy = (-x)y. – Qiaochu Yuan Nov 12 '10 at 2:24
It is not an uncommon question, and it's never not easy to show. – J. M. Nov 12 '10 at 4:06
J.M. is pointing out that logical negation works the same way as multiplying negative numbers (two negatives make a positive), not belittling your question. You understood his double negative statements, and so you have another intuitive path to see that negative*negative=positive. – Jonas Kibelbek Nov 12 '10 at 4:48
Somewhat relevant... – J. M. Sep 3 '11 at 9:08
Does someone have a text about the history of negative numbers multiplication? – Lucas Zanella Jun 29 '14 at 21:48

18 Answers 18

up vote 40 down vote accepted

Well if I were to explain this in an intuitive way to someone (or at least try), I would like to think of an analogy with walking over the real line, by agreeing that walking left will be walking in the negative direction and walking right in the positive direction.

Then I will try to convey the idea that if you are multiplying to numbers (let's suppose they are integers to make things easier to picture) then a product as $2*3$ would just mean that you have to walk right (in the positive direction) a distance of $2$ (say miles for instance) three times, that is, first you walk $2$ miles, then another $2$ miles and finally another $2$ miles to the right.

Now you picture where you're at? Well, you're at the right of the origin so you are in the positive section. But in the same way you can play this idea with a negative times a positive.

With the same example in mind, what would $-2*3$ mean? First, suppose that the $-2$ just specifies that you will have to walk left a distance of $2$ miles. Then how many times you will walk that distance? Just as before $3$ times and in the end you'll be $6$ miles to the left of the origin so you'll be in the negative section.

Finally you'll have to try to picture what could (-2)*(-3) mean. Maybe you could think of the negative sign in the second factor just making you change direction, that is, it makes you turn around and start walking the specified distance. So in this case the $-2$ tells you to walk left a distance of $2$ miles but the $-3$ tells you to fist turn around and then walk $3$ times the $2$ miles in the other direction, so you'll end up walking right and end in the point that is $6$ miles to the right of the origin, so you'll be in the positive section, and $(-2)*(-3) = 6$.

I don't know if this will help but it's the only way I can think of this in some intuitive sense.

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Informal justification of $\text{positive} \times \text{negative} = \text{negative}$

Continue the pattern:

$$ \begin{array}{r} 2 & \times & 3 & = & 6\\ 2 & \times & 2 & = & 4\\ 2 & \times & 1 & = & 2\\ 2 & \times & 0 & = & 0\\ 2 & \times & -1 & = & ? & (\text{Answer} = -2 )\\ 2 & \times & -2 & = & ? & (\text{Answer} = -4 )\\ 2 & \times & -3 & = & ? & (\text{Answer} = -6 )\\ \end{array} $$

The number on the right-hand side keeps decreasing by 2.

Informal justification of $\text{negative} \times \text{negative} = \text{positive}$

Continue the pattern:

$$ \begin{array}{r} 2 & \times & -3 & = & -6\\ 1 & \times & -3 & = & -3\\ 0 & \times & -3 & = & 0\\ -1 & \times & -3 & = & ? & (\text{Answer} = 3 )\\ -2 & \times & -3 & = & ? & (\text{Answer} = 6 )\\ -3 & \times & -3 & = & ? & (\text{Answer} = 9 )\\ \end{array} $$

The number on the right-hand side keeps increasing by 3.

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Simple and just what I needed. Thank you. – John H Feb 18 '12 at 21:07
And very informal :) But I like intuitive approaches. Separates us from the nerds who live their lives without intuition. – Alfe Nov 12 '13 at 11:28
@Alfe I would say the opposite. "Nerds" might not talk "intuitive" to you, but I think their intuition is much better than "ours" and that might be what makes them nerds.:-) – quapka Nov 20 '14 at 0:25
I find it nice to extend this to all the reals by drawing a graph. – Michael Anderson Jun 12 at 7:41
This is by far the best answer, +1. – Zubin Mukerjee Jun 13 at 3:25

Hint $\rm\ \ x\, y\ =\ (-x)\, (-y)\ $ by rewriting $\rm\ x\, y + x\ (-y) + (-x)\, (-y)\ $ in two different ways.

Or: $\rm\ x\ y,\ (-x)\ (-y)\ $ are inverses of $\rm\ x\ (-y)\ $ so are equal by uniqueness of inverses.

As I often emphasize, uniqueness theorems provide powerful tools for proving equalities.

