# People don't know how to math



## AvidCuber (Sep 18, 2010)

Okay. So in math class today, we proved some very interesting things.

From the book written by Mr. Charles Seife (_Zero: The Biography of a Dangerous Idea_, ISBN #978-0140296471), on p. 2, he says: “Zero is powerful because it is infinity’s twin. They are equal and opposite…”

Assuming this statement is true, we can prove that 0 = ∞, and also that 0 = -∞, since they are equal *and* opposite. Therefore, from Euclid’s common notion 1, from his first book of the elements, ∞ = -∞.

Therefore, the number line is actually a number circle. Infinity is at the top, and zero at the bottom. They are on the same vertical line, since they are equal. So, you can go around from zero to infinity, and from negative infinity (since infinity equals negative infinity) to zero. The circumference of the circle is 2∞, because zero to infinity is infinity, and negative infinity to 0 is negative infinity, but it’s also positive infinity since they are equal. Therefore, the circumference is 2∞.

From the equation for the circumference of a circle, C=∏d. Therefore, ∏=C/d. The diameter must be ∞. So, since the circumference is 2∞, then ∏=2∞/∞. The infinities cancel. Clearly, ∏=2.

----

From one simple statement, we have proved that ∞=-∞=0, that the number line is actually a number circle, and that ∏=2.

What do you all think?


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## Weston (Sep 18, 2010)

Infinity divided by infinity isn't 1.


I think.


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## AvidCuber (Sep 18, 2010)

Weston said:


> Infinity divided by infinity isn't 1.
> 
> 
> I think.


 Well, infinity equals infinity, and any equal value divided by itself is 1...


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## pappas (Sep 18, 2010)

Everything you said makes sense but I still don't believe it.


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## Weston (Sep 18, 2010)

AvidCuber said:


> Weston said:
> 
> 
> > Infinity divided by infinity isn't 1.
> ...



That would mean 0/0 = 1.


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## aronpm (Sep 18, 2010)

AvidCuber said:


> Weston said:
> 
> 
> > Infinity divided by infinity isn't 1.
> ...



0

Also the reason this 'proof' 'works' is because of the 0 = infinity crap

0 = infinity
add 1 to both sides
1 = infinity
but infinity = 0 so
1 = 0
add 1 to both sides
2 = 1
therefore I am the king of earth, as the set containing me and the king of earth contains 2 people, but 2 is 1, therefore it contains 1 person, so I must be the king of earth.


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## Weston (Sep 18, 2010)

AvidCuber said:


> *From one simple statement, we have proved that ∞=-∞=0*, that the number line is actually a number circle, and that ∏=2.



Actually, that was your assumption. And from that assumption, you calculated that ∏ = 2. I don't see how that makes you arrive at the conclusion that ∏=2. That tells me that ∞ =/= -∞ =/= 0.


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## ShadenSmith (Sep 18, 2010)

Infinity isn't the finite number you're treating it to be.


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## AvidCuber (Sep 18, 2010)

aronpm said:


> Also the reason this 'proof' 'works' is because of the 0 = infinity crap


 Well, I did say it was based off of that statement.


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## hawkmp4 (Sep 18, 2010)

2*∞=∞. ∞/∞=/=1. Like someone already said, infinity is not a finite number, you cannot treat it as such. Normal arithmetic does not work with infinity.

Which of course means that this is not a proof.

EDIT: http://en.wikipedia.org/wiki/Indeterminate_form


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## ThatGuy (Sep 18, 2010)

Math beatdown here.


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## Forte (Sep 18, 2010)

\( \infty+2=\infty \)

\( 0=2 \)

OMG?!?!


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## LewisJ (Sep 18, 2010)

If you actually knew much about math, you'd know that this proof is merely a for fun proposition, almost a demonstration of what you get when you think of infinity incorrectly. 

If you treat infinity the way this proof treats it, all kinds of ridiculous things can be demonstrated.

