I used to think of it like this: negative is "not", positive is "yes". Negative times positive is "not yes", which is "no" which is negative. And for negative times negative, it's "not no" which is yes, which is positive.
In Vietnamese maths textbooks for grade 6, the intuitive explanation goes like this: (-5) ×3 = (-5) + (-5) + (5) = -15 (-5) ×2 = (-5) +(-5) = -10 (-5) ×1 = (-5) (-5) ×0 = 0 There are clear arrows to indicate that as the 2nd factor decreases by 1, the product goes up by 5. And then students are asked to guess what happens when the factor keeps decreasing to negative values. I think it's intuitive enough for most students, haven't seen anyone struggle with this.
This pattern recognition technique also works well for teaching students how negative integer exponents work. They can see how decreasing the exponent by one is the same as dividing by the base, and continue that pattern into the negative integer exponents.
I like the 3rd proof, because it is intuitive and sets up a foundation for thinking of numbers in terms of directions, which is incredibly useful when learning immaginary numbers. If a negative means a 180° rotation, and rotations add up, then it is intuitive that i=sqrt(-1) represents a 90° rotation.
Very nice observation... an operator operating on a number set, some combine values while other encode rotation.. the way I saw it was the general idea behind all of the many concrete examples worked through in school. And life
but there is nothing "directiony" about real numbers other than order. it feels like something act on another because an operation sends 2 elements to a single and it could be interpreted as a set of actions by currying.
Thinking of the complex plane as a vector makes things way easier. Negation and multiplying by I is rotation, adding values is translation, and multiplying by positive reals is scaling
as a math major, my favourite one was still the number line explanation. if you thing of multiplying by negative as a reflection, reflecting twice just gives the original.
As a non-math person, it made sense to me too ... but only once I started to think about imaginary numbers. Because a follow up question would be "Well, what if I do a half rotation?"
Which is what I'd use to explain it. If you view multiplication by a negative number as a 180° rotation on the pos/neg number line, suddenly it doesn't seem so weird that i exists, that i² is -1 or that multiplying by i is an orthogonal movement.
Imaginary numbers being dubbed “imaginary” will never not be one of the greatest unsung tragedies of modern mathematics, science, and philosophy! There should have been a greater push to re-dub them as “lateral” numbers and introduce the lateral number line and number/rotational plane far earlier in math instruction (or at all since many don’t even get to that point!)
The multiplication of two negative numbers being positive clicked with me during my complex analysis course, akin to the level 3 you presented. In particular, representing negative numbers with euler’s formula and showing that multiplication leads to an angle of 2pi, which corresponds to the positive side of the real number line.
Wow, I feel like I've finally leveled up. This helped me actually understand how a proof for simple concepts works. After spending the last several months studying modular arithmetic and rings, it's finally clicking.
Sir, I'm a 10th grade student from The Philippines and I've been watching your videos since last month. Looking forward for more interesting videos and thank you!
Got randomly recommended this and ended up watching a bunch of your videos. Hope ya keep making them! I majored in math but I think your videos are easy enough to understand and just about anyone can appreciate them. Easiest subscribe in a long time
If you imagine the number line as a sphere, the operation of multiplying by negative one is equivalent to rotating the sphere downwards (if using i, negative i is upwards) passing through the complex modulus a la e**iπ. Factoring out (-2)(-3) to -1(2)•-1(3), you end up at positive six, and travel through both poles in the complex plane to arrive at positive 6.
I was actually thinking of this exact question earlier today when a student asked me what happens when you multiply negative numbers. I got up to level 3 on my own, but level 4 was my sweet spot. The concept of rings was interesting, but really only useful for math majors.
I always thought it was a function with 2 inputs and with that you can define any list of relationships you want with a two input function. So it was because we say it is so.
Even as an engineering student who has a ton of math experience I still think about it as a (-) x (-) = + since one negative sign rotates and combines with the other - to physically create a + sign. Than a (-) x (+) = - since the - erases a line from the plus leaving you with a -. Does it make any sense…. no not really. But did it make sense to 8 year old me… yup!
Very interesting video! Your explanation of the ring is incredibly clear, and most importantly, it makes clear a very important thing in all of mathematics: apparently, all mathematical operations are determined by their properties. The same can be said about physical quantities - they are all nothing more than numbers, but their key difference from random numbers is that they have properties that reflect processes in the physical world. Definitely like!
when I was in elementary school, I thought of it as flipping the whole number line, so flipping it twice is the same as the original if you apply the inverse of negative to the number line, it's the same as flipping in the other direction ends up looking the same as applying the negative to it, which explains why 1÷(-1)=-1 after learning about complex numbers, I thought of it is rotating the number line instead of flipping
Here's an approach from the set theoretic construction of integers built on top of set theoretic construction of natural numbers, natural number addition and natural number multiplications (no subtractions yet): You can define an equivalence relation ~ between a pairs of natural numbers: (a, b) ~ (c, d) if and only if a + d = b + c. To prove that it is in fact an equivalent relation I will prove its 3 defining properties: 1. Reflexivity: a + b = a + b, therefore (a, b) ~ (a, b) 2. Symmetry: a + d = b + c if and only if b + c = a + d, therefore (a, b) ~ (c, d) if and only if (c, d) ~ (a, b) 3. Transitivity: Assume a + d = b + c (a, b) ~ (c, d) Assume c + e = d + f (c, d) ~ (e, f) Therefore a + d + c + e = b + c + d + f a + e + c + d = b + f + c + d a + e = b + f (a, b) ~ (e, f) (This last line is based on x = z x + y = z + y which is trivial) Then, you can define equivalence classes under the ~ relation. Let's use the notation [(a, b)] to represent the equivalence class with (a, b) in it. So, for example, the equiv. class that contains the pair (1, 3) will also contain (2, 4) because 1 + 4 = 3 + 2, will also contain (0, 2) because 1 + 2 = 3 + 0, etc. The underlying motivation here is that if you swap a + d = b + c around, you get a - b = c - d, and [(0,2)] represents 0 - 2 = 3 - 4 = -2, but without having to have subtraction and negative numbers already defined. Then you can define positive integers as any equivalence class that contains (x, 0) where x is a non zero natural number. You can also negative integers as any equivalence class that contains (0, x) where x is a non zero natural number. The proof for how the set of positive and negative integers don't intersect as a little sanity check for these definition, by contradiction: Assume A is a positive integer that is also a negative integer. Therefore A can be represented as (x, 0) where x is a non zero natural number. Therefore A can be represented as (0, y) where y is a non zero natural number. That is, A = [(x, 0)] = [(0, y)]. That is, (x, 0) and (0, y) is in the same equivalent class A. That is, (x, 0) ~ (0, y), which happens if and only if x + y = 0 + 0 = 0, which is a contradiction since both x and y are non zero natural numbers. Therefore A cannot exist. Since natural number addition and multiplication is defined, you can define integer addition as [(a, b)] + [(c, d)] = [(a + c, b + d)] and integer multiplication as [(a, b)] . [(c, d)] = [(ac + bd, ad + bc)]. As far as I know, this is the canon way to construct the set of integers along with integer addition and integer multiplication. Then, from this definition, you can have two integers A = [(0, x)] and B = [(0, y)] where x, y are non zero natural numbers, therefore A and B is negative. Multiplying integers A and B, you'd get the integer [(0*0 + xy, 0*y + x*0)] = [(xy, 0)]. This integer fits in the definition of positive integers that we've defined above. The rational numbers are built from equivalence classes of pairs of integers, the same way integers are built from equivalent classes of natural numbers, but with a different equivalent relation. The proof for that from definition wouldn't be as exciting because it would mostly use the result proven above in one way or another, and for real numbers (which are btw constructed in an entirely different way as Dedekind cuts) it's just an identity in real number multiplication's definition, which is not really exciting to be honest. Hope I haven't made mistakes lol.