Remark $\ $ More generally the Law of Signs holds for any odd functions under composition, e.g. polynomials with all terms having odd power. Indeed we have

$\qquad\rm\ \ f(g)\ =\ (-f)\ (-g)\ = -(f(-g))\ \iff\ f(-g)\ = -(f(g)) $

$\qquad\rm \phantom{\ \ (-f)\ (-g)\ = -(f(-g))\ =\ f(g)\ }\!\!\overset{\ \large g(x)=x}\iff \ f(-x)\ = -f(x),\ $ ie. $\rm\:f\:$ is odd

Generally such functions enjoy only a weaker near-ring structure. In the above case of rings, distributivity implies that multiplication is linear hence odd (viewing the ring in Cayley-style as the ring of endormorphisms of its abelian additive group, i.e. representing each ring element $\rm\ r\ $ by the linear map $\rm\ x \to r\ x,\ $ i.e. as a $1$-dim matrix).

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+1 thanks for bringing uniqueness theorems into this – Sev Nov 14 '10 at 20:02
+1. The simplest way to prove why is by using the group axioms and thier consequences. – Mathemagician1234 Jun 13 at 3:18

The elementary intuition behind the product of two negatives can be thought of as follows. You have a bank account. You pay 3 bills for 40 dollars each, $3 \cdot (-40) = -120$ is added to your account.

The opposite of being billed would be billing someone else.

So, if you bill 3 people for 40 dollars each, $(-3) \cdot (-40)$ is added to your account. This value should be positive since it results in you receiving money.

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Here's a proof. First, for all $x$, $x\cdot 0=x\cdot(0+0)=x\cdot 0 +x\cdot 0$. Subtracting $x\cdot0$ from each side, $x\cdot0=0$. Now, for all $x$ and $y$, $0=x\cdot0=x\cdot(-y+y)=x\cdot(-y)+x\cdot y$. Subtracting $x\cdot y$ from both sides, $x\cdot(-y)=-(x\cdot y)$. Applying this twice along with the identity $-(-a)=a$, $(-x)\cdot(-y)=-(-x)\cdot y=-(-(x\cdot y))=x\cdot y$.

Your proof implicitly uses the fact that $-xy=(-x)y$, and assumes that there are only two possibilities, $xy$ or $-xy$, then shows that the latter is impossible. These seem like plausible assumptions, but I tried to be very careful in my proof above (thus using $-(x\cdot y)$ rather than simply $-xy$ to not be confused with $(-x)\cdot y$).

I only have a vague intuitive notion that I probably can't explain well, but I sometimes think of a negative number like $-5$ as being "$5$ in the other direction", and so multiplying by $-5$ means "multiply by $5$ and switch direction", i.e., sign. This means if you multiply $-5$ by a negative number, you should switch its direction back to positive.

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+1 thanks for the math as well as the intuition here – Sev Nov 14 '10 at 20:02

Simple Answer: $$ (-a)b + ab = (-a)b + ab $$ $$(-a)b + ab = b(a-a) $$ $$(-a)b + ab = b(0) $$ $$(-a)b + ab = 0 $$ $$(-a)b = -ab $$ $$(-a)(-b) + (-ab) = (-a)(-b) + (-a)b $$ $$(-a)(-b) + (-ab) = (-a)(b-b) $$ $$(-a)(-b) + (-ab) = (-a)(0) $$ $$(-a)(-b) + (-ab) = 0 $$ $$*(-a)(-b) = ab $$ Hope this helps (Credit to Michael Spivak's Calculus)

~ Alan

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Well,I wouldn't exactly call that a SIMPLE answer. A DIRECT and STRAIGHTFORWARD one,yes,absolutely. But not simple. – Mathemagician1234 Nov 20 '14 at 0:30
Easily the best and most satisfying answer here. – Derek 朕會功夫 Aug 17 at 20:42

This is really one of those important questions that leads many people to say "Math sucks!". In fact, for many students, mathematics stopped making sense somewhere along the way. Either slowly or dramatically, they gave up on the field as hopelessly baffling and difficult, and they grew up to be adults who — confident that others share their experience — nonchalantly announce, “Math was just not for me” or “I was never good at it.” Usually the process is gradual, but for Ruth McNeill, the turning point was clearly defined. In an article in the Journal of Mathematical Behavior, she described how it happened:

What did me in was the idea that a negative number times a negative number comes out to a positive number. This seemed (and still seems) inherently unlikely — counter intuitive, as mathematicians say. I wrestled with the idea for what I imagine to be several weeks, trying to get a sensible explanation from my teacher, my classmates, my parents, any- body. Whatever explanations they offered could not overcome my strong sense that multiplying intensifies something, and thus two negative numbers multiplied together should properly produce a very negative result. I have since been offered a moderately convincing explanation that features a film of a swimming pool being drained that gets run back- wards through the projector. At the time, however, nothing convinced me. The most commonsense of all school subjects had abandoned common sense; I was indignant and baffled.