Sidenote: There is a more believable proof by construction/demonstration of sorts by closing in a wave-like string on a line to show that 2 = pi. Unfortunately it is hard to describe in words - maybe someone else here knows the proof I am referring to.


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## ColdFactor (Sep 18, 2010)

http://en.wikipedia.org/wiki/Mathematical_fallacy#Fundamental_theorem_of_algebra

false proof any number equals any another number?


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## nck (Sep 18, 2010)

Infinity isn't a number but a direction.
Nevertheless I love fallacy proves.
http://www.jimloy.com/geometry/obtuse.htm


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## Kenneth (Sep 18, 2010)

AvidCuber said:


> “Zero is powerful because it is infinity’s twin. They are equal and opposite…”
> 
> Assuming this statement is true, we can prove that 0 = ∞, and also that 0 = -∞, since they are equal *and* opposite.



I had this idea for years but never came up with any proof, to me it's kinda intuitive it must be like so.

Look at this 2 digit binary system: 00 = zero, 01 = "mid" between zero and ∞, 11 = mid of zero and -∞, 10 = ∞

Then just add digits to get higer resolution. I have no idea what this "mid" should be thought :confused:

Edit: noramlly when doing maths you just grasp numbers from nowhere assuming there are countless of infinitys but the real world is not like so, you must take from somewhere to add to your sum or whatever you are doing. So if you start from zero (no apples), then you will have a set that is equal to infinity (all apples) to get your numbers from, but as soon as you add 1 to your set you must also subtract 1 from the other set.

Like yin&yan.

To twist one corner clockwise you must also twist another counter clockwise


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## nck (Sep 18, 2010)

AvidCuber said:


> Zero is powerful because it is infinity’s twin. They are equal and opposite…


To my understanding, they are 'equal and opposite' as in zero is infinitely small and infinity is infinitely large. (NOT as in 0 = -∞ as this is quite absurd)Therefore you cannot directly compare them as the ratio 0/∞ can be anything.

Just as infinity is not a number, zero is not really a 'number' rather than the origin in space. For example, when x is approaching to 0, the ratio x/sinx is approaching to one. In this first case, 0/0=1 as the linear and sine function are increasing at the same rate at x approaching to 0. However this is not always true as 0/0 technically can be anything.

This is also why division by 0 is also invalid, just as division by infinity.


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## Siraj A. (Sep 18, 2010)

Infinity is equal to zero in the sense that 1/∞ = 0 because 1/∞ is infinitesimally small. (like when doing limits)


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## Stefan (Sep 18, 2010)

AvidCuber said:


> What do you all think?



I think you just posted a lot of rubbish.


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## qqwref (Sep 18, 2010)

It's a nice idea to start with infinity being equal to 0, but you can do a lot more from that.

Specifically I really don't like how you state that, because you have a circle with diameter ∞ and circumference 2∞, π must be 2. This is false because infinity multiplied by *any* number is the same, even in normal math. So the circumference is 2∞ and also π∞, because those two are equal.

Did you know:
- Napoleon's army had an infinite number of war elephants?
- The number of atoms in the universe is almost zero?
- A typical teenager has committed suicide infinity times?
- You cannot make a strictly decreasing sequence of positive integers that has length at least zero?
- If I solve zero random scrambles on a Rubik's Cube in competition, it is certain that I will break 7.08 seconds?


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## riffz (Sep 18, 2010)

I'm with Stefan. This thread has a lot of silliness in it.


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## irontwig (Sep 18, 2010)

The author was just expressing the interesting properties of the concepts of zero and infinity poetically ***.


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## hawkmp4 (Sep 18, 2010)

Siraj A. said:


> Infinity is equal to zero in the sense that 1/∞ = 0 because 1/∞ is infinitesimally small. (like when doing limits)



Wait. You're saying that ∞=0 because 1/∞=0?
:fp


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## CharlesOBlack (Sep 18, 2010)

infinity / infinity = 0 / 0

that's why this works, right?