I like the expansion of understanding of mulitplication which also holds for complex numbers. A number can be expressed as a length and an angle. A positive number has angle zero, while a negative number has angle pi (or 180 degrees if you don't like radians). I.e. the number 1 can be expressed as length = 1, angle = 0. And -2 could be expressed as length = 2, angle = pi. Multiplication is then an operator which multiply the lengths (which are positive and purely real) and add together the angles. So (-2)(-3) yields length = 2*3 = 6 and angle = pi+pi = 2*pi (which is 0 since angles are mod 2"pi). Interestingly, this holds for complex numbers as well. I.e. i is length = 1, angle = pi/2. With that, it's pretty simple to intuitively perform multiplication within the complex space.
I feel it's easier to understand intuitively through division. Negative number divided by a negative number is positive because thats how many times it can be divided. And division can be easily shown as multiplication.
I thought he was going to talk about the complex plane, multypling two numbers with angle π (negative) give us a number with angle 2π (positive) because we add the angles.
Using the properties of imaginary multiplication in the 2d plane, where multiplying two numbers looks like adding their angles (from the x-axis) and multiplying their distances from the origin, multiplying two negative numbers looks like multiplying their distances, then adding 180°+180°, for 360°, or 0°
It's good to get pupils to derive it themselves. Create a a coordinate system and at each point in the first quadrant get them to enter a number which is equal to the product of the x and y coordinate. They will then see arithmetic sequences (times tables) when considering constant x and constant y. Ask them to extrapolate sensibly into the 2nd and 4th quadrants, and then again into the 3rd quadrant. To keep the structure of arithmetic sequences they will see that all the entries in the 3rd quadrant are positive.
I used to think of the multiplication by -1 as rotating the location vector to the number I’m multiplying on the number line with -1 by 180 degrees around the 0 point. Thus 1 times -1 times -1 yields a rotation by 360 degrees, putting me back to where I started, positive 1. This visualization also comes in handy when trying to get a grasp on the complex plain and understanding why the square root of -1 is orthogonal to the number line. Because instead of rotating by 180 degrees, you just rotate by 90 degrees. And Sqrt -1 times Sqrt -1 is rotating by 90 degrees two times giving you the 180 degrees of the multiplication by -1. There you go.
I like making a times table for 1 to 9 and pointing out how the rows and columns increase regularly. then start extending the table to the other quadrants going from (x,y) to (x-1,y) and (x,y-1) a step at a time
An attempt at a proof by contradiction (I have never taken a formal proofs class, so I won’t be surprised if I have made a mistake somewhere)- Let a, b, and c be arbitrary, strictly positive numbers > 0. Assume that (-a)*(-b)=(-c) for some a, b, and c. We can factor out a negative one from some of these numbers. Rewrite (-b) as (-1)*(b) and (-c) as (-1)*(c) : (-a)*(-1)*(b) = (-1)*(c) divide both sides by negative 1: (-a)*(b) = (c) (does this step implicitly assume that negative times negative is positive? not sure) an this point, we can see that the left side must be negative, and the right side is positive, since a, b, and c were chosen as strictly positive. this is a contradiction, meaning that negative times negative cannot equal negative!
So I've studied a little bit on formal proof's but I'm not a mathematician (just studying Comp Sci) so could also be wrong here. You can use this proof to show -1/-1 = 1 without assuming negative * negative is positive: let a be a non-zero number: a/a = (a)^1 * (a)^(-1) = (a)^(1 -- 1) =a^0 = 1 Thinking about your proof also made me think of a proof, although I may be doing something I can't do: let x,y be any positive (>0) real numbers: assume -x * -y 0) By our above proof: (-x)/(-x) = 1 (-x) * 1/(-x) = 1 we know 1/n != 0 for any non-zero real n, so 1/(-x) is not zero. Since -x is negative, if 1/(-x) were positive then: (-x) * (1/-x) < 0 Thus: by exhaustion 1/(-x) must be negative Thus 1 0
In the intuitive example, I think the person is left with 6 cookies, because it is equivalent to say 6 cookies are put into my 🍪 jar or left in my possession outside of a jar. I think of multiplication and division as the same as adding full containers of the same size. For example, 12 eggs is the same as a carton of a dozen eggs. Multiplication is just adding up cartons: 1/2 of a carton is splitting the carton into two, each with 6 eggs, but I still possess all 12 eggs. The trick is that we are implying that changing the size of a group also means we lose possession of some of the contents and groups. But perhaps we could interpret x/0 to mean the information possessed is ungrouped...and so has to be counted individually.
i feel kinda stupid for having to look up what distributive propertys are but I feel really good for understanding it now.. I cant believe I went so long trying to rearange formulas with this pretty basic concept but Im glad I got it now
Awesome. I'm relearning math to get better for engineering. This helps so much, although simple it helps me actually understand the math vs just learning by memorizing formulas. Can you make one on why we set use zero property rule for quadratics? I can't understand the purpose of making the equation equal to zero.
I'm trying to research logic systems, and I'm running up against constructivism and paraconsistent logic. I know that YT isn’t the ideal place to ask in-depth questions, but perhaps you can point me toward some logic literature in the field? Specifically, (how) are double-negation and apartness related, and are double-negation and equivalence a problem for binary logic/LEM?
It is cool to see Level 5 after just learning about Vector Spaces in my linear algebra class. It was not taught to me as 'rings' the first 7 properties at 5:26 were the same as my textbook, however, the 8th one was '1v = v'. Anyways, cool to see and nice video.
My favorite is multiplication as rotation of a vector in the complex plane. Multiplying by a negative is multiplying by 180°. To multiply a negative by a negative is to rotate by 360°, which is identical to 0°, or positive.
The rules for rings are defined that way because we want them to be useful and generalize the idea of integers, i.e. counting stuff like sheep and yes, debts too.
