>a negative times a negative equals a positive
What's the logic behind that?
Isn't multiplication supposed to be a glorified summation ?
>>7908964
>What's the logic behind that?
You rotate something backwards and then rotate that backwards, you get something right side up
>>7908964
I don't know, but it would pretty much break all of math to do anything differently. Math that we know corresponds to real world phenomena.
>>7908969
I ask because imaginary numbers seem like damage control to me.
>>7908964
>being this stupid
Life must be really difficult for you
>>7908996
no, ur a moran
>>7908977
They're not. They represent stable oscillatory values. Basically sine and cosine are a more user-friendly way of using imaginary numbers.
>>7908964
it's consequential of some popular axioms
>>7908964
but anon 1+2+3+4+5... = -1/12
sums of positive can be negative too
negative force times negative distance gives you positive work
b8 b8 b8 b8 b8
Multiplying something by -1 flips it across the number line
Multiplying something by -1 twice takes it to the original spot, so (-1)*(-1)=1
Think of the first negative number as a debt and the second as an act of taking away. For example, I will take your debt of 5$ 3 times. You are now 15$ richer. (-5 x -3=15)
Think about it. You have no apples. Let's say you owe the Mafia 1 apple, so you now have -1 apples. Then you owe the Mafia -1 multiplied by the current number of apples you have, -1. How many apples do you now have? 1 apple. Understand now?
>>7908964
Well anon imagine a line where one side is negative and the other positive, with 0 being in the middle.
If something moves to the left, its speed and direction is negative. To the right, positive.
So if an object is moving with negative speed towards the positive side, it will actually move backwards into negative. And if an object is moving with negative speed to the negative side, it's actually moving to the positive side.
Get it?
Negative numbers are imaginary quantities
Only the natural number excluding zero are inherent
>>7908964
>Isn't multiplication supposed to be a glorified summation ?
It still works as "glorified summation".
3 * 4 = "3 plus itself 4 times" = 12
-3 * 4 = "-3 plus itself 4 times" = -12
-3 * -4 = "-3 plus itself -4 times" = 12
>>7909130
>>7908964
Time for some fucking IMAGINARY SCIENCE
Negative numbers aren't negative numbers, they're tuples in the Cartesian product of {true, false} and the reals.
1 = (true, 1)
-1 = (false, 1)
Certain operations on these tuples are just implemented in a fun and quirky way.
>>7909093
>>7909116
I'll chip in to these guys.
If you have a note worth 100 USD, you have +100 USD (100 dollar note, duh).
If you have a note saying you have to pay 100 USD, you have -100 USD (e.g. parking ticket)
Before the 'parking police' issue give you the note, they are in possession of it, but they still owe you the note. They owe you a note saying you owe 100 USD.
-(-100 USD) = + 100 USD, what they will obtain from the situation.
>>7908964
You are forced to have it like that because
>x + (-x) = 0
>1*x = x
>0*x = 0
Proof:
0 = -1*0 = -1*(1 + (-1)) = (-1)*1 + (-1)*(-1)=(-1) + (-1)*(-1)
so
0 = (-1)*(-1) + (-1)
but
0 = 1 + (-1)
so
(-1)*(-1) = 1
>>7908964
>glorified addition
I'll have to respond to that with a "sort of, but not really." It just helps kids grasp it to teach it like that.
Why is -1 * -1 = 1? I know it is an axiom we take for granted. It seems at some point we came up with bad notation, and this should be expressed or explained better.
>>7908964
Why not? It works in physics. If you push something to the right you assume the force has positive value, if something moves right you assume speed or change in distance has positive value.
If you have negative force and negative speed you are doing positive work because you help dragging some along, while with positive force/negative speed you are working against some other guy pushing it from the other side doing negative work.
>>7909382
Modulo the equivalence relation that identifies (true,0) and (false,0)?
>>7909569
Right, so {x + (-x) = 0,1*x = x, 0*x = 0} are the axioms. That is correct, I haven't seen this proof in school.
Formal:
Because negatives are inverses, and they way inverses works is defined by our axioms. Multiplication of inverses has been proved with axioms that it equals one.