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## mcciff2112 (Sep 18, 2010)

Go ahead and use 2 the next time you use pi for anything and I'll use 3.14 and we'll see whose answer is closest.


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## Joker (Sep 18, 2010)

StefanPochmann said:


> AvidCuber said:
> 
> 
> > What do you all think?
> ...


Agreed...if infinity = 0 and negative infinity = 0, then that would mean negative infinity = positive infinity...


hawkmp4 said:


> Siraj A. said:
> 
> 
> > Infinity is equal to zero in the sense that 1/∞ = 0 because 1/∞ is infinitesimally small. (like when doing limits)
> ...



Yes he is...there is a big difference between the 2.


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## hawkmp4 (Sep 18, 2010)

Joker said:


> StefanPochmann said:
> 
> 
> > AvidCuber said:
> ...


Somewhat related, though not quite the same and certainly not what the OP meant- http://en.wikipedia.org/wiki/Real_projective_line


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## blah (Sep 18, 2010)

How infinity works:

Let \( x=1-1+1-1+\cdots \).
Then \( x=1-(1-1+1-1+\cdots)=1-x \).
Adding \( x \) to both sides, we have \( 2x=1 \) and so \( x=\frac{1}{2} \), which means that \( 1-1+1-1+\cdots=\frac{1}{2} \).

Infinity is awesome.


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## AvidCuber (Sep 18, 2010)

irontwig said:


> The author was just expressing the interesting properties of the concepts of zero and infinity poetically ***.


 Thank you. I based it all off of that statement, so in a way none of this is my own work.


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## Siraj A. (Sep 19, 2010)

hawkmp4 said:


> Wait. You're saying that ∞=0 because 1/∞=0?
> :fp


 
Erm, did you read what I wrote?


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## blade740 (Sep 19, 2010)

I think this was the best part:



AvidCuber said:


> From the equation for the circumference of a circle, C=∏d. Therefore, ∏=C/d. *The diameter must be ∞.* So, since the circumference is 2∞, then ∏=2∞/∞. The infinities cancel. Clearly, ∏=2.


 
I can use any equation to prove anything if I pull numbers out of my ass like that.


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## vcuber13 (Sep 19, 2010)

blade740 said:


> I think this was the best part:
> 
> 
> 
> I can use any equation to prove anything if I pull numbers out of my ass like that.


 
its true though, one end is infinty and the other is 0, infinity - 0 = infinity


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## uberCuber (Sep 19, 2010)

vcuber13 said:


> its true though, one end is infinty and the other is 0, infinity - 0 = infinity


 
this is a circle, not a line. 0 is at one end of the circle, and infinity is at the other end of the circle. Which means that a semicircle is length infinity, not the diameter.

Of course this is all assuming you can treat infinity normally which you can't..


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## Stefan (Sep 19, 2010)

AvidCuber said:


> I based it all off of that statement, so in a way none of this is my own work.



Nah. It was entirely your work.



AvidCuber said:


> he says: “Zero is powerful because it is infinity’s twin. They are equal and opposite…”



Interesting. Why did you stop there? Let's continue just a bit further:
_"Zero is powerful because it’s infinity’s twin. They are equal and opposite, yin and yang. They are equally paradoxical and troubling.”_

*That* is how he meant they're equal. They're equal *in some sense*. Not literally 0 = ∞. Similar to Newton's third law about "equal and opposite" forces - equal *in magnitude* and opposite in direction. Major facepalm for reading that as 0 = ∞, which the author never said.

So already 0 = ∞ was your work, not that author's, and then the other stuff you added doesn't follow from it, you further made that all up yourself.



AvidCuber said:


> Assuming this statement is true, we can prove that 0 = ∞



Nah. Even if that was the case, you're not "assuming" and "proving", you're just *reading* it like that.


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## BigSams (Sep 19, 2010)

This is some real epic stuff here. First the OP writes a jumble of randomness, then people take it seriously and go on to divide infinity by infinity to get 1.. this is better than celebrity millionaire.