Meanwhile, the curriculum kept rolling on, and I could see that I couldn’t stay behind, stuck on nega- tive times negative. I would have to pay attention to the next topic, and the only practical course open to me was to pretend to agree that negative times nega- tive equals positive. The book and the teacher and the general consensus of the algebra survivors of so- ciety were clearly more powerful than I was. I capitu- lated. I did the rest of algebra, and geometry, and trigonometry; I did them in the advanced sections, and I often had that nice sense of “aha!” when I could suddenly see how a proof was going to come out. Underneath, however, a kind of resentment and betrayal lurked, and I was not surprised or dismayed by any further foolishness my math teachers had up their sleeves.... Intellectually, I was disengaged, and when math was no longer required, I took German instead.

I will show in this answer that: negative $\times$ negative $=$ positive, is in fact not counter-intuitive at all! There are many ways that we can use to show that result, but I'd like to show my personal way of thinking about the latter.

Let's imagine we're sitting near a road, and there is a car that is moving with a constant speed. We also have a clock, and so we can measure time.

Before going any further, we should first specify some assumptions like if the car is moving in the right, then its velocity will be positive, and if it's moving in the left direction, then its velocity will be negative.

enter image description here

Imagine now that you have a video of the above scene, and time is positive if you play the video normally but is negative if you play it backwards. We also know the following: $$\rm Velocity=\dfrac{\rm Distance}{\rm Time}.$$ Solving for distance we get: $$\rm Velocity\times{\rm Time}={\rm Distance}.$$ Here the important part comes, if the car is moving in the $+$ direction and the time the video is played is positive, i.e. the video is played normally, then you'll see that the car moves along the $+$ direction and you'll calculate that it moves "a positive distance". Therefore the following holds: $$\rm positive\times positive=positive.$$

If the car moves along the $-$ direction and the time the video is played is positive, i.e. the video is played normally, then you'll see the car going along the $-$ direction, you'll then calculate that it moves "a negative distance". Thus: $$\rm negative\times positive=negative.$$

If however, the car moves along the $+$ direction but the time the video is played is negative, i.e. the video is played backwards, then you'll see that the car is moving in the $-$ direction, and you'll calculate that it moves "a negative distance". Thus: $$\rm positive\times negative=negative.$$

If the car moves along the $-$ direction but the time the video is played is negative, i.e. the video is played backwards, you'll see that the car moves along the $+$ direction! Thus it moves "a positive distance". And therefore: $$\color{grey}{\boxed{\color{white}{\underline{\overline{\color{black}{\displaystyle\rm\, negative\times negative=positive.\,}}}}}}$$ As you have seen, it takes only a little bit of imagination for it to make sense.

I hope this helps.
Best wishes, $\mathcal H$akim.

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The material you quoted contains the interesting statement "Whatever explanations they offered could not overcome my strong sense that multiplying intensifies something". I wonder whether anyone tried the very short answer "yes, but negative intensification is attenuation"? – Andreas Blass Jun 29 '14 at 21:43
@AndreasBlass Or in another form: "negative multiplication is reverse intensification". – Hakim Feb 22 at 19:56

I think the x and y get in the way a bit; you can see the crucial steps using just 1 and -1. What you've really shown is that (-1)(-1)=-1 leads to a contradiction. If we divide by -1, we get -1=1, which is not true!

Getting the right answer, (-1)(-1)=1, uses a couple more steps: First, you must agree that (1)+(-1)=0, (1)(-1)=-1, and (0)(-1)=0.

Now, we multiply the first equation by (-1) and use the distributive property to get (-1)(-1)+(-1)(1)=(-1)(0). Now, we simplify the parts we know to get (-1)(-1)+(-1)=0. Solve for (-1)(-1), and you get (-1)(-1)=1.

So, we must have (-1)(-1)=1 if we accept basic rules of arithmetic: 0 is the additive identity, 1 is the multiplicative identity, -1 is the additive inverse of 1, and multiplication distributes over addition.

