why does multiplying two negative numbers equal a positive number?
>>17071519Multiplying by -1 is a 180 degree rotation.
>>17071519because miggers keep using axioms that say it does>>17071522migger
>>17071522>I multiply by -2 and walk away
>>17071519Its just magic math bullshit. There is no real world use case for multiplying two negative numbers. No one is saying "Oh bro, -100,000,000 people have a debt of -100,000,000, the economy is going to be raped from inflation". Its all worthless junk
The symbol on the numbers only means direction.
>>17071534
>>17071540Good to know.
You have a debt of 200. You haven't paid for 2 months. That means you have (-200)(-2)=400 in your account.
Multiplying two like signs is what yields a positive number. Why does multiplying two positive numbers give a positive number? Because they have like signs.
>>17071519It's the only way multiplication make sense.
>>17071519since in any ring including the real number fieldwhen a and b are positive(-a)*(-b)=-((a)*(-b)) (since the sum of the both sides is zero)=-(-((a)*(b))) (since the sum of the both sides is zero)=a*b (since the inverse element of addition is unique)>0 (because of the axion of the real number field)
>>17071519>(-2) * (-2) = -4>2 * (-2) = -4>2 = -2>negative numbers are positive and the minus doesn't mean anythingThat's why.
>>17071522Wdym 180°? I multiple on 3D not 2D?
>>17071519Multiplying your walking speed by 2 means walking twice as fast. Multiplying it by –1 means walking backwards. But if you go backwards relative to an already backwards direction (–1 × –1) you end up going forwards.
>>17071750Sir, I am wheelchair-bound and I don't understand your answer. I am not going in any direction.
Creators of the system decided so.
>>17071519Forget about negative numbers for a second. Actually, forget about numbers. Let's say you have three possible values: {'a', 'b', 'c'}. Now suppose you want to define multiplication between them. The values in and of themselves don't mean anything (yet), so you can't intuit it. You have to go by the properties of multiplication. You need the zero and identity properties. Let's say 'a' is the zero element and 'b' is the identity element:a*X = ab*X = XFor c*X, commutativity gives you:c*a = ac*b = cBut what about c*c?c*c =/= a because only multiplying by the zero element should produce the zero element.c*c =/= c because then c*c = c*b, implying c=b.c*c = b is your only option.{a, b, c} are analogous to {0, 1, -1}. So without assuming anything, going just by what it means to multiply, you can see (-1)*(-1) = 1 is the only way it can work.
>>17071622Think, Anon. Think long and hard about what rotating a vector through 180 degrees in R^3 would look like.Take your time.
>>17071760>I am not going in any direction.Downhill is a direction.
>>17071760Multiplying your wheeling speed by 2 means wheeling twice as fast. Multiplying it by –1 means wheeling backwards. But if you wheel backwards relative to an already backwards direction (–1 × –1) you end up wheeling forwards.
>>17071982I still don't understand what wheelchairs have to do with math and I don't think your reasoning is sound.
>>17071519I know this is a troll thread but I really wonder where the "joke" is supposed to be with all these shitposts. It's like a competition who can pretend to care the least while still posting.To answer the question though: It follows from asserting that distributivity holds for negative numbers, too.6*6 = 36, 7*7=49. 7*7=(6+1)(6+1) = 6*6 + 2*6 +16*6=(7-1)(7-1) = 7*7 - 2*7 +1
>>17072421Meh. You could have an alien arithmetic that doesn't even HAVE addition and (-)*(-)=(+) would still be the only option, so long as multiplication has the zero, identity, commutativity and cancellation properties.
>>17072434That's because you think of - from a product group perspective while I describe it as the additive inverse of a field. Technically you can even drop the zero in your case. Take point group symmetries, where you can have different kinds of parities (horizontal and vertical mirror planes). In those cases you have basically two different kinds of minus sign and no sensible definition of + at all.
>>17071519Any type of math starts with a set of arbitrarily decided-upon axioms. They only need to be internally self consistent. Example: Non-Euclidean geometry as opposed to plane geometry. They do not agree. Things in one are nonsensical in the other. However, both are internally self-consistent, and thus valid.By this logic, it may be possible to have a type of math where two negative numbers, multiplied, always have a negative result.My own work on this was merely for an audio distortion algorithm: I needed for two negative numbers to have a negative result when multiplied.Over time, though, while I have not done a complete nor rigorous analysis, I've yet to find an inconsistency. In fact, having two negative numbers always multiply to be negative means that division has to work in a similar way, which results in the square root of negative one being negative one. This completely removes any perceived need for complex numbers resulting from operations on Real numbers!