Would it be possible to use complex numbers to prove it? Because multiplying by i is the same as a 90 degree rotation, and i*i = -1. 4 i’s is 4 90 degree rotations which is 360 degrees. i*i*i*i = 1, and i*i = -1, so (i*i)*(i*i)=1 and (-1)*(-1)=1
Here is one using the construction of the integers: Integers are defined as equivalence classes on N^2 where (a,b)~(c,d) if a+d=c+b. All equivalence classes can be written in the form [(a,0)] or [(0,a)]. Ones in the form [(a,0)] are just denoted as a and ones in the form [(0,a)] are denoted as -a Multiplication is defined as [(a,b)] * [(c,d)] = [(ac+bd,ad+bc)]. We have -1*-1=[(0,1)]*[(0,1)]=[(0*0+1*1,0*1+1*0)]=[(1,0)]=1
Fields needs to have multiplicative identity (1) that is not equal to 0. And all non zero elements has a multiplicative inverse. So fields are also rings but with additional restriction.
I use the last method with all students from 11 years old up. I don't talk about ring theory just algebra after they have learnt to expand brackets and to factorise.
I thought you would use the line method for level 1. Just counting the number of lines and if the number is a multiple of 2 it is positive. - is 1 line and + is 2. You could only count the number of - in general.
I find that it is easier to understand with the complex numbers. If you multiply a complex number by i, it is shifted by π/2(90°): 1 * i = i i * i = -1 -1 * i = -i -i * i = 1 And if you now multiply by -1 instead of multiplying by i, the angle of rotation doubles, i.e. π/2 becomes π(180°).
I like to think of this from a statistical standpoint. You can combine + and - in 4 different ways. (+)x(+), (+)x(-), (-)x(+) and (-)x(-). If you threw a dart randomly at the number line then your answer would have a 50/50 to be (+) or (-). Each one of these 4 combinations accounts for %25 of the answers. Starting with the axiom that a +x+=+ which is %25 of your answers. Then that +x- must also be %25 of your answers and that -x+ is the same so together count for %50 of the answers therefore must be -. Leaving the last %25 (-)x(-) which must be +.
when any complex number (real numbers are a subset of the complex numbers) is multiplied by another complex number their magnitudes are multiplied (this will always be a positive real number so no need to fight the negative multiplication basis here) and their phases are added (phase ranges from 0 to 2*pi with the modulus of any phase greater or equal to 2pi to bring it back) since the positive numbers have a phase of 0 0+0 = 0 and the negative numbers have a phase of pi pi+pi = 2pi since this is equal to or greater than 2pi then when we do the modulus it goes back to 0 its as simple as that we just gotta take a trip to the world of complex numbers
here I've got a question do u not have even a plank length worth of space to move why are u standing is such a tight space? it almost feels uncanny idk also idk if there's already a video on it yet but i want one on imaginary numbers make em feel as natural as breathing air then quantum physics then axioms then ok actually anything math or science related and I'll watch lol
You've described the ring theory and the twelve hours ring, but you've used the same proof from step three again in step five, ie you didn't need to use the ring theory at all, but thanks for explaining the interesting theory though 😊😊
I was literally in the shower thinking „what is the actual mathematical proof of neg x neg = pos” and after i came out of the shower and open yt this showed up.
Very good video, u are really good at explaining. The only critique i have is that i would not call level 5 math major as i did the concept of rings in uni in like the first semester. I study computer science
Haven’t seen level 4/5 done like that before. I would go more like 1. -(-a) = a. Any definition I can think of is obviously symmetric, e.g. if b is called -a when a + b = 0 then a is also -b. 2. (-a)(-b) = ? = -(-a)(b) = ab if you can justify moving the -… 3. -a = (-1)a. No trick: Just check that (-1)a is -a. (-1)a + a = (-1 + 1)a = 0a = 0. 4. 0a = 0. Same way in video :-)
You can build so many proof of this law, using different level in math's. Saying that's exponential is a group morphism from C to C* with 2piZ as a kernel. Or using the Euclidean division in the polynomial ring, writing X^2=(X-1)(X+1)+1 and that - 1 is the unique root of X+1 in Z[X]. Here are the proof that comes immediately at my mind.
It's not a proof, but intuitively works: If you can agree that multiplying bij -1 is just flipping a number's sign. Which I think makes sense. If you also agree that (-a)(b) =-ab. Which also makes sense, because this is just adding -a b times: (-2)(3)= (-2)+(-2)+(-2). I.e. shifting three time 2 to the left on the number line. Then it follows: (-a)(-b) = (-1)(a)(-b) = (-1)(-ab) = ab
Well if the n is the square root of a real number m then n has 2 important properties: 1. n^2=m 2. Both the real part and imaginary part of n are nonnegative For all a and b (sqrt(a)*sqrt(b))^2=sqrt(a)^2*sqrt(b)^2=a*b. So the first property is met no matter what The 2nd propert is where things get complicated. If a and b are both nonnegative then sqrt(a) and sqrt(b) are both nonnegative so sqrt(a)sqrt(b) is nonnegative. So 2nd property holds when a and b are both nonnegative. If a is nonnegative and b is negative then sqrt(a) is positive and sqrt(-b) is positive so sqrt(a)*sqrt(b)=sqrt(a)*isqrt(-b)=i(sqrt(a)*sqrt(b)) has a nonnegative imaginary part (and a real part of 0) so the 2nd property still holds when a is nonnegative and b is negative. By symmetry, when a is negative and b isn't the 2nd property also holds. Now when a and b is negative you get something interesting. sqrt(-a) and sqrt(-b) are both positive so you would have sqrt(a)*sqrt(b)=i sqrt(-a)*i sqrt(-b)=i^2*sqrt(-a)*sqrt(-b)=-sqrt(-a)*sqrt(-b). This is a negative so the 2nd property does not hold. Overall this means that sqrt(a)*sqrt(b)=sqrt(a*b) if and only if a and b are not both negative.
I would use -1 = e^(pi*i). So (-1) * (-1) = (e^(pi*i)) * (e^(pi*i)) = e^(2*pi*i). And by looking at the unit circle we can see that e^(2*pi*i) = 1. Although I guess complex numbers already rely on (-1) *( -1) = 1 so this might be circular.
Yeah, the way I think about negative operations is basically via boolean algebra, as it’s easier to reason about values encoded on a single bit. You can draw a 2×4 table with all possible bit combination, and realize that this negative operation is basically the same as the NOT boolean operator. A true value is +P, and a false value is -P.. Negating simply flips the sign. So -(+P) is -P, -(-P) is +P. Then, -a * -b is the same as first doing a * -b, and flipping the result sign, which ends up positive.
what about charges where you get more of an xor effect -> - * - = - , - * + = + , + * - = +, + * + = - ... pretty sure you can also define vector spaces where - * - is positive as well
Level 6: The positive reals P are a subgroup of the multiplicative group of reals. The reals are then partitioned into two cosets, P and -P (negative reals). The quotient group is then the cyclic group of order 2, and negative * negative = positive. By definition of multiplication in the quotient group, this means that a negative real times another negative gives a positive. I believe this isn't circular because the only things you have to know about the group is that the negative reals and the positive reals are disjoint and their union gives R, and that the positive reals are stable.
