Let x be a real number s.t. x/x=x & x-x=x

>let x be a real number s.t. x/x=x & x-x=x

Reminder LWAfags are the new jojofags

you're still the same brainletfag as before at least

>brainletfag
It's like I'm being insulted by a 12 year old in 2005

how would you know what 2005 was like if you weren't even born yet?

0 is what's going on

>0/0=0

Try again bra-
>0/0=0
>0=0*0

At last i see the light. Thank you kind brainling.

0=0*any real number

b-b-but 0/0 is undefined

X cant be 0
Then
X =/= X - X

Actually x can be 0

X/X = X
X^2 = X
X = 0

X - X = X
0 - 0 = 0

I believe [math]0/0[/math] since division is basically advanced subtraction, say 20 divided by 4 to get 5 you would continuosly subtract 4 from 20 5 times to get 5. 20-4=16-4=12-4=8-4=4-4=0 which grants you 5. So if you were to do 0/0 you would subtract 0-0=0 1 time to get 0. for 0 times 0=0

x can be 0 for 0 is equal to 0 - 0 = 0

x/x=x, therefore x=x^2

x(x-1)=0 => x=1 or x=0

x-x=x, therefore x=0

Thats the problem. You cant subtract something 0 times.

>not using the extended hyperreals
brainlet; the answer should be apparent

[math]x=1-0.\bar9[/math]

>x(x-1)=0
You assume x is not 0 in order to do this, and it leads to x = 0 which is a contradiction. Wrong assumption boi

x=0, it's usually accepted that 0 kills everything, including infinity.

Does 0/0=0 lead to anny contridictions?

You basically said that a0=0 only if a=0 doing that

Consider lim A->0, B->0 A/B

As A approaches 0, A/B goes to 0

As B approaches 0, A/B goes to infinity

If you look at the graph of Z=X/Y and look along the X and Y axes you will see that if you approach the origin from the X axis it is near zero, but from the Y axis towards + or - infinity. This is why you can't assume 0/0 = 0, it is undefined at that point.

0-0=0 & 0/0 = opinion

3/3 = 1
2/2 = 1
1/1 = 1
0.5 / 0.5 = 1
x/x = 1

Post yfw you realize math is just a bunch of opinions.

>mfw 3/0 = undefined, 2/0 = undefined 1/0 = undefined 0.5/0 = undefined x/0 = undefined

[eqn]
1 ≠ 2
\\
1a ≠ 2a
\\
1 * 0 ≠ 2 * 0
\\
0 ≠ 0
[/eqn]
0 is made up, just like the rest of math. The entire field is impure.

Oh I get it, we just subtract the 0 out of everything to become pure

x-0=0-0

qed; x

but 1 * 0 does = 2 * 0

no you have to subtract the 0s

1*0=2*0
1*0-0=2*0-0
1*=2*
1-2*=*
-1*=*
-1=*-*

x/x=x
So f(x) = x/x- x = 0.
x = 0 is not in the domain of this function.
So no value of the domain of this function satisfies the equation.

He doesn't assume that at all

x(x-1) does =0 when x=0

>systems of equations do not always have a solution
Who knew?

>0=0*any real number
Does that mean i*0!=0?

let x be a real number such that x =/= x

i*0 factorial = i

No. You are missing one step
a0=0
a0/0=0/0
If 0/0=0, as we assumed, we again get
a0=0, no contridictions
But to conclude that a=0 you need to have 0/0=1

> prove something
> well, we can see it from the graph

x, y, z are natural numbers
x^1+y^1=z^1 has solutions
x^2+y^2=z^2 has solutions
x^3+y^3=z^3 doesnt have a solution
> Post yfw you realize math is just a bunch of opinions.

It's 0. You can't divide by 0 but anything divided by 0 is 0.
Source: I just made that up and I hope it's right

But 0 divided by anything is 0, fix'd

He is assuming it because he is factoring by x and you can't divide by 0

I can see it now

Meant not equal but whatever. Why thoes i*0=i tho?

Sure you can. Just don't subtract.

>"mr. user it appears in your assignment you attempted to subtract two matrices, one of size 2x3 and 3x2. given they're different you aren't able to get a meaningful result. you assumed that the size of the matrix were equal but with zero entries, but these are entirely different structures.
>just don't subtract
>excuse me?
>just don't subtract if you do that it'll stay the same
>i don't understand...
>think about it. the entries don't exist right? then you're doing nothing. it's really simple.
>i see...

>factoring is dividing

For [math]x,y\in\mathbb{C}[/math]
[math]x+y[/math] has [math]1[/math] solution.
[math]x-y[/math] has [math]1[/math] solution.
[math]x\cdot y[/math] has [math]1[/math] solution.
[math]\frac{x}{y}[/math] has [math]0[/math] solutions if [math]x\neq0,y=0[/math]
[math]\frac{x}{y}[/math] has [math]1[/math] solution if [math]y\neq0[/math]
[math]\frac{x}{y}[/math] has [math]\infty[/math] solutions if [math]x=0,y=0[/math]

>let
>not assume
Gtfo

x doesn't give a flying fuck about what a brainlet like you assumes

Not subtracting is the same as subtracting something 0 times.

x/x is x times its multiplicative inverse
x-x is x plus its additive inverse
so we are looking for a structure such that x(x^-1) = x and x+(-x) = x

i can think of a way to make this happen - if x belongs to a zero object where multiplication and addition are defined (the zero ring, zero module, etc)