>>17072444Yeah, I understand. All I'm saying is that if you only had "unsigned" numbers and multiplication, but then for some reason you wanted to extend your number system by attaching a "direction" component, you could work out that (-)*(-)=(+) just by requiring that multiplication work the same way on that extra component and your nerfed arithmetic would work just like normal arithmetic for the things it can do.
>>17071519Let's say I owe you 2 donuts. Then let's say I multiply those 2 donuts I owe you by an owance of 2. If I were to multiply the donuts I owe you by 2, I would owe you 4. But if I multiply it by 2 * the concept of owing you a donut. Well, now you owe ME 4 donuts, of course.
ITT: Everyone who fails to carry the units gets the wrong answer and an invalid interpretation. Physics laughs at Maths for their sloppiness.
>>17071519negative numbers do not even exist.Give me -1 apple please.
>>17072711*steals your apple*
>initial rules[math] \begin{aligned} \forall & (a,b,c) \in R_{ing} & a + b & = b + a \\\exists 0: 0 & = 0a = a0 & \exists(-a)& :\ 0 = (-a) + a = a + (-a)\\a & = 0 + a \\(a+b)+c & = a+(b+c) & (ab)c & = a(bc) \\a(b+c) & = ab + ac & (a+b)c & = ac + bc \\\end{aligned} [/math]>play the game[math] \begin{array}{c c c c c c} 0 & = & (-a)0 & = & -a(-b + b) & = & (-a)(-b) + (-a)b \\0 & = & 0b & = & (-a + a)b & = & (-a)b + ab \\[10pt]\end{array} [/math][math] \Rightarrow\ \begin{aligned}(-a)(-b) + (-a)b & = (-a)b + ab = ab + (-a)b\\(-a)(-b) + \underline{ (-a)b + (-((-a)b)) } & = ab + \underline{ (-a)b + (-((-a)b)) } \\[5pt] (-a)(-b) + 0 & = ab+ 0 \\ (-a)(-b) & = ab\end{aligned} [/math]
>>17071519>>17072857forgot to @ OP
>>17071519it's a consequence of the field axioms>>17071612or perhaps the more general ring axioms, but i haven't studied those as much
>>17071519I've heard it explained a few different ways and none of them clocked for me.The only way that made sense was to think of it in terms of cardinal directions.
>>17072711First, give me 1 apple. Exactly 1 apple, and nothing extra.
>>17072857interesting
>>17072904Do the following things make sense to you:1) "a times b is the same as adding up a copies of b"2) "-1 plus 1 is 0"3) "Adding 0 does nothing"If -1 times -1 is not 1, at least one of these is false.
>>17073003>1) "a times b is the same as adding up a copies of b"That already doesn't make sense.
>>17073005so you are saying that b+b=2b makes no sense to you?
>>17073007I'm saying "add -pi copies of 2" makes no sense to me. You have a kindergarten idea of arithmetic.
>>17073011>You have a kindergarten idea of arithmetic.Not really, it is just you who fails understanding math at the kindergarden level if you struggle with -1*-1 = 0. >piOh you're one of those troglodytes who have hot opinions about the real numbers.Christ, you are a complete fucking moron. The reals inherit their arithmetic from the rationals, which inherit most from the integers, which inherit most from the natural numbers. That multiplication with an irrational confuses you is because you don't understand how it is the unique extension of rational arithmetic. Entirely unrelated to the subject of the thread.
-1*-1 = 1, too annoyed with your brainletism to even proofread. Christ you smug reddit faggots need to get a grip.
>>17073013I have no idea what your psychotic breakdown is about. I'm just asking: how do you do something -pi times?
>>17073014You have to be told 0.999r = 1 or negative numbers become positive when multiplied. You don't reach these conclusion naturally...Why? Fake money rules/cheating/conspiracy.
>>17073016Math is even scarred with 'Pi'- a pseudo infinite string- apparently you can know an infinite directly
>>17073017It's just some guys feel. You feel more 'Pi' into existence. There's a normal way to do this, non-invasive of maths.
Most/if not all but me, on this planet, are super dumb to some degree.If you check my thread- on head/brain- that's why I stick out above the rest.