At my university i was taught since you can prove -x=(-1)x=x(-1), the fact that multiplication in a ring is associative means that you can "migrate" the minus signs in the expression (-x)(-y) to the front, and get (-x)(-y)=(-1)(-1)xy. Which can then be shown to be equal to xy.
The first meaning that we give to multiplication is adding a number repeatedly. The two numbers involved in the multiplications have very different meaning, the first one indicating which number will be added repeatedly and the second one indicating how many times is this going to happen. From the significance point of view factors are not interchangeable. Negative numbers are introduced to manage situations that are hard to manage using only natural numbers, in fact you would need to use two sets of natural numbers and quantities of one of these sets cancel out with corresponding quantities from the other one. So i think that before answering what -ax-b means we should answer what meaning can be given to repeating addition a negative number of times. But it turns out that repetitions are also quantities that can be sometimes considered of two different kinds that cancel each other: in your second example you mentioned a friend who repeatedly gives or repeatedly takes away the same quantity, there can also be repetitions from this point in time to the future vs repetitions in the past to this point in time: form example if my account now is 0 and a friend has deposited 5 dollars in my account for the past 3 days, it means that my account was -15 dollars 3 days ago. Using the same example if my account was debited 5 dollar for the las three days and now is 0, it means that 3 days ago y had 15 dollars, thus -5x-3=15. My point here is that you basic example is more basic than the first because the first involves continuity and the second one doesn't, but also because it better addresses the questions young students have when they try to reconcile this new "rule" with what they previously knew about this operation.
The way I see it is this: On a number line, each sides “negative” direction is towards 0 and through the other end of 0. 2-3 being 2 towards 0 and 1 through 0 from the right side. And each sides positive direction is farther away from 0. Thus, 2 x 3 = 2 + 2 + 2 Adding 2 in the positive direction (right) 3 times, getting further away from 0. -2 x 3 = -2 + -2 + -2 Adding 2 in the negative direction (left) 3 times. Think of it as starting on the right side of 0 and going through 0 in the negative direction (towards 0). -2 x -3 = -2 - -2 - -2 Subtracting 2 in the “negative”direction 3 times. Think of it as starting from the left (negative) side of 0, and going towards and through 0, in the left sides “negative” direction, which is really the right sides positive direction. Thus meaning that it really is just going positively
The way how I think of negative numbers is this: I picture a graph of a number line in my head. Positive numbers are at the right while negative numbers are at the left and zero is at the middle. When I add a number with positive value, the position of said value moves to the right. When I add a number by a negative value, the direction is reversed. When I multiply a number by a positive number, the position of the number that is multiplied remains the same, then the value gets multipled and moved. When I multiply a number by a negative value, the position of the number being multiplied gets "mirrored", that being the number being converted into a negative value. When the multiplier is greater than 1 or less than -1, the value moves away from 0. When the multiplier = (-1 , 1), excluding zero, the value is moved towards 0. When the multiplier is zero, the value moces straight to 0. For division, the direction is reversed compared to multiplication. However, in the case where the divisor is 0, if the multiplied number is not 0, it cannot be any value. If the multiplied nunber is 0, it "technically" can be any number. However, that would be useless and a worthless excuse to use this number to answer every mathematical problems so it is written as "undefined" instead. For exponentiation, specifically when the exponents are even numbers, "mirrored" values (Negative values) and "normal" values (Positive values) are presented. Roots are the reverse of exponentiations except that if the exponent is an even number, it can only represent one of the possible value (Example: √4 = 2 and only 2, while -√4 = -2 and only -2). I might have explained the concept of negative values as the literal traditional definition of it but personally, I understand them this way. There might be some flaws in them though.
multiplying by i in the complex plane is a rotation of 90 deg. multiplying by i twice is the same as multiplying by -1. Do that again and we rotate a full 360 getting back to a positive number!
Level 0: 4chan
>turn around
>turn around again
>wtf im facing the same direction
real
>don't turn around
>don't turn around again
>wtf i'm facing the same direction
>be me
@@dedede5586 That's positive times positive.
Turn 90°.
Turn 90° again.
Wtf, I'm facing backwards from how I was originally!
And that is why i² = -1.
I used to think of it like this: negative is "not", positive is "yes". Negative times positive is "not yes", which is "no" which is negative. And for negative times negative, it's "not no" which is yes, which is positive.
exactly how i think of it
i use turn around and dont turn around
@@djfghjif that also works
So negative x negative = F in English.
So negative x negative = F in English.
In Vietnamese maths textbooks for grade 6, the intuitive explanation goes like this:
(-5) ×3 = (-5) + (-5) + (5) = -15
(-5) ×2 = (-5) +(-5) = -10
(-5) ×1 = (-5)
(-5) ×0 = 0
There are clear arrows to indicate that as the 2nd factor decreases by 1, the product goes up by 5. And then students are asked to guess what happens when the factor keeps decreasing to negative values. I think it's intuitive enough for most students, haven't seen anyone struggle with this.
This pattern recognition technique also works well for teaching students how negative integer exponents work. They can see how decreasing the exponent by one is the same as dividing by the base, and continue that pattern into the negative integer exponents.
This also works for showing why factorial(0)=1
the first one is wrong
@@imjustagirl18 it was prob just a typo
The fact that Negative × Negative = Positive doesn't not make sense, obviously.
why
@@Fire_Axus Re-read his statement carefully xD
amazing, +2
This doesn't make a unreasonable reply, -(-1) @@quixotix9540
i don't think this comment isn't the best
I like the 3rd proof, because it is intuitive and sets up a foundation for thinking of numbers in terms of directions, which is incredibly useful when learning immaginary numbers.
If a negative means a 180° rotation, and rotations add up, then it is intuitive that i=sqrt(-1) represents a 90° rotation.
Very nice observation... an operator operating on a number set, some combine values while other encode rotation.. the way I saw it was the general idea behind all of the many concrete examples worked through in school. And life
That's funny, I actually thought the 3rd proof was by far the hardest to understand 😅
@@Erlewyn That proof was my reasoning that led me to realize that -1x-1=1... In Primary 3, by myself, without any input from my pretty bad teacher.
but there is nothing "directiony" about real numbers other than order. it feels like something act on another because an operation sends 2 elements to a single and it could be interpreted as a set of actions by currying.
Thinking of the complex plane as a vector makes things way easier. Negation and multiplying by I is rotation, adding values is translation, and multiplying by positive reals is scaling
I like how straight to the point you are
as a math major, my favourite one was still the number line explanation. if you thing of multiplying by negative as a reflection, reflecting twice just gives the original.
