Why isn't 0/0 defined? Isn't it just equal to one?

You can't have that a(x/y)=(ax)/y with 0/0=1. It's like the contradictions that come from using square roots of complex numbers incorrectly.

0/0 = 0 * (any number/0) = any number

>doesn't understand calculus
brainlets OUT

0 is a placeholder symbol for nothing. You cant divide nothing with nothing and have it equal to one.

>Since 1+0+1+0+1... is equal to 1/2
Your currymath is not valid here.

No, 0/0 = 1
Therefore 2*(0/0) = 2
Everything works.

By 0/0, I mean the function x/x. Which will always equal 1 even in x is equal to zero.

0/0 = 1
therefore 0/0 * any number = any number

How narrow-minded can you get? People used to think negative roots were impossible, hell people used to even think negative numbers were wrong hence the term, "negative".

I'm talking about the function x/x which will always be one which technically would allow for 0/0.

Examine the function x/x as x approaches 0. Oh right, it's equal to one. By allowing this, we can use this property to simplify certain equations.

Isn't the symbol for 1/0 just the "90 degrees 8" symbol?

But 2*(0/0) = (2*0)/0 = 1

Then let K = 0/0 = 1

2*K = 2

That identity doesn't work in the 0/0=1 number system.

Because then Integers wouldn't be a ring, and Rationals and Reals wouldn't be a field.

en.wikipedia.org/wiki/Ring_(mathematics)
en.wikipedia.org/wiki/Field_(mathematics)

en.wikipedia.org/wiki/Wheel_theory

>limit of 1/x as x approaches 1 is 1
>1/1 is 1

>limit of 1/x as x approaches 2 is 0.5
>1/2 is 0.5

>limit of 1/x as x approaches 0 is infinity
>HURR DURR 1/0 IS UNDEFINED

the answer is staring right at you, you fucks

>>limit of 1/x as x approaches 0 is infinity

wolframalpha.com/input/?i=plot[1/x]

really tickles those brain cells

assuming 0/0=1 then it follows 2=2*0/0=(0*2)/(0*1)=0/0
so we have 1=2.

This is a contradiction.

>why not just make a new symbol for 1/0.
Because then [math]\mathbb R[/math] is not a field anymore.

>equal to 1/2
WRONG

>>limit of 1/x as x approaches 0 is infinity
WRONG. The limit does not exist.
Infinity is NOT A NUMBER and especially not a REAL NUMBER.

I am NOT A NUMBER, I AM A FREE MAN.

>>limit of 1/x as x approaches 0 is infinity
>WRONG. The limit does not exist.
It's not unreasonable to say "the limit of a function at a point is infinity" if the function increases without bound on both sides of that point. 1/x doesn't do that, though.

>Infinity is NOT A NUMBER and especially not a REAL NUMBER.
But it's an element of the closure of the set of real numbers.

if 1/0 would make reals not a field anymore, you could say that 1/0 is an unreal number (Я)

Infinity is NOT A NUMBER, IT'S A FREE MAN

We have a word for that
"The function is unbounded"
More concise and succinct than the misleading (if not outright false, according to the rule of universal instantiation) claim that
"the limit of the function (exists and) is infinity"

You can't just plug in numbers, find the limit as the function approaches infinity.

This is wrong.
The limit exists, it is infinity.
By convention, we say the limit diverges to infinity.

Compare this to limit as x-> infinity of (-1)^x
This does not converge to any number
It does not diverge to infinity
This limit does not exist

1/x doesn't have a limit at x->0. approaching from left, it goes to negative infinity. approaching from right, it goes to positive infinity. so, it has no limit there.

Depends on how you extend the real numbers. The extended real line distinguishes [math]+\infty[/math] and [math]-\infty[/math] so 1/x is discontinuous at 0, while the projective real line has [math]\infty = -\infty[/math] so 1/x is continuous and equal to [math]\infty[/math] at 0.

If you say 0/0=1 then 0/0=63 also as:
0=63x0
0=0
Therefore 0/0 can be defined as any part of the universal set by this logic.

This board is so mathematically illiterate that it took until now for someone to mention the extended reals.

And this person continues to complain about a problem that has already been addressed.