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## Rinfiyks (Sep 19, 2010)

I love threads like this. So many lulz :>


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## nck (Sep 19, 2010)

vcuber13 said:


> its true though, one end is infinty and the other is 0, infinity - 0 = infinity


 
One important rule my math professor taught me in messing around with calculus is never do arithmetic calculation with infinity. Frankly, it just doesn't work that way. 

As I said, infinity is not a number, it's a direction. Depending on the context, one infinity may be larger than another. 

An easy easy example would be (X^2)/X when X is approaching to 0. It's obviously equal to zero, however, if you see 0 as a number than 0/0=1...which just doesn't work.


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## riffz (Sep 19, 2010)

CharlesOBlack said:


> infinity / infinity = 0 / 0
> 
> that's why this works, right?


 
No. Just no.


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## irontwig (Sep 19, 2010)

AvidCuber said:


> Thank you. I based it all off of that statement, so in a way none of this is my own work.


 
"Thank you"? I basically called you a moron.


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## qqwref (Sep 19, 2010)

A lot of teachers will explicitly tell you not to do this, but I find it really useful to think of infinity as a number. It's got some strange properties just like zero does, so you have to be careful with it, but there are quite a few situations (such as inversions, and limits) where you can get a much better intuitive feel for an idea by treating infinity as a special number rather than a completely different type of thing (a direction, an idea, a way of notating divergence, etc.).


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## hawkmp4 (Sep 19, 2010)

Siraj A. said:


> Erm, did you read what I wrote?


I did. And I just did again, and it's equally nonsensical this time.


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## Siraj A. (Sep 19, 2010)

hawkmp4 said:


> I did. And I just did again, and it's equally nonsensical this time.


 
I didn't say infinity is equal to zero. Only _in the sense_ that 1/infinity = 0 when doing LIMITS.

Edit: I guess I should have said that 1/infinity = 0 for convenience, but it's really just REALLY CLOSE to zero. I guess I have to explain everything out to you, sorry.


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## hawkmp4 (Sep 19, 2010)

Siraj A. said:


> I didn't say infinity is equal to zero. Only _in the sense_ that 1/infinity = 0 when doing LIMITS.
> 
> Edit: I guess I should have said that 1/infinity = 0 for convenience, but it's really just REALLY CLOSE to zero. I guess I have to explain everything out to you, sorry.


I know perfectly well how \( \frac{1}{\infty} \) behaves in the context of limits.
Your argument is that you could say that 2 is equal to .5 in the sense that 1/2 = .5.

Which is, like I said, nonsense.


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## Siraj A. (Sep 19, 2010)

hawkmp4 said:


> I know perfectly well how \( \frac{1}{\infty} \) behaves in the context of limits.
> Your argument is that you could say that 2 is equal to .5 in the sense that 1/2 = .5.
> 
> Which is, like I said, nonsense.



I'm NOT saying 0 = infinity. You're just too stupid to interpret my post correctly. But if you knew, then you knew what I meant, so you were just intentionally being an ass? I understand.


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## Andrew Ricci (Sep 19, 2010)

blade740 said:


> I think this was the best part:
> 
> 
> 
> I can use any equation to prove anything if I pull numbers out of my ass like that.


 
LOLOL

That pretty much sums up his whole theory.


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## RyanReese09 (Sep 19, 2010)




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## cyoubx (Sep 19, 2010)

Infinity does not equal infinity...

Your statement assumes infinity as a number, but it is not, it is a concept, a limit. It has no value, so you can't multiply/divide/add/subtract two infinities together.


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## Joker (Sep 19, 2010)

Siraj A. said:


> I'm NOT saying 0 = infinity. You're just too stupid to interpret my post correctly. But if you knew, then you knew what I meant, so you were just intentionally being an ass? I understand.