One physical explanation people often like for negative*negative=positive is multiplying rates. You can film someone walking forwards (positive rate) or walking backwards (negative rate). Now, play the film back, but in reverse (another negative rate). What do you see if you play a film backwards of someone walking backwards? You see the person walking forwards, because negative*negative=positive!

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Quite a good explanation is that one wants the distributive law to work in general with positive quantities when you add (smaller) negative ones: If $x>a\ge0$ and $y>b\ge0$ then $$ (x-a)(y-b)=(x+(-a))(y+(-b))=xy+(-a)y+x(-b)+(-a)(-b) $$ For this to always work, one needs $(-a)y=-(ay)$ in case $b=0$, $x(-b)=-xb$ in case $a=0$, and $(-a)(-b)=ab$.

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One thing that must be understood is that this law cannot be proven in the same way that the laws of positive rational and integral arithmetic can be. The reason for this is that negatives lack any "external" (external to mathematics, ie. pre-axiomatic, intuitive, conceptuel, empirical, physical, etc.) definition.

For example. Without even getting into the Peano axioms, I can prove that, where $a$ and $b$ are positive integers, $ab=ba$. Indeed, $ab$ is just the process of taking $a$ sets of $b$. Take one element from each of these sets, thus forming a set of $a$ elements. Repeat this $b$ times: you will clearly use up exactly all of the elements and obtain $b$ sets of $a$ elements, in other words, $ba$. Similar informal (but entirely convincing, reasonable, and I would say irrefutable) reasoning can be used to demonstrate the rules for manipulating positive fractions, say.

Notice that in the above paragraph I used the fact that both positive integers and positive integer multiplication have pre-axiomatic, "physical" definitions.

Ask someone why the product of two negatives is positive, and the best they can do is explain, not prove. "Well, negative kind of means 'opposite', so doing the opposite twice means doing the usual, ie positive" does not constitute a proof, but merely an explanation serving to make the accepted mathematical axiom less surprising. Another common one begins with "we would like the usual properties of arithmetic to hold, so assume they do...", but then it remains to be explained why it's so important that the usual laws of arithmetic hold. Euler himself, in an early chapter of his textbook on algebra, gave the following supremely questionable justification. After justifying $(-a)b=-(ab)$ by analogies with debts, he writes:

It remains to resolve the case in which - is multiplied by -; or, for example, -a by -b. It is evident, at first sight, with regard to the letters, that the product will be ab; but it is doubtful whether the sign + or the sign - is to be placed before it, all we know is, that it must be one or the other of these signs. Now, I say that it cannot be the sign -: for -a by +b gives -ab, and -a by -b cannot produce the same result as -a by +b...

With no disrespect to Euler (especially consdiering this was intended as an introductory textbook), I think we can agree that this is a pretty philosophically dubious argument.

The reason it is impossible is because there is no pre-axiomatic definition for what a negative number or negative multiplication really is. Oh, you could probably come up with one involving opposite "directions", and notions of symmetry, but it would be quite artificial and not at all obviously "the best" definition. In my opinion, negatives are ultimately best understood as purely abstract objects. It so happens - and this is quite myseterious - that these utterly abstract laws of calculation lead to physically meaningful results. This was nicely expressed in 1778 by the mathematician John Playfair, when addressing the then controversial issues of negative and complex numbers:

Here then is a paradox which remains to be explained. If the operations of this imaginary arithmetic are unintelligible, why are they not altogether useless? Is investigation an art so mechanical, that it may be conducted by certain manual operations? Or is truth so easily discovered, that intelligence is not necessary to give success to our researches?

Quoted in Negative Math: How Mathematical Rules Can be Positively Bent by Alberto A. Martínez.

One way of approching the problem is with the idea that negative numbers are a different name for subtraction. The differences between subtraction and addition force us, if we reject negatives, to create many different rules covering all the different possibilities ($a - b$, $b - a$, and $a + b$, and if the particular theorem or problem involves more than two variables, the difficulty is compounded further...). The idea of negatives could be described as the insight that rather than having two operations and one type of number, we can have one operation and two types of number. Indeed, if you start with some perfectly physically meaningful axioms about subtraction, you will find that the $(-1)(-1)=1$ law seems to be implicit within them. Hint: starting from the very reasonable axioms $a(b-c)=ab-ac\ ,\ a - (b - c) = a - b + c$, consider the product $(a-b)(c-d)$.

But even that explanation doesn't altogether satisfy me. I've become convinced that my education cheated me on how deep an idea negative numbers are, and I expect to remain puzzled by them for many years. Anyway, I hope some of the above is useful to someone.

