# ∞

This has been bugging me for a while. Yes, I’m the kind of person who is bothered by mathematical mysteries.

x/0 = ∞

Think of that for a second.

Let x = 100.

100/5 = 20
100/4 = 25

So, 100/y, where y is between 5 and 4, should equal to something between 20 and 25, correct? I plug in any number between the two, and yes; it’s correct.

100/1 = 100
100/-1 = -100

Therefore, 100/y, where y is between 1 and -1, should equal to something between 100 and -100, correct? This is where the problem arises.

If x/0 = ∞, then 100/0 = ∞.

So, therefore ∞ is between 100 and -100. I do further calculations, and yes, it is correct. See below:

0/x = 0, correct? 0/5 = 0, 0/4 = 0, et al.

Let’s do this:

0/1 = 0
0/-1 = 0

x/0 = ∞, if x=0, then 0/0=∞

This is true, right?

0/1 = 0
0/0 = ∞
0/-1 = 0

Therefore, ∞ is between 0 and 0. What number is between 0 and 0? Nothing. Meaning that ∞ cannot exist. Which is true, entirely. Correct?

Please, fix/change anything. This is really bothering me.

No, and you can easily see this if you let y = 0.5. Try graphing the function

f(x) = 100/x

and you should see how it behaves when x gets close to zero.

God damnit. Topic destroyed. >: (

100/y, if y>1, or if y<-1, then the answer would be below 100, but if -1< y >1, it’s above 100?

It’s all to do with limits. For 1/x, as the value of x approaches 0, the value of 1/x approaches infinity. So we basically just say for convenience’s sake that 1/0 is infinity.

50000000 hours in MSPaint

That would be what it looks like, correct? But where does it quickly transition from +∞ to -∞, if going from left to right?

You should flip it about the y-axis, but yeah, that’s about right. It doesn’t “transition,” because it isn’t continuous at zero. The limits on the left and right sides are different.

0/1 = 0
0/0 = ∞
0/-1 = 0

0<∞>0?

0/0 is defined as being, well, undefined. Its status as undefined is different from any other x/0. That’s due in part because any number divided by itself is 1; however 0/0 can’t really be 1.

x/0 = ∞
0/x = 0
x/x = 1

0/0 = ∞, 0, and 1?

Hence why you never divide by zero. You end up being able to do funky stuff such as say ∞ doesn’t exist and 3=4.

You’re treating infinity as a number, which is wrong. 0/0 does not equal infinity. In fact, nothing equals infinity, except a limit.

Well, while we’re on the topic of infinity and shit…

How is it possible to move a distance of x-meters, if x can be divided into infinite numbers? Technically, to pass the x-meter line, you need to pass the x/2-meter line, to pass that you need to pass the x/4 line, then the x/8 line, etcetera…?

That’s an infinite series problem where each incremental step is a half of the previous step. If you graphed it, it would look like the ones you drew earlier where the line approaches 0 but never reaches it because you keep dividing x by a larger and larger number.

Edit. DIdn’t quite answer the question, did I?

Anyway, if you did that in real life you’d find yourself taking really small steps eventually as the denominator got really big. The distance you’re covering isn’t that much.

(Infinite number of divisions) * (Infinitely small divisions) = a normal, reasonable distance. It’s the principle of calculus.

Yeah, 0/0 can really be anything. For instance, if you take the limit of the function

f(x) = sin(x) / x

as x goes to zero, that limit equals 1. But the limit of

f(x) = x^2 / sin(x)

as x goes to zero is zero. In both cases, f(0) is 0/0.

Assuming you’re going at a constant speed of x meters/second, and that you need some amount of time <i>t</i> to run x meters, then the time that you take to run x/2 meters is t/2, the time that it takes you to run x/4 meters is t/4, etc. The amount of time for you to run each successive interval is also getting smaller and smaller in direct proportion to the amount of distance you’re running, so even though you’re making smaller and smaller subdivisions in space, you’re also making smaller and smaller subdivisions in time.

It’s infinity, not 0. The limit of f(x) = x^2 / sin(x) as x approaches zero is zero.

Yeah as you take the limit of x to approach infinity, then sin(x)/x^2 tends to infinity.

I can’t believe all this stuff from Calculus, last term is coming useful on here nowadays.

My bad, I meant to write x^2 / sin(x). Thanks.

No no no! My calculus proofreading sense has detected another mistake!!! One of the infinities should be replaced by a zero to make the statement correct.