 
OMG :fp
He's saying you said 1/infinity = 0. He never said you said infinity = 0. He's saying something like 
1/[999999999999^999999999999999999999999999999999999999999999999999999999999999999999999999999999999999] has value, and isn't complete 0. Your saying 1/infinity = 0 with limits. That's your debate, now continue lol.


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## hawkmp4 (Sep 19, 2010)

Joker said:


> OMG :fp
> He's saying you said 1/infinity = 0. He never said you said infinity = 0. He's saying something like
> 1/[999999999999^999999999999999999999999999999999999999999999999999999999999999999999999999999999999999] has value, and isn't complete 0. Your saying 1/infinity = 0 with limits. That's your debate, now continue lol.


No, you have it wrong. But I'm done arguing with Siraj because he's just flaming now and not listening or saying anything constructive.


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## ThatGuy (Sep 19, 2010)

Siraj A. said:


> *Infinity is equal to zero in the sense* that 1/∞ = 0 because 1/∞ is infinitesimally small. (like when doing limits)


It doesn't seem like infinity is zero in that sense...


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## Igora (Sep 19, 2010)

AvidCuber said:


> “Zero is powerful because it is infinity’s twin. They are equal and opposite…”
> 
> Assuming this statement is true, we can prove that 0 = ∞, and also that 0 = -∞, since they are equal *and* opposite . . . Clearly, ∏=2.


 
Assuming the statement is true gives you a false answer. Therefore, the statement isn't true.


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## Siraj A. (Sep 19, 2010)

Joker said:


> OMG :fp
> He's saying you said 1/infinity = 0. He never said you said infinity = 0. He's saying something like
> 1/[999999999999^999999999999999999999999999999999999999999999999999999999999999999999999999999999999999] has value, and isn't complete 0. Your saying 1/infinity = 0 with limits. That's your debate, now continue lol.



Um, no, actually he IS saying that I said infinity = 0. I said 1/infinity = 0 in the context of limits. That is the same as saying 1/999999999....Now with limits, that is = 0 for convenience. He knows this as well. And now you are trying to tell me what I am saying when you clearly don't know what this "debate" is even about. 



hawkmp4 said:


> No, you have it wrong. But I'm done arguing with Siraj because he's just flaming now and not listening or saying anything constructive.


 
I like how when you have no reply you just go and report the post. That's real mature of you. You KNOW exactly what I said in my original post, and if it was unclear, I know you KNOW what I meant. But I guess you know what I'm saying better than I do.


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## ThatGuy (Sep 19, 2010)

Siraj A., I think your problem was that you put


Siraj A. said:


> *Infinity is equal to zero *in the sense that 1/∞ = 0 because 1/∞ is infinitesimally small. (like when doing limits)


 instead of


Siraj A. said:


> Um, no, actually he IS saying that I said infinity = 0. *I said 1/infinity = 0 in the context of limits*. That is the same as saying 1/999999999....Now with limits, that is = 0 for convenience.


 just putting that.


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## Siraj A. (Sep 19, 2010)

@Thatguy

Yeah, you're right. But I said "in the sense that." But I guess I should have been more clear, even though that if anyone knew what I was referring to, they should have known what I meant, and there was no reason to argue with me in the first place. >_>


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## nitrocan (Sep 19, 2010)

AvidCuber said:


> ....
> 
> Assuming this statement is true, we can prove that 0 = ∞, and also that 0 = -∞, since they are equal *and* opposite. Therefore, from Euclid’s common notion 1, from his first book of the elements, ∞ = -∞.


 
This is the stupidest thing I've ever heard. How about I come up and say "27 is powerful, because it is 18's twin. They are equal and opposite."? Now go have fun with that.
It's obvious that he didn't actually mean "0 equals infinity".


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## hawkmp4 (Sep 19, 2010)

Siraj A. said:


> Um, no, actually he IS saying that I said infinity = 0. I said 1/infinity = 0 in the context of limits. That is the same as saying 1/999999999....Now with limits, that is = 0 for convenience. He knows this as well. And now you are trying to tell me what I am saying when you clearly don't know what this "debate" is even about.