>logic
Well, in our system defined by axioms it is logic, what is isn't logic is our axioms. They are made up and is inside our imaginary world alas it still describes our nature very well.
>>7909591
>that it equals 1
That it equals a positive*
>>7909591
Then they are not made up.
>>7909597
I see them as made up.
Becuase we have other systems with different axioms that can describe reality but not in the broad of sense that euclidian and axiomatic axioms do..
>>7909607
So you are saying they are incomplete, not that they are made up. incomplete =! made up
>>7908964
Imagine if it wasn't the case
-1 * -1 = -1 (according to this new system)
(-1 * -1) * n = -1 * n
-1 * (-1 * n) = (-1 * n)
-1 = (-1 * n)/(-1 * n)
-1 = 1
Which just ain't so
>>7909627
You could just divide both sides of the first row by -1
>>7909639
That is not a definition of something not-made up, you ape
>>7908964
Ladies and gentlemen, I present to you... Common Core! We're doomed.
>>7909434
/thread
>>7909722
yes, it is
>>7908964
This is why Abstract Algebra should be taught in elementary school.
>>7908964
>What's the logic behind that?
Because it makes the operation invertible and we can do algebra with negative numbers
I get given two $5 bills. +2 * +5 = +10 profit
I give away two $5 bills. -2 * +5 = -10 profit
I get given two $5 debts. +2 * -5 = -10 profit
I give away two $5 debts. -2 * -2 = +10 profit
>>7910616
>-2 * -2 = +10 profit
and
>lose debt = profit
I would still need to have the 10 dollars
>>7910875
>-2 * -2 = +10
>/sci/
>>7908964
2 times 3 is 2 3's or 3 2's. -1 times 1 is 1 -1 or -1 1's. -1 * -1 is -1 -1's.
To say that a number is negative is to say it's the "opposite" of a given positive number. So the opposite of a given negative number, or the negative of a negative number, is positive, as negative and positive are "opposite."
>>7909130
Or just say distance = speed * time
Negative speed = moving left
Negative time = measure the position before it went through zero
Object should be on the right (positive distance).
>>7908964
Multiplications in this way are essentially rotations. This is easier to understand with complex numbers, where you have multiple directions you can point and so it's obvious that multiplying by -1 rotates by 180°, multiplying by i rotates +90°, etc.
If you rotate 180 twice, you're back to where you started.
>>7908964
>Isn't multiplication supposed to be a glorified summation ?
That is the definition for natural numbers: m*n = result of adding n to zero m times.
But you can extend it to negative integers pretty easily if you take "do thing -n times" to mean "undo thing n times." And un-adding (subtracing) -3 four times from zero gives you positive 12.
>>7909627
>reductio ad absurdum
This and only this senpai.
>>7908964
>Isn't multiplication supposed to be a glorified summation ?
it's *chained* summation, retard
and negation is an order-two operation
Lrn2mathematics some time
>>7909050
>negative distance
I hope you mean displacement senpai
>>7908964
Because if they didn't, the distributive property would not hold and we would not have a useful algebraic group.
Example assuming a negative times a negative results in a negative number:
-1(1 + -1) = -1*1 + -1*-1 = -1 + -1 = -2
Which is different than what we get if we do not distribute the -1:
-1(1 + -1) = -1 (0) = 0
>>7909021
What the fuck is this?
>>7908964
>nobody gave the simple answer
[math] a \cdot b [/math] means we add [math] a [/math] to [math] 0 [/math] [math] b [/math] times.
[math] 5 \cdot -2 [/math] means we add [math]5[/math] to [math]0[/math] [math] -2 [/math] times, which means we subtract [math]5 [/math] from [math]0[/math] [math]2[/math] times.
Therefore [math] -2 \cdot -3 [/math] means we subtract [math] - 2 [/math] from [math] 0 [/math] [math] 3 [/math] times. Thus [math] -2 \cdot -3 = 0 - (-2) - (-2) - (-2) = 6[/math].