As a non-math person, it made sense to me too ... but only once I started to think about imaginary numbers. Because a follow up question would be "Well, what if I do a half rotation?"
@@PhotonBeastthen things get complex
As a high-school math teacher, that's what I used in class.
@@PhotonBeast its complex but i can do it
The proof for non conmutative rings is the same, except we'd factor that -b at the right 😁
Thanks, I accidentally did it like that the first time and got very confused about where I allegedly used the commutative property😂😅
i was 100% expecting for you to pull out the complex plane.
Which is what I'd use to explain it. If you view multiplication by a negative number as a 180° rotation on the pos/neg number line, suddenly it doesn't seem so weird that i exists, that i² is -1 or that multiplying by i is an orthogonal movement.
Imaginary numbers being dubbed “imaginary” will never not be one of the greatest unsung tragedies of modern mathematics, science, and philosophy! There should have been a greater push to re-dub them as “lateral” numbers and introduce the lateral number line and number/rotational plane far earlier in math instruction (or at all since many don’t even get to that point!)
@@jemm113 ok that makes sense
The reversed curse technique
Level 6: category theory. Guaranteed to make any simple fact mindbreakingly difficult
You definitely deserve a lot more views. Love these '5 levels' of math topics 👏
The multiplication of two negative numbers being positive clicked with me during my complex analysis course, akin to the level 3 you presented. In particular, representing negative numbers with euler’s formula and showing that multiplication leads to an angle of 2pi, which corresponds to the positive side of the real number line.
Does this concept apply to complex numbers as well?
Nice intro to rings also.
If you turn backward and walk backward, you are walking forward in the original direction
Great video, not at all surprised you blew up so quickly. Hopefully the begining of a long RUclips career!
Thanks so much!
Wow, I feel like I've finally leveled up. This helped me actually understand how a proof for simple concepts works. After spending the last several months studying modular arithmetic and rings, it's finally clicking.
The enemy of your enemy is your friend
machiavelli would be in awe with your IQ
Still my enemy.
😅😅
Soon you will find out it cuts both ways
"I enjoyed your breakdown of the different types of rings, like commutative vs. non-commutative. Very informative!"
Sir, I'm a 10th grade student from The Philippines and I've been watching your videos since last month. Looking forward for more interesting videos and thank you!
I'm glad you like them!
Got randomly recommended this and ended up watching a bunch of your videos. Hope ya keep making them! I majored in math but I think your videos are easy enough to understand and just about anyone can appreciate them. Easiest subscribe in a long time
Thanks! I really appreciate that
@@DrSeanGroathouse Surprised to see such a young channel, first video is only 2 weeks ago like DAMN
Really a nice explanation..i like math a lot, now teaching my middle schooler and elementary...your videoa are invaluable
If you imagine the number line as a sphere, the operation of multiplying by negative one is equivalent to rotating the sphere downwards (if using i, negative i is upwards) passing through the complex modulus a la e**iπ. Factoring out (-2)(-3) to -1(2)•-1(3), you end up at positive six, and travel through both poles in the complex plane to arrive at positive 6.
keep up the good work...
May the mighty bless you..
I was actually thinking of this exact question earlier today when a student asked me what happens when you multiply negative numbers. I got up to level 3 on my own, but level 4 was my sweet spot. The concept of rings was interesting, but really only useful for math majors.
Level 4 is level 5 with numbers.
Thank you so so much! I
You teach very good logic at math and thats what i have been craving for years!
I always thought it was a function with 2 inputs and with that you can define any list of relationships you want with a two input function. So it was because we say it is so.
Even as an engineering student who has a ton of math experience I still think about it as a (-) x (-) = + since one negative sign rotates and combines with the other - to physically create a + sign. Than a (-) x (+) = - since the - erases a line from the plus leaving you with a -. Does it make any sense…. no not really. But did it make sense to 8 year old me… yup!
Very interesting video! Your explanation of the ring is incredibly clear, and most importantly, it makes clear a very important thing in all of mathematics: apparently, all mathematical operations are determined by their properties. The same can be said about physical quantities - they are all nothing more than numbers, but their key difference from random numbers is that they have properties that reflect processes in the physical world. Definitely like!
Thanks so much, I'm glad you liked it!
when I was in elementary school, I thought of it as flipping the whole number line, so flipping it twice is the same as the original
if you apply the inverse of negative to the number line, it's the same as flipping in the other direction ends up looking the same as applying the negative to it, which explains why 1÷(-1)=-1
after learning about complex numbers, I thought of it is rotating the number line instead of flipping
Here's an approach from the set theoretic construction of integers built on top of set theoretic construction of natural numbers, natural number addition and natural number multiplications (no subtractions yet):
You can define an equivalence relation ~ between a pairs of natural numbers: (a, b) ~ (c, d) if and only if a + d = b + c.
To prove that it is in fact an equivalent relation I will prove its 3 defining properties:
1. Reflexivity: a + b = a + b, therefore (a, b) ~ (a, b)
2. Symmetry: a + d = b + c if and only if b + c = a + d, therefore (a, b) ~ (c, d) if and only if (c, d) ~ (a, b)
3. Transitivity:
Assume a + d = b + c (a, b) ~ (c, d)
Assume c + e = d + f (c, d) ~ (e, f)
Therefore a + d + c + e = b + c + d + f
a + e + c + d = b + f + c + d
a + e = b + f (a, b) ~ (e, f) (This last line is based on x = z x + y = z + y which is trivial)
Then, you can define equivalence classes under the ~ relation. Let's use the notation [(a, b)] to represent the equivalence class with (a, b) in it.
So, for example, the equiv. class that contains the pair (1, 3) will also contain (2, 4) because 1 + 4 = 3 + 2, will also contain (0, 2) because 1 + 2 = 3 + 0, etc.
The underlying motivation here is that if you swap a + d = b + c around, you get a - b = c - d, and [(0,2)] represents 0 - 2 = 3 - 4 = -2, but without having to have subtraction and negative numbers already defined.
Then you can define positive integers as any equivalence class that contains (x, 0) where x is a non zero natural number.
You can also negative integers as any equivalence class that contains (0, x) where x is a non zero natural number.
The proof for how the set of positive and negative integers don't intersect as a little sanity check for these definition, by contradiction:
Assume A is a positive integer that is also a negative integer.
Therefore A can be represented as (x, 0) where x is a non zero natural number.
Therefore A can be represented as (0, y) where y is a non zero natural number.
That is, A = [(x, 0)] = [(0, y)]. That is, (x, 0) and (0, y) is in the same equivalent class A.
That is, (x, 0) ~ (0, y), which happens if and only if x + y = 0 + 0 = 0, which is a contradiction since both x and y are non zero natural numbers.
Therefore A cannot exist.
Since natural number addition and multiplication is defined, you can define integer addition as [(a, b)] + [(c, d)] = [(a + c, b + d)] and integer multiplication as [(a, b)] . [(c, d)] = [(ac + bd, ad + bc)].