Yeah. Ignore him.
My point is not about the terminology that you used. I find it perfectly acceptable to say that \( \frac{1}{\infty}=0 \), since using \( \infty \) in that sense makes it clear you're already talking about limit behaviour. You don't need to say it twice. So that's not what my point is; everyone who is saying so needs to read my posts again.

My issue is your statement "Infinity is equal to zero in the sense that 1/∞ = 0..."
Yes, \( \frac{1}{\infty}=0 \), I agree. But that doesn't mean that "infinity is equal to zero [in that sense]", it means that 1 divided by infinity is equal to zero. They're very different statements. \( \frac{1}{\infty}=0 \Rightarrow \infty=0 \) is just false. Bad logic.

And before you say it, YES. You _did_ say that \( \infty=0 \).


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## blah (Sep 19, 2010)

\( \aleph_0<\aleph_1 \)

kthxbai


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## nitrocan (Sep 19, 2010)

hawkmp4 said:


> My issue is your statement "Infinity is equal to zero in the sense that 1/∞ = 0..."
> Yes, \( \frac{1}{\infty}=0 \), I agree. But that doesn't mean that "infinity is equal to zero [in that sense]", it means that 1 divided by infinity is equal to zero. They're very different statements. \( \frac{1}{\infty}=0 \Rightarrow \infty=0 \) is just false. Bad logic.[/math].


 
Can anyone explain to me the good/bad logic behind this? I don't get how if 1/x = y, then x = y


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## Christopher Mowla (Sep 19, 2010)

Siraj A. said:


> Infinity is equal to zero in the sense that 1/∞ = 0 because 1/∞ is infinitesimally small. (like when doing limits)


 So this is the statement that hawkmp4 is trying to tell you is false. Well, how about we analyze it piece by piece.

Infinity is equal to zero
This is definitely false.

in the sense that
This wording attempts to justify the previous piece by giving the restriction of the last piece "(like when doing limits)". But you will see (hopefully) how you do not understand that restriction when I explain the remaining pieces of your statement.

1/∞ = 0 because 1/∞ is infinitesimally small (like when doing limits)
Being infinitesimally small doesn't make it 0.0, no matter how small it gets. Unlike calculators, there is no round off point with the concept of infinity.

When doing limits and we write \( \underset{x\to \infty }{\mathop{\lim }}\,\left[ \frac{1}{x} \right]=0 \), we are NOT saying "\( \frac{1}{\infty }=0 \) in the terms of limits". The equal sign used in that arithmetic can be misleading.
\( \underset{x\to \infty }{\mathop{\lim }}\,\left[ \frac{1}{x} \right]=0 \) basically says "the function \( f\left( x \right)=\frac{1}{x} \) converges to (but never is equal to) zero as the x-value increases without bound".

In algebra terms, there is an asymptote about the line y=0. Hence, no matter how large the x value gets, the graph of \( f\left( x \right)=\frac{1}{x} \) will *never* touch the x-axis. (Hence, the value of zero is not in the range of the function).

Another important fact about calculus you need to remember is that there are three conditions (all of which must be true) for a function \( f\left( x \right) \) to be continuous at \( x=a \).

For a function to be continuous at a value of a means that \( \underset{x\to a}{\mathop{\lim }}\,\left[ f\left( x \right) \right]=f\left( a \right) \).

This is crucial for this argument because you are claiming that, in terms of limits, 
\( \underset{x\to \infty }{\mathop{\lim }}\,\left[ \frac{1}{x} \right]=0=\frac{1}{\infty } \), which is false as you shall soon see.