>>17073011kek. bodied that freak>>17073013holy seethe, imagine getting this mad about being wrong
>>17073015>psychotic breakit wasn't difficult to understand what I said, it is just that you are an inbred who thinks real numbers have something to do with integer arithmetic. Let me try again for the mouth breathers:1) multiplying in the domain of integers should uphold that a times b is adding a copies of b.2) How this is generalized to rationals is that "a/b times c/d should be (ac)/(bd)", which is the only way to generalize the integer products to fractions.3) Generalizing this to reals is done with limits (for example cauchy completion) and it turns out that because the product in the rationals is unique this is unique too. But 2-3 are not necessary to understand OP's question
>>17073034>have a provable true statement (in the integers)>somehow "wrong"of course I am annoyed, dysgenic vermin annoy me
>>17073154>multiplying in the domain of integers-pi is not in the domain of integers.>Generalizing this to reals is done with limits (for example cauchy completion)So why don't you show me how you "generalize" to doing something a negative, irrational number of times? You can start from the "negative number of times" part.
>>17073155>have a provable true statement (in the toddler playpen)
>>17072711*kicks you in the crotch*
>>17071519Because it has the same effect as multiplying by the imaginary number i 4 times.
>>17072857(0 = 0a = a0) is a redundant rule since distribution is already a rulea0 = a(0 + 0) = a0 + a0a0 + [-(a0] = a0 + a0 + [-(a0)] = a0 + 0 0 = a0and you can do the same for 0aAssociative multiplication also isn't used here either so it isn't needed for (-a)(-b) = ab
>>17071538Just warning, you might actually be mentally retarded, so try to be safe while learning to drive or cook etc. Anyway, probably you play a lot of games, and the coordinate system that is used is encoded by vector coordinates, and transformations to these vector coordinates very often involves multiplication of two negative numbers, so for people who aren't mentally retarded, you learn that multiplication of negative numbers is used in actual applications that you learn when you're in early highschool.
>>17075138didn't read
>>17075165haha I love that, thanks for sharing!
>>17075138>its totally useful bro, video game developers (which everyone is of course) uses it to [jargon] and [jargon] in their video games. Bro EVERYONE does this, youd have to mentally retarded to have missed thisNot as convincing of an argument as it probably sounded in your head
>>17075426You realize that the fundamental building blocks of our universe is very similar to code, right?
>>17075426Imagine thinking negative number multiplication isn't real when it is used in every game you've ever played, you'd have to be literally retarded (sub 70 iq)
>mfw I skipped highschool mathematics when they explained reflection matrices>mfw ignored any circuit diagram modules in highschool physics>mfw actually retarded for life
>>17071519can you give a definition of negative and positive and multiplying? This seems interesting. I'll provide my owna positve number is an abstraction of a discrete quantity, e.g. 5 sheep and 5 dolphinstaking: if every time we kill a sheep we kill the same quantity of dolphin til none remain then we can be sure after killing a sheep 5 times then the dolphines are gone too. times: I used that word in the definition of taking and I will make it more abstract. we will say that a positive number a times another positive number b is b summed a times where sum is the opposite of take.a negative number is that positive number which is taken from 0Since negative numbers are just positive numbers but being taken from the origin rather than summed it stands to reason that a double negative gives the positive.
>>17077500Said more succinctly, there is no new information provided by negatives, they are just the positives with another relation. So if you apply that relation in a way that cancels out you will be left with the positives.
>>17071519I'd like to multiply with Bocchi s numbers
>>17077500What the fuck is this retard attempt at philosophy?
>>17071534>>I multiply by -2 and walk awayDo you mean -1 times 2?
I mean why would multiplying two numbers equal a negative number?
>>17077632That’ll fly when the number is *i*.We all cry, “Complexifly!!!”—Please buy my demo tape. 0% AI Promise.
>>17073159Assuming you understand multiplying negative numbers from the integers and rational numbers (where it follows from the unique generalization of the natural numbers), then the hard part is showing how real multiplication relates to rational. When defining reals you necessarily deal with infinite sequences of rational numbers, but assuming you understand equivalence classes you can explicitly construct the real number multiplication by defining a product on sequences:[eqn]\odot : \mathbb{Q}^{\mathbb{N}}\times\mathbb{Q}^{\mathbb{N}}\to \mathbb{Q}^{\mathbb{N}}\\(a_n)\odot(b_n) = (a_n\cdot b_n)[/eqn]Where the product on the right side is the multiplication defined on the rationals.Real number multiplication arises from the fact that his product of sequences is well-behaved. So in other words you would represent -pi as a rational sequence converging to -pi, and whatever other real as a rational sequence converging to that number. Since for these sequences the product of their limits equals the limit of their products, this is the complete definition. The rest is just writing down the details in your preferred number of real number representation. But you can choose a representation such that you can bootstrap it uniquely and constructively from natural numbers.