As far as I know, this is the canon way to construct the set of integers along with integer addition and integer multiplication.
Then, from this definition, you can have two integers A = [(0, x)] and B = [(0, y)] where x, y are non zero natural numbers, therefore A and B is negative.
Multiplying integers A and B, you'd get the integer [(0*0 + xy, 0*y + x*0)] = [(xy, 0)]. This integer fits in the definition of positive integers that we've defined above.
The rational numbers are built from equivalence classes of pairs of integers, the same way integers are built from equivalent classes of natural numbers, but with a different equivalent relation. The proof for that from definition wouldn't be as exciting because it would mostly use the result proven above in one way or another, and for real numbers (which are btw constructed in an entirely different way as Dedekind cuts) it's just an identity in real number multiplication's definition, which is not really exciting to be honest.
Hope I haven't made mistakes lol.
Near the end of your second paragraph, you wrote a+e=c+f instead of a+e=b+f
@@globglogabgalabyeast6611 thx for the correction i've fixed it now
Can it be proven by expressing the real numbers as complex numbers and then completing the process of multiplying the numbers?
This is pretty neat, definitely will show to my younger siblings!
Thanks Dr. S, I wish you luck in your growing channel 🥳
Id love to see a video like this on the idea of mathamaticl proofs. Thank you for this.
I like the expansion of understanding of mulitplication which also holds for complex numbers. A number can be expressed as a length and an angle. A positive number has angle zero, while a negative number has angle pi (or 180 degrees if you don't like radians). I.e. the number 1 can be expressed as length = 1, angle = 0. And -2 could be expressed as length = 2, angle = pi. Multiplication is then an operator which multiply the lengths (which are positive and purely real) and add together the angles. So (-2)(-3) yields length = 2*3 = 6 and angle = pi+pi = 2*pi (which is 0 since angles are mod 2"pi).
Interestingly, this holds for complex numbers as well. I.e. i is length = 1, angle = pi/2. With that, it's pretty simple to intuitively perform multiplication within the complex space.
I feel it's easier to understand intuitively through division. Negative number divided by a negative number is positive because thats how many times it can be divided. And division can be easily shown as multiplication.
Not sure why you think that. The question from a class would be. 9 divided by negative three how can you get a negative answer in division?
@@johnmaguire2185 I think it in terms that negative fraction is a positive fraction multiplied by -1.
yaaay keep the series going
I thought he was going to talk about the complex plane, multypling two numbers with angle π (negative) give us a number with angle 2π (positive) because we add the angles.
That's one way I immediately thought about it
Using the properties of imaginary multiplication in the 2d plane, where multiplying two numbers looks like adding their angles (from the x-axis) and multiplying their distances from the origin, multiplying two negative numbers looks like multiplying their distances, then adding 180°+180°, for 360°, or 0°
It's good to get pupils to derive it themselves. Create a a coordinate system and at each point in the first quadrant get them to enter a number which is equal to the product of the x and y coordinate. They will then see arithmetic sequences (times tables) when considering constant x and constant y. Ask them to extrapolate sensibly into the 2nd and 4th quadrants, and then again into the 3rd quadrant. To keep the structure of arithmetic sequences they will see that all the entries in the 3rd quadrant are positive.
Thank you so much for this, I came upon a version of the last Ring proof in an OU maths course in the 1990's but forgot it.
I'm glad it was helpful!
I used to think of the multiplication by -1 as rotating the location vector to the number I’m multiplying on the number line with -1 by 180 degrees around the 0 point. Thus 1 times -1 times -1 yields a rotation by 360 degrees, putting me back to where I started, positive 1. This visualization also comes in handy when trying to get a grasp on the complex plain and understanding why the square root of -1 is orthogonal to the number line. Because instead of rotating by 180 degrees, you just rotate by 90 degrees. And Sqrt -1 times Sqrt -1 is rotating by 90 degrees two times giving you the 180 degrees of the multiplication by -1. There you go.
That level 4 one is great!
I like making a times table for 1 to 9 and pointing out how the rows and columns increase regularly.
then start extending the table to the other quadrants going from (x,y) to (x-1,y) and (x,y-1) a step at a time
An attempt at a proof by contradiction (I have never taken a formal proofs class, so I won’t be surprised if I have made a mistake somewhere)-
Let a, b, and c be arbitrary, strictly positive numbers > 0. Assume that (-a)*(-b)=(-c) for some a, b, and c.
We can factor out a negative one from some of these numbers. Rewrite (-b) as (-1)*(b) and (-c) as (-1)*(c) :
(-a)*(-1)*(b) = (-1)*(c)
divide both sides by negative 1:
(-a)*(b) = (c)
(does this step implicitly assume that negative times negative is positive? not sure)
an this point, we can see that the left side must be negative, and the right side is positive, since a, b, and c were chosen as strictly positive. this is a contradiction, meaning that negative times negative cannot equal negative!
So I've studied a little bit on formal proof's but I'm not a mathematician (just studying Comp Sci) so could also be wrong here.
You can use this proof to show -1/-1 = 1 without assuming negative * negative is positive:
let a be a non-zero number:
a/a = (a)^1 * (a)^(-1) = (a)^(1 -- 1) =a^0 = 1
Thinking about your proof also made me think of a proof, although I may be doing something I can't do:
let x,y be any positive (>0) real numbers:
assume -x * -y 0)
By our above proof:
(-x)/(-x) = 1
(-x) * 1/(-x) = 1
we know 1/n != 0 for any non-zero real n, so 1/(-x) is not zero.
Since -x is negative, if 1/(-x) were positive then:
(-x) * (1/-x) < 0
Thus:
by exhaustion 1/(-x) must be negative
Thus
1 0
In the intuitive example, I think the person is left with 6 cookies, because it is equivalent to say 6 cookies are put into my 🍪 jar or left in my possession outside of a jar. I think of multiplication and division as the same as adding full containers of the same size. For example, 12 eggs is the same as a carton of a dozen eggs. Multiplication is just adding up cartons: 1/2 of a carton is splitting the carton into two, each with 6 eggs, but I still possess all 12 eggs. The trick is that we are implying that changing the size of a group also means we lose possession of some of the contents and groups. But perhaps we could interpret x/0 to mean the information possessed is ungrouped...and so has to be counted individually.
i feel kinda stupid for having to look up what distributive propertys are but I feel really good for understanding it now.. I cant believe I went so long trying to rearange formulas with this pretty basic concept but Im glad I got it now
Awesome. I'm relearning math to get better for engineering. This helps so much, although simple it helps me actually understand the math vs just learning by memorizing formulas. Can you make one on why we set use zero property rule for quadratics? I can't understand the purpose of making the equation equal to zero.
Very helpful, thank you. 🎉
Glad you liked it. Thanks for the idea!