The three conditions for continuity are:

1) \( f\left( a \right) \) is defined.
2) \( \underset{x\to a}{\mathop{\lim }}\,\left[ f\left( x \right) \right] \) exists.
3) \( \underset{x\to a}{\mathop{\lim }}\,\left[ f\left( x \right) \right]=f\left( a \right) \)

Hence, \( f\left( x \right)=\frac{1}{x} \) violates the first of the three conditions because this function is not defined at 0 since its graph never touches the x-axis. (And remember we have \( f\left( a \right)=0 \) since \( \underset{x\to \infty }{\mathop{\lim }}\,\left[ \frac{1}{x} \right]=0,\text{ where }f\left( x \right)=\frac{1}{x}\text{ and }f\left( a \right)=0 \)).

Again, since it only takes one of the three conditions to not have continuity, this function is not continuous everywhere.

Comparing this fact with the general equation \( \underset{x\to a}{\mathop{\lim }}\,\left[ f\left( x \right) \right]=f\left( a \right) \), we have for our case:
\( \underset{x\to \infty }{\mathop{\lim }}\,\left[ \frac{1}{x} \right]=0\ne \frac{1}{\infty } \)


All of this disproves the portion of your statement:
1/∞ = 0 because 1/∞ is infinitesimally small. (like when doing limits)
which is the condition for your argument
Infinity is equal to zero 
to be true since you linked the two together with the phrase
in the sense that

*Conclusion*
If you really understood all of the material I presented, then you must agree that your statement was false. That's okay because the equal sign in \( \underset{x\to \infty }{\mathop{\lim }}\,\left[ \frac{1}{x} \right]=0 \) is misleading. However, your statement's claim was false and the conditions under which it was supposed to be true violate the fundamentals of calculus.

hawkmp4 wasn't trying to deliberately skip over what you claimed to be true (and the restrictions under which it was true), but instead he attempted to use the fact that the inverse of a number is not equal to itself (with the unique exception of the number 1). That was my first thoughts to disprove the validity of your statement too (since it is very obvious) but I instead geared this post towards the concepts of basic calculus which were a little unclear to you (or so it seems). If they weren't unclear to you, then the only other possibility is that you wrote something without really thinking about it.


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## Siraj A. (Sep 19, 2010)

@hawkmp4 and cmowla

I understand now. I guess my original post was incorrect. My wording was stupid. I didn't mean to say infinity = 0, and I definitely *don't* think that infinity = 0 (because we all know that's absurd), but I see how what I presented does say that. Sorry for the confusion (and frustration) and thanks for the calc lesson, lol.


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## Joker (Sep 19, 2010)

Siraj A. said:


> @hawkmp4 and cmowla
> 
> I understand now. I guess my original post was incorrect. My wording was stupid. I didn't mean to say infinity = 0, and I definitely *don't* think that infinity = 0 *(because we all know that's absurd), *but I see how what I presented does say that. Sorry for the confusion (and frustration) and thanks for the calc lesson, lol.


Seems like some people here disagree...


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## Siraj A. (Sep 19, 2010)

Joker said:


> Seems like some people here disagree...


 
Well um if you read anything they just wrote, or know some math, then you shouldn't disagree.


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## blah (Sep 19, 2010)

cmowla said:


> The three conditions for continuity are:
> 
> 1) \( f\left( a \right) \) is defined.
> 2) \( \underset{x\to a}{\mathop{\lim }}\,\left[ f\left( x \right) \right] \) exists.
> 3) \( \underset{x\to a}{\mathop{\lim }}\,\left[ f\left( x \right) \right]=f\left( a \right) \)


\( \forall x_0 \in D(f), \forall \varepsilon > 0, \exists \delta \) such that \( |x-x_0|<\delta\Rightarrow|f(x)-f(x_0)|<\varepsilon \).


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## Andrew Ricci (Sep 19, 2010)

This thread got an epic new title.


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## nitrocan (Sep 19, 2010)

theanonymouscuber said:


> This thread got an *appropriate* new title.


Fixed


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## hawkmp4 (Sep 19, 2010)

blah said:


> \( \forall x_0 \in D(f), \forall \varepsilon > 0, \exists \delta \) such that \( |x-x_0|<\delta\Rightarrow|f(x)-f(x_0)|<\varepsilon \).