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A bit late, but I have written a formal version of Jonas' proof here. The "applying this twice" was the only tricky part.

See Lemma 7 starting at line 224 in

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Perhaps some intuition can be gained by plotting each number's position on the number line. Taking the inverse of any number is visualized by taking the mirror-image of the original plot. So the inverse of a positive number (a point to the right of zero) is a negative number (a point to the left of zero, at the same distance from zero). Likewise, the inverse of a negative number is a positive number. If we agree that multiplying a number by -1 is the same as finding the inverse, then we can see that the product of two negatives must be a positive, because the mirror-image of a mirror-image is the original image.

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Extend reals to the complex plane. Multiplication by $-1$ is a rotation by $\pi$ radians. When you multiply by two negatives, you rotate by $2\pi$. :-)

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As for the product of two negatives being a positive, simply consider the multiplicative inverse:

$$-a\cdot -b$$ $$(-1)a\cdot (-1)b$$ $$(-1)(-1)a\cdot b$$

Note that $(-1)^{-1} = -1$. Any number times its multiplicative inverse is $1$.

$$(-1)(-1)^{-1}a\cdot b$$ $$(1)a\cdot b$$ $$=a\cdot b$$

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I prefer the explanation by my favorite mathematician , V. I. Arnold (physicist really, since in his own words, "mathematics is a part of physics" and "an experimental science"). I believe it's the most natural (yet totally mathematical) explanation of a basic notion like multiplication of negative numbers.

This is an excerpt from Arnold's wonderful memoir "Yesterday and Long Ago" (3d ed., available in English from Springer), full of world history, drama and ingenious storytelling. The 2007 translation into English, I believe, is not of best quality, but it's the only one so far.

from the short story The Arnold Family

I faced a real difficulty with school mathematics several years after the multiplication table: it was necessary to leam that “minus multiplied by minus is plus” I wanted to know the proof of this rule; I have never been able to leam by heart what is not properly understood. I asked my father to explain the reason why (—1) • (—2) = (+2). He, being a student of great algebraists, S. O. Shatunovsky and E. Noether, gave the following “scientific explanation”: “The point is,” he said: “that numbers form a field such that the distributive law (x+y)z=xz+yz holds. And if the product of minus by minus had not been plus, this law would be broken”.

However, for me this “deductive” (actually juridical) explanation did not prove anything - what of it! One can study any axioms! Since that day I have preserved the healthy aversion of a naturalist to the axiomatic method with its non-motivated axioms.

The axiomophile Rene Descartes stated that “neither experimental tests that axioms reflect a reality, nor comparison of theoretical results with reality should be a part of science” (why should results correspond to reality if the initial principles do not correspond to it?).

Another thesis of Descartes’ theory and methods of education is even more peculiar and contemporary: “It is necessary to forbid all other methods of teaching except mine because only this method is politically correct: with my purely deductive method any dull student can be taught as successfully as the most gifted one, while with other methods imagination and even drawings are used unavoidably, and for this reason geniuses advance faster than dunces”.

Contrary to the deductive theories of my father and Descartes, as a ten year old, I started thinking about a naturally-scientific sense of the rule of signs, and I have come to the following conclusion. A real (positive or negative) number is a vector on the axis of coordinates (if a number is positive the corresponding vector is positively directed along this axis).

A product of two numbers is an area of a rectangle whose sides correspond to these numbers (one vector is along one axis and the other is along a perpendicular axis in the plane). A rectangle, given by an ordered pair of vectors, possesses, as a part of the plane, a definite orientation (rotation from one vector to another can be clockwise or anti-clockwise). The anti-clockwise rotation is customarily considered positive and the clockwise rotation is then negative. And lastly, the area of a parallelogram (for example, a rectangle) generated by the two vectors x and у (taken in a definite order) emanating from one point in the plane is considered to be positive if the pair of vectors (taken in this order) defines positive orientation of the plane (and negative if the rotation from the direction of the vector x to the direction of the vector у is negative).

Thus, the rule of signs is not an axiom taken out of the blue, but becomes a natural property of orientation which is easily verified experimentally.

from the short story Axiomatic Method

My first trouble in school was caused by the rule for multiplication of negative numbers, and I asked my father to explain this peculiar rule.

He, as a faithful student of Emmy Noether (and consequently of Hilbert and Dedekind) started explaining to his eleven-year-old son the principles of axiomatic science: a definition is chosen such that the distributive identity a(b+c)=ab+ac holds.