I'm trying to research logic systems, and I'm running up against constructivism and paraconsistent logic. I know that YT isn’t the ideal place to ask in-depth questions, but perhaps you can point me toward some logic literature in the field? Specifically, (how) are double-negation and apartness related, and are double-negation and equivalence a problem for binary logic/LEM?
Another video from the goat! Just solved one of IMO questions and here is my reward video!
Thanks, and nice work with the IMO question!
So Level 5 is just a generalization of Level 4, or Level 4 is an example in the ring of integers of Level 5?
Im loving these videos so much!!
I'm glad you like them!
The first thing i thought of was rotations in the complex plane. Two 180 rotations just end up at the same place
Same
It is cool to see Level 5 after just learning about Vector Spaces in my linear algebra class. It was not taught to me as 'rings' the first 7 properties at 5:26 were the same as my textbook, however, the 8th one was '1v = v'. Anyways, cool to see and nice video.
My favorite is multiplication as rotation of a vector in the complex plane. Multiplying by a negative is multiplying by 180°. To multiply a negative by a negative is to rotate by 360°, which is identical to 0°, or positive.
The rules for rings are defined that way because we want them to be useful and generalize the idea of integers, i.e. counting stuff like sheep and yes, debts too.
Complex numbers in exp form?
Would it be possible to use complex numbers to prove it? Because multiplying by i is the same as a 90 degree rotation, and i*i = -1. 4 i’s is 4 90 degree rotations which is 360 degrees. i*i*i*i = 1, and i*i = -1, so (i*i)*(i*i)=1 and (-1)*(-1)=1
Removing sadness leaves you happier
I'd love to see a "subtracting a negative" is like adding video, please!!
Here is one using the construction of the integers:
Integers are defined as equivalence classes on N^2 where (a,b)~(c,d) if a+d=c+b.
All equivalence classes can be written in the form [(a,0)] or [(0,a)]. Ones in the form [(a,0)] are just denoted as a and ones in the form [(0,a)] are denoted as -a
Multiplication is defined as [(a,b)] * [(c,d)] = [(ac+bd,ad+bc)].
We have -1*-1=[(0,1)]*[(0,1)]=[(0*0+1*1,0*1+1*0)]=[(1,0)]=1
Is it just me, or is a ring just a vector space? I swear that’s what I learned is the definition of a vector space
So what is the difference between rings and fields and whatever other abstract objects there are?
Thanks!
Fields needs to have multiplicative identity (1) that is not equal to 0. And all non zero elements has a multiplicative inverse. So fields are also rings but with additional restriction.
@@kazedcat Gotcha, thanks!
I use the last method with all students from 11 years old up. I don't talk about ring theory just algebra after they have learnt to expand brackets and to factorise.
I thought you would use the line method for level 1. Just counting the number of lines and if the number is a multiple of 2 it is positive. - is 1 line and + is 2. You could only count the number of - in general.
Visualization, realization, numericization, abstraction, arbitration
I find that it is easier to understand with the complex numbers. If you multiply a complex number by i, it is shifted by π/2(90°):
1 * i = i
i * i = -1
-1 * i = -i
-i * i = 1
And if you now multiply by -1 instead of multiplying by i, the angle of rotation doubles, i.e. π/2 becomes π(180°).
I like to think of this from a statistical standpoint. You can combine + and - in 4 different ways. (+)x(+), (+)x(-), (-)x(+) and (-)x(-). If you threw a dart randomly at the number line then your answer would have a 50/50 to be (+) or (-). Each one of these 4 combinations accounts for %25 of the answers. Starting with the axiom that a +x+=+ which is %25 of your answers. Then that +x- must also be %25 of your answers and that -x+ is the same so together count for %50 of the answers therefore must be -. Leaving the last %25 (-)x(-) which must be +.
When we divided both sides with a.0 was it actually valid?
when any complex number (real numbers are a subset of the complex numbers) is multiplied by another complex number their magnitudes are multiplied (this will always be a positive real number so no need to fight the negative multiplication basis here) and their phases are added (phase ranges from 0 to 2*pi with the modulus of any phase greater or equal to 2pi to bring it back) since the positive numbers have a phase of 0 0+0 = 0 and the negative numbers have a phase of pi pi+pi = 2pi since this is equal to or greater than 2pi then when we do the modulus it goes back to 0
its as simple as that we just gotta take a trip to the world of complex numbers
here I've got a question do u not have even a plank length worth of space to move why are u standing is such a tight space? it almost feels uncanny idk
also idk if there's already a video on it yet but i want one on imaginary numbers make em feel as natural as breathing air then quantum physics then axioms then ok actually anything math or science related and I'll watch lol
You've described the ring theory and the twelve hours ring, but you've used the same proof from step three again in step five, ie you didn't need to use the ring theory at all, but thanks for explaining the interesting theory though 😊😊
The level 5 brought back memories of Spivak's Calculus.
I was literally in the shower thinking „what is the actual mathematical proof of neg x neg = pos” and after i came out of the shower and open yt this showed up.
1:05 This is what made me understand! Thanks! but I'm gonna watch the rest.
Very good video, u are really good at explaining. The only critique i have is that i would not call level 5 math major as i did the concept of rings in uni in like the first semester. I study computer science
Haven’t seen level 4/5 done like that before. I would go more like
1. -(-a) = a. Any definition I can think of is obviously symmetric, e.g. if b is called -a when a + b = 0 then a is also -b.
2. (-a)(-b) = ? = -(-a)(b) = ab if you can justify moving the -…
3. -a = (-1)a. No trick: Just check that (-1)a is -a. (-1)a + a = (-1 + 1)a = 0a = 0.
4. 0a = 0. Same way in video :-)
You can build so many proof of this law, using different level in math's.
Saying that's exponential is a group morphism from C to C* with 2piZ as a kernel.
Or using the Euclidean division in the polynomial ring, writing X^2=(X-1)(X+1)+1 and that - 1 is the unique root of X+1 in Z[X].
Here are the proof that comes immediately at my mind.
It's not a proof, but intuitively works:
If you can agree that multiplying bij -1 is just flipping a number's sign. Which I think makes sense.
If you also agree that (-a)(b) =-ab. Which also makes sense, because this is just adding -a b times: (-2)(3)= (-2)+(-2)+(-2). I.e. shifting three time 2 to the left on the number line.
Then it follows: (-a)(-b) = (-1)(a)(-b) = (-1)(-ab) = ab
It's one of the simpliest problems to solve and understand
what about sqrt(a*b) = sqrt(a) * sqrt(b) and why it doesn't hold for when a and b are both negative?
Well if the n is the square root of a real number m then n has 2 important properties:
1. n^2=m
2. Both the real part and imaginary part of n are nonnegative
For all a and b (sqrt(a)*sqrt(b))^2=sqrt(a)^2*sqrt(b)^2=a*b. So the first property is met no matter what
The 2nd propert is where things get complicated.