Take your analysis and your deltas and epsilons somewhere else. You're not welcome here.

(Kidding!)



nitrocan said:


> Can anyone explain to me the good/bad logic behind this? I don't get how if 1/x = y, then x = y


You don't get it because it just doesn't follow. Take (for example) 2. 1/2 = .5 doesn't imply 2=.5. Like cmowla said, the only number that statement holds for is 1.


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## blah (Sep 19, 2010)

So talking about limits is okay but talking about something more intuitive (and rigorous) isn't?


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## Joker (Sep 19, 2010)

Siraj A. said:


> Well um if you read anything they just wrote, or know some math, then you shouldn't disagree.


 
I find it stupid that some people just assume random things. I don't disagree.


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## hawkmp4 (Sep 19, 2010)

blah said:


> So talking about limits is okay but talking about something more intuitive (and rigorous) isn't?


 
It was a joke. I know lots of people hate the epsilon-delta proofs. I don't. I agree with you; they make more sense to me.


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## qqwref (Sep 19, 2010)

cmowla said:


> In algebra terms, there is an asymptote about the line y=0. Hence, no matter how large the x value gets, the graph of \( f\left( x \right)=\frac{1}{x} \) will *never* touch the x-axis. (Hence, the value of zero is not in the range of the function).


It is if you allow infinity in the domain.



cmowla said:


> Comparing this fact with the general equation \( \underset{x\to a}{\mathop{\lim }}\,\left[ f\left( x \right) \right]=f\left( a \right) \), we have for our case:
> \( \underset{x\to \infty }{\mathop{\lim }}\,\left[ \frac{1}{x} \right]=0\ne \frac{1}{\infty } \)


I have no clue what you were getting at with the continuity thing, since proving a function is not continuous does not in fact prove that any one of the three conditions you listed is false, merely that SOME one of them is. Proving 1/x is continuous does not in any way imply that the limit as x approaches infinity of 1/x is NOT 1/infinity. Even if you believe that, since infinity is not a real number, it is not possible to plug it into the function 1/x, I'd say that in some contexts this is completely possible and gives a well-defined result (that is, 0).



cmowla said:


> If you really understood all of the material I presented, then you must agree that your statement was false.


Unless there was a mistake in your proof, of course. If your proof is convincing and true, why would this disclaimer be necessary?


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## cmhardw (Sep 19, 2010)

blah said:


> How infinity works:
> 
> Let \( x=1-1+1-1+\cdots \).
> Then \( x=1-(1-1+1-1+\cdots)=1-x \).
> ...


 
I say this line of reasoning is false.

The Cauchy Convergence Criterion for series says that a series \( \sum_{k=1}^\infty a_k \) converges iff for each positive number \( \epsilon \) there is a natural number \( N \) such that:

\( |a_{n+1} + \cdots + a_{n+k}| < \epsilon \) for all integers \( n \geq N \) and all natural numbers \( k \).

I don't know if the following line of reasoning is completely and thoroughly rigorous. I would argue, though, that it is at least a good start toward being so while still presenting my point.

Let 
\( \epsilon > 0 \)
\( N = 2m \) where m is a natural number. Notice that N is an even natural number.
\( k \) be any even natural number.

\( |a_{2m+1} + a_{2m+2} + \cdots + a_{2m+k}| = |1+ (-1) + \cdots + 1 + (-1)| = 0 < \epsilon \)

now let k be any odd natural number
\( |a_{2m+1} + a_{2m+2} + \cdots + a_{2m+k}| = |1+ (-1) + \cdots + 1| = 1 \not< \epsilon \) for all \( \epsilon > 0 \)

So when N is even, and k is odd then you have not satisfied the criterion for all \( \epsilon > 0 \)

Now suppose that:
\( \epsilon > 0 \)
\( N = 2m+1 \) where m is a natural number. Notice that N is odd.
\( k \) be any even natural number.

\( |a_{(2m+1)+1} + a_{(2m+1)+2} + \cdots + a_{