Thus, subtracting a negative from [math]0[/math] a negative number of times is the same as adding the corresponding positive to [math]0[/math] the corresponding positive number of times.
You can prove it from ring axioms, too.
Let [math]1[\math] be the multiplicative identity of some ring [math] R [\math], and let [math]-1[\math] denote its additive inverse. Then we see
[math]-1 + 1 = 0 [\math]
[math] -1(-1 + 1) = 0 [\math]
[math] (-1)^2 + (-1)\cdot 1 = 0[\math]
Since [math]1[\math] is unity, hence [math] (-1)\cdot 1 = -1[\math]. Adding [math]1[\math] to both sides gives
[math] (-1)^2 = 1 [\math]
>>7912219
seriously how the fuck do i do latex
>>7912222
chekt
also, it's [/math], not [\math]
>>7912222
[math] \mathbb{G}e\tau \: \mathbb{g}_{\underline{oo}}d \: \mathbb{b}\iota\tau\textbf{ch}. [/math]
>>7909406
i didnt understand that at all
>>7912604
>[/math], not [\math]
i'm too used to actual latex, where all commands start with \
let's try this again
Let [math]1[/math] be the multiplicative identity of some ring [math] R [/math], and let [math]-1[/math] denote its additive inverse. Then we see
[math]-1 + 1 = 0 [/math]
[math] -1(-1 + 1) = 0 [/math]
[math] (-1)^2 + (-1)\cdot 1 = 0[/math]
Since [math]1[/math] is unity, hence [math] (-1)\cdot 1 = -1[/math]. Adding [math]1[/math] to both sides gives
[math] (-1)^2 = 1 [/math]
>>7908964
z>=0
|z|=z
|-z| =z
a * 1 = a
a *-1 = -a
so if a*b=c and |a|=|c| then b = 1 or -1
-1 * x = 1
-1 * 1 = -1
x = -1
-1 * -1 = 1
What about -1/-1?
If a*b=c, then c/a=b and c/b=a
1* -1=-1, so -1/1=-1 and -1/-1=1
-----
A direct answer to your question is that multiplication can be summation and subtraction.
>>7911954
Nothing to special
https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF
It just looks amazing at first glance
>>7908964
>>a negative times a negative equals a positive
Not true in finite fields.
>>7914057
There is no absolute notion of negative and positive in finite fields.
>>7908964
I always imagine the absolute values of numbers.
3 x 4 = 12
Three groups of 4 sheep is 12 sheep if you start from 0.
-3 x 4/3 x -4 = -12
Negative three groups of 4 or 3 groups of negative 4 sheeps. If you start from 0, either of these groups of sheep leave negative zones where sheep could be. -12 sheep to be exact.
-3 x -4 = 12
3 negatives groups of negative sheep. You take away 3 groups of those negative sheep so you're left with 3 groups of positive sheep.
It's an elementary way of thinking/explaining but yeah. I always think of them as absolute values. 5 + (-3) is 5 that had 3 suicide bombers.
>>7914919
Two negatives cancel out.
>>7916800
Are you insinuating that somewhere in my post I had misled somebody into believe that a negative times a negative equals a negative?
>>7916991
I am insinuating you did just that in the 2nd equation.
>>7917203
Actually, I just misunderstood what it was saying. To be fair, "/" usually means division.
>>7908964
1-1 = 0
(-1)(1-1) = (-1)(0)
(-1)(1)+(-1)(-1) = 0
-1+(-1)(-1) = 0
(-1)(-1) = 1
(-a)(-b)
(-1)(a)(-1)(b)
(-1)(-1)(a)(b)
(a)(b)
>>7912222
Quads cfm: sci has fucked up latex
If 20 × -10 is -200, then the (20 × 10) becomes negative. It flips polarity.
If -20 × -10 is 200, then the (20 × 10) is positive. It flips polarity twice. That means the (-) signs cancel.
-1 * 3 = -3
-1 * 2 = -2
-1 * 1 = -1
-1 * 0 = 0
-1 * -1 = ?
Just follow the pattern, it's perfectly intuitive if you think of it like this