If a and b are both nonnegative then sqrt(a) and sqrt(b) are both nonnegative so sqrt(a)sqrt(b) is nonnegative. So 2nd property holds when a and b are both nonnegative.
If a is nonnegative and b is negative then sqrt(a) is positive and sqrt(-b) is positive so sqrt(a)*sqrt(b)=sqrt(a)*isqrt(-b)=i(sqrt(a)*sqrt(b)) has a nonnegative imaginary part (and a real part of 0) so the 2nd property still holds when a is nonnegative and b is negative.
By symmetry, when a is negative and b isn't the 2nd property also holds.
Now when a and b is negative you get something interesting.
sqrt(-a) and sqrt(-b) are both positive so you would have sqrt(a)*sqrt(b)=i sqrt(-a)*i sqrt(-b)=i^2*sqrt(-a)*sqrt(-b)=-sqrt(-a)*sqrt(-b). This is a negative so the 2nd property does not hold.
Overall this means that sqrt(a)*sqrt(b)=sqrt(a*b) if and only if a and b are not both negative.
I would use -1 = e^(pi*i). So (-1) * (-1) = (e^(pi*i)) * (e^(pi*i)) = e^(2*pi*i). And by looking at the unit circle we can see that e^(2*pi*i) = 1.
Although I guess complex numbers already rely on (-1) *( -1) = 1 so this might be circular.
I remember learning about multiplying by negatives and thought "if I multiply them together do you get positive" yea I was kinda a child prodigy
Yeah, the way I think about negative operations is basically via boolean algebra, as it’s easier to reason about values encoded on a single bit. You can draw a 2×4 table with all possible bit combination, and realize that this negative operation is basically the same as the NOT boolean operator. A true value is +P, and a false value is -P.. Negating simply flips the sign. So -(+P) is -P, -(-P) is +P.
Then, -a * -b is the same as first doing a * -b, and flipping the result sign, which ends up positive.
In complex number plane -1 = e^(pi*i) is rotating 180 degrees. So (-1)^2 = e^(2*pi*i) = e^0 = 1
1:40 if my friend takes 3 of my depths It's not the same as multiplying -3 by -5
I absolutely love this channel and the way you explain topics in 5 levels. You just earned a sub
Thanks so much! I'm glad you liked it
what about charges where you get more of an xor effect -> - * - = - , - * + = + , + * - = +, + * + = - ... pretty sure you can also define vector spaces where - * - is positive as well
Level 6: The positive reals P are a subgroup of the multiplicative group of reals. The reals are then partitioned into two cosets, P and
-P (negative reals). The quotient group is then the cyclic group of order 2, and negative * negative = positive. By definition of multiplication in the quotient group, this means that a negative real times another negative gives a positive.
I believe this isn't circular because the only things you have to know about the group is that the negative reals and the positive reals are disjoint and their union gives R, and that the positive reals are stable.
At my university i was taught since you can prove -x=(-1)x=x(-1), the fact that multiplication in a ring is associative means that you can "migrate" the minus signs in the expression (-x)(-y) to the front, and get (-x)(-y)=(-1)(-1)xy. Which can then be shown to be equal to xy.
Yep. You can also prove -x=-1x=x(-1) from (-a)(-b) = ab as well.
The first meaning that we give to multiplication is adding a number repeatedly. The two numbers involved in the multiplications have very different meaning, the first one indicating which number will be added repeatedly and the second one indicating how many times is this going to happen. From the significance point of view factors are not interchangeable. Negative numbers are introduced to manage situations that are hard to manage using only natural numbers, in fact you would need to use two sets of natural numbers and quantities of one of these sets cancel out with corresponding quantities from the other one. So i think that before answering what -ax-b means we should answer what meaning can be given to repeating addition a negative number of times. But it turns out that repetitions are also quantities that can be sometimes considered of two different kinds that cancel each other: in your second example you mentioned a friend who repeatedly gives or repeatedly takes away the same quantity, there can also be repetitions from this point in time to the future vs repetitions in the past to this point in time: form example if my account now is 0 and a friend has deposited 5 dollars in my account for the past 3 days, it means that my account was -15 dollars 3 days ago. Using the same example if my account was debited 5 dollar for the las three days and now is 0, it means that 3 days ago y had 15 dollars, thus -5x-3=15. My point here is that you basic example is more basic than the first because the first involves continuity and the second one doesn't, but also because it better addresses the questions young students have when they try to reconcile this new "rule" with what they previously knew about this operation.
I think I might have a grasp of the last method actually
The way I see it is this:
On a number line, each sides “negative” direction is towards 0 and through the other end of 0. 2-3 being 2 towards 0 and 1 through 0 from the right side. And each sides positive direction is farther away from 0.
Thus,
2 x 3 = 2 + 2 + 2
Adding 2 in the positive direction (right) 3 times, getting further away from 0.
-2 x 3 = -2 + -2 + -2
Adding 2 in the negative direction (left) 3 times. Think of it as starting on the right side of 0 and going through 0 in the negative direction (towards 0).
-2 x -3 = -2 - -2 - -2
Subtracting 2 in the “negative”direction 3 times. Think of it as starting from the left (negative) side of 0, and going towards and through 0, in the left sides “negative” direction, which is really the right sides positive direction. Thus meaning that it really is just going positively
The way how I think of negative numbers is this:
I picture a graph of a number line in my head.
Positive numbers are at the right while negative numbers are at the left and zero is at the middle. When I add a number with positive value, the position of said value moves to the right. When I add a number by a negative value, the direction is reversed.
When I multiply a number by a positive number, the position of the number that is multiplied remains the same, then the value gets multipled and moved. When I multiply a number by a negative value, the position of the number being multiplied gets "mirrored", that being the number being converted into a negative value.
When the multiplier is greater than 1 or less than -1, the value moves away from 0. When the multiplier = (-1 , 1), excluding zero, the value is moved towards 0. When the multiplier is zero, the value moces straight to 0.
For division, the direction is reversed compared to multiplication. However, in the case where the divisor is 0, if the multiplied number is not 0, it cannot be any value. If the multiplied nunber is 0, it "technically" can be any number. However, that would be useless and a worthless excuse to use this number to answer every mathematical problems so it is written as "undefined" instead.
For exponentiation, specifically when the exponents are even numbers, "mirrored" values (Negative values) and "normal" values (Positive values) are presented. Roots are the reverse of exponentiations except that if the exponent is an even number, it can only represent one of the possible value (Example: √4 = 2 and only 2, while -√4 = -2 and only -2).
I might have explained the concept of negative values as the literal traditional definition of it but personally, I understand them this way. There might be some flaws in them though.
multiplying by i in the complex plane is a rotation of 90 deg. multiplying by i twice is the same as multiplying by -1. Do that again and we rotate a full 360 getting back to a positive number!
The fact that this needs to be explained to more than a single-digit percentage of grown adults ought to _terrify_ us.
"Throughout heaven and earth, I alone am the honored one."