The axiomatic method requires that one should accept any axiom with a hope that its corollaries would be fruitful (probably this could be understood by the age of thirty when it would be possible to read and appreciate “Anna Karenina”). My father did not say a word either about the oriented area of a rectangular or about any non-mathematicai interpretation of signs and products.

This “algebraic” explanation was not able to shake either my hearty love for my father or a deep respect of his science. But since that time I have disliked the axiomatic method with its non-motivated definitions. Probably it was for this reason that by this time I got used to talking with non-algebraists (like L. I. Mandel’shtam, I. E. Tamm, P. S. Novikov, E. L. Feinberg, M. A. Leontovich, and A. G. Gurvich) who treated an ignorant interlocutor with full respect and tried to explain non-trivial ideas and facts of various sciences such as physics and biology, astronomy and radiolocation.

Negative numbers I came to understand a year later while deriving an “equation of time”, which takes into account a correction for the length of a day corresponding to the time of year. It is not possible to explain to algebraists that their axiomatic method is mostly useless for students.

One should ask children: at what time will high tide be tomorrow if today it is at 3 pm? This is a feasible problem, and it helps children to understand negative numbers better than algebraic rules do. Once I read from an ancient author (probably from Herodotus) that the tides "always occur three and nine o'clock". To understand that the monthly rotation of the Moon about the Earth affects the tide timetable, there is no need to live near an ocean. Here, not in axioms, is laid true mathematics.

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2 very good and insightful anecdotes-although I wouldn't really call either of them PROOFS, would anyone here?They're more intuitive explanations why it's so.Of course, the natural next question is:Can either of these arguments be precisely reformulated as proofs? – Mathemagician1234 Jun 13 at 3:21
@Mathemagician1234 you know, I agree with Arnold on the idea that explanations or intuition like this are much much more powerful than any proofs or definitions, because they are real and without artificial restrictions of formal systems. Using them, one can come up with a correct system of definitions in no time, if they have to. However, having the definitions in the first place does not add anything to one's true understanding of the subject. – level1807 Jun 13 at 7:36

It might be easiest to explain using whole numbers. Suppose $P$ is some positive number. Then $-P$ is negative. Now $-2P$ is $P$ subtracted from $-P$, so is still negative. Subtract another $P$ and you get $-3P$, which is still negative. Similarly for $-4P, -5P$, and so on. So negative times positive is positive. Same idea for positive times negative.

When it comes to negative times negative, it's a little harder... But how about... $-P$ is negative, so $-(-P)$ is now positive, flipping around $0$. So $-2(-P)$, which is $-(-P)$ added to itself, is still positive. In general adding $-(-P)$ to itself $Q$ times gives $(-Q)(-P)$, which is therefore positive as well.

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We know that $\mathbb{R}\subset \mathbb{C}$.

Let $x,y\in \mathbb{R}$ and $x<0, y<0$

Any number $n$ can be written in polar form as $n=|n|e^{i\theta}$ where $\theta$ is the angle made by the line joining origin and $n$ with positive $x$ axis.

Therefore $x=|x|e^{i\pi}$ and $y=|y|e^{i\pi}$


$x$ and $y$ being negative but there product is $|x||y|$ which is positive!


$e^{im}=\cos m +\iota \sin m$

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-1: This is an incredibly over-complicated argument about real numbers that actually holds in far more generality anyway. – Zev Chonoles Jun 12 at 7:42
@ZevChonoles Sir though it may feel complicated it is technically correct and provides and alternative approach to the given problem. I request you to kindly elaborate your comment. – Singh Jun 12 at 7:56
@Zev I agree ,but not only this, proving it's true in the complex plane does NOT necessarily imply it's true for real numbers only. To give a simple counterexample, the square roots of negative numbers can be multiplied in C, which does not carry over by inclusion to R. – Mathemagician1234 Jun 13 at 3:17

One way to picture this is to imagine a number line. Then rotate it $180^{\circ}$. Each number will now be superimposed over its negative: $-1$ will be where $+1$ was; $+2$ will be where $-2$ was. Rotation of the number line by $180^{\circ}$ is the equivalent of multiplying by $-1$.

Now do the rotation twice. The number line is unchanged. So, multiplying by $-1$ twice is the same as multiplying by $+1$.

This approach has applications with Complex numbers. In these scenarios, the number line is rotated $90^{\circ}$ counter clockwise to multiply by $i$.

But that's another story.

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