5/5=1

you're all morons

0 and infinity aren't numbers.

0 is a constant and infinity is a variable.

>0 isn't a number
Nice one

when is a number not constant?

>this thread
Everyone is wrong. % = 1/100.

I don't disagree with that breathtaking unveil, but who the fuck are you responding to.

2/4 = 2 * 1/4
1/2 = 1 * 1/2
0/0 = 0 * 1/0

This thread.

when it's a function.
[math] \mathbb{R} \subset (\mathbb{N} \to \mathbb{Q}) / \equiv [/math]

ass

>0 is a constant and infinity is a variable
wat
0 is a number

infinity is the size of any group closed under the binary operation a*b = a+b

do u even math, bro?

???

0 is a number because it obeys the properties of real numbers: if you take 0 + a, you get a. If a =/= 0 then 0+a =/= 0.

infinity is not a number because infinity + a = infinity for any number a. Infinity is used to define the size of obnoxiously large sets (like the whole numbers).

One way of defining this size is to say the collection of all numbers such that for any a and b a+b is also in the set. You can create a unitary set {0} and 0+0 is in the set, but this isn't infinite. So we say that for any a+b where a=/=b, then a+b is in the set, then the size of the set is infinity.

0/0=Φ Nullity
1/0=∞
0/1=∞

a-cubed.info/Publications/PerspexMachineVIII.pdf

I mean, I understand what you're saying, but why did you make the same post twice?

browser froze. when it unfroze there were two

...

no one cares how floats work

So ∞*∞ = (0/1)*(1/0)= (1/1)*(0/0) = 1*Φ?

Which is why these "equalities" only work in one direction. Which is why this system is useless bullshit for everything except floating point. Which is why

If you get 0/0, you need L'hospital's rule to find the real limit.

One should not have to use special case differentials to get an arithmetic result.

Ah. I'm not seeing how they'd be useful for floating point, but w/e

We have the axiom that for all [math]a \in \mathbb{C}[/math], [eqn]0 \times a = 0.[/eqn] Observe that if we have [eqn]a \times {1 \over a} = 1,[/eqn] letting [math]a = 0[/math] gives us [eqn]0 \times {1 \over 0} = 1.[/eqn] Suppose that [math]b = {1 \over 0} \in \mathbb C[/math]. Then we get [eqn]0 \times b = 1.[/eqn] This is a contradiction to our earlier axiom.

So there is no complex number [math]b[/math] (and therefore real number by extension since [math]\mathbb{R} \subset \mathbb{C}[/math]) such that [math]0 \times b = 1[/math] (That is, [math]0[/math] has no multiplicative inverse in [math]\mathbb{C} [/math])

but proof by contradiction is a meme

>math should be different than it is
No one cares what you feel.

Because ~nullity~ is basically error state equivalent to NaN, the wiki page for this shitty system draws a direct 1-to-1 comparison, be less lazy.

No one cares what you feel either.

I'm not quite sure I follow your construction there. You said 0 was in the set and the set is closed under addition. That's not necessarily infinite. It's could just have zero in it.

This is not the standard construction of the inductive set.

0/3 = 0
0/2 = 0
0/1 = 0
0/0 = 0
and
3/0 = infinity
2/0 = infinity
1/0 = infinity
0/0 = infinity

>No one cares what you feel.
It was a statement about the elegance/logical integrity of the axiomatic system.

Why use a system that makes me swat a fly with a sledgehammer?

indeed, the limit as x goes to zero of x/x is 1.

Did you know that the limit as x goes to zero of 2x/x is 2?

Did you further know that the limit as x goes to zero of x/(x^3) is infinity?

Really makes you think.

It's undefined not infinity you brainlet

> So we say that for any a+b where a=/=b, then a+b is in the set, then the size of the set is infinity.

So if we're saying a=/=b, then we're saying our "seed" set has at least two elements. Because the identity for any binary operation is unique this set has three operations (assuming commutative): a*b = a b*b = b and a*a = ???

To preserve closure under the operation, we say a*a = c but then we find a*c = ???

This goes on until you've got a set isomorphic with the whole numbers. You can also show this works for any set with a binary operation and two specified elements.

>This is not the standard construction of the inductive set.

I figured, I learned this from one of my favorite professors and I like it better than any of the others I've heard.

I was making a rhetorical point.

Say Johnny has two oranges and wants to share with pondu. Johnny takes two oranges and splits them between two people. 1 orange for each person say Johnny has nothing and wants to share it with nobody. Not even himself because he doesn't exist. There is nothing there.

That's not what my calculator says.

forgot pic

nan-defined ?

You're a brainlet that doesn't know basic math

uhuh, sure thing. not damage control at all.

Boy look at the quality of Veeky Forums these days. Can never be sure if it's bait or the average visitor is actually a twelve year old with zero reading comprehension.

division by any number is perfectly
well-defined, you retard
division by zero is indeterminate
because it can't be done

with Scientist !!ThFjnJh4EkH you can be sure it's the latter.

Thanks. It's ok. Don't worry about it.

not a number, dumbass.

[eqn] \lim _{ x \to 0 } \frac { x } { x } [/eqn] So clearly 0/0 = 1.

0/1 = 0
0/2 = 0
0/3 = 0
0/4 = 0
0/5 = 0
0/0 = wat

[eqn]\lim_{x \to 0} \frac{x}{x} = \lim_{x \to 0} \frac{2x}{x} = \lim_{x \to 0} \frac{3x}{x} = ... \\
\lim_{x \to 0} \frac{x}{x} = \lim_{x \to 0} \frac{x^2}{x} = \lim_{x \to 0} \frac{x^3}{x} = ...[/eqn]

Literally all wrong. [eqn] \lim _{ x \to 0 } \frac { 2 x } { x } =2 [/eqn] As you'd expect if 0/0 =1

Exactly, yet all tend to 0/0 as x approaches 0.

So you can't just say 0/0 = 1 because I can just as easily say that 0/0 = 2 or 3 or whatever.

I don't think thats right about your closure.

It says IF a =/= b, then a+b is in the set.

This does not imply there are at least two elements. It doesn't say that there exist at least two elements. You can't say that just because you have a rule that requires two elements that you somehow have two elements.

Nothing is nothing.

...

What about the limit of x^2/x? Surely that tends to 0/0 but it also tends to 0? Oh wait, I guess you could distinguish 0^2/0 from 0/0, saying that 0^2/0=0*0/0=0*1=0.

0/0 SHOULD be equal to 1 but due to some inconsistencies in math they had to make an exception to the rule to fit other rules and so on.

Ok.
Let a,b be real numbers in S, such that a=/=b. If we require S to be closed under addition, then |S| = aleph null.

What about y=ax+c

kek