>>17071519Let (G,+), (H,+) be two commutative groups and f: G^2 -> H a bilinear map (i.e. f(x+y,z) = f(x,y)+ f(x,z) and f (x,y+z) = f(x,y)+f(x,z) for every x,y,z in G). Then for every t in G, f(0,t) = 0 = f(t,0) and for every x,y, f(x,y) = f(-x, -y).Indeed, (i) for any t in G, 0 = f(0,t) - f(0,t) = f(0+0,t) - f(0,t) = f(0,t) + f(0,t) - f(0,t) = f(0,t); the equality f(t,0) = 0 is proven in a similar fashion.(ii) let x,y in G; by (i) w have 0 = f(0,y) = f(-x+x,y) = f(-x,y) + f(x,y), hence f(-x,y) = - f(x,y) (*). Also, we have 0 = f(x,0) = f(x,-y+y) = f(x,-y) + f(x,y) hence f(x,-y) = -f(x,y) (**).From (*) and (**) we get for every x,y, f(-x,-y) = -f(x, - y) = - (- f(x,y)) = f(x,y) (rules of sign).
>>17076394video games are not real, retard. Arbitrary abstractions in code compiling to be interpreted in a computer is not tangible.
>>17077712That's nice. However, nothing is being done an irrational number of times anywhere in your post.
By trichotomy, a*b < 0, a*b = 0 or a*b > 0. (Exactly one of these occurs.)Since a<0 and b<0, -a>0 and -b>0. It can be proved that (-a)*(-b)=a*b. The product of two positives is positive. Thus, a*b>0.
>why does rotating by π twice in a plane equal rotating by 2π radiansIt's a mystery
>>17079399It seems you do not understand what a real number is in this construction. We will first define [math] \mathcal{C}_\mathbb{Q}\subset\mathbb{Q}^\mathbb{N}[/math] using [math] \mathbb{Q}^+ = \{x\in\mathbb{Q}|x>0\} [/math] specifically[eqn] \mathcal{C}_\mathbb{Q} = \left\{(a_n)\in\mathbb{Q}^\mathbb{N}\middle| \forall\varepsilon\in\mathbb{Q}^+\exists N(\varepsilon):\forall n,k>N(\varepsilon):|a_n-a_k|<\varepsilon\right\}[/eqn]This is the set of rational Cauchy sequences.The set of rational Null sequences is:[eqn] \mathcal{N}_\mathbb{Q} = \left\{(a_n)\in\mathbb{Q}^\mathbb{N}\middle| \forall\varepsilon\in\mathbb{Q}^+:\exists N(\varepsilon):\forall n>N(\varepsilon):|a_n|<\varepsilon\right\}[/eqn]With that, [math] \mathbb{R} = \mathcal{C}_\mathbb{Q}/\mathcal{N}_\mathbb{Q} [/math]. This is true up to isomorphism, if you don't like that definition of the reals, you can easily create an isomorphism to this definition. In this definition irrational numbers are equivalence classes of sequences of rational numbers and all arithmetic on reals simplifies to rational, element-wise arithmetic. Yes, pi for example is a an equivalence class in this definition.. which I also explained previously. This is just the long version for slow learners. Any question?
>>17079413>Any question?Can you quote the specific part of your post that implies doing something an irrational number of times? Maybe you're just confused wrt. what the disagreement you stumbled into was actually about?
>>17079422Your explicit example was -pi. Pick any rational sequence that converges to pi (see https://en.wikipedia.org/wiki/Pi), call it [math] (a_n) [/math]. Take any other real numner of your choosing. Let [math] (b_n) [/math] be a rational sequence that converges to it. Given the above definitions[math] \pi\cdot x := [(a_n)\odot(b_n)] = [(a_n\cdot b_n)] [/math]. That is what it means to multiply two real numbers. For rational numbers this can be simplified (mapping to a constant sequence), for irrationals it cannot.
>>17079429>That is what it means to multiply two real numbersSo you agree with me that what is meant is not repeating addition -pi times.
>>17079462that concept is true for integers and uniquely generalizes from integers to all other numbers. You can write it down in arbitrarily many ways, but because of reals and rationals are ultimately equivalence classes of integer vectors (one finite, one countable) and their respective multiplication is just element-wise integer multiplication, the picture holds at its core. It's why distributivity holds for all of them.
>>17079522>that concept is true for integers and uniquely generalizesAt this point I'm forced to conclude you're very, very severely retarded and don't understand what the words you use actually mean. Like a chatbot.
>>17072726How does this work if you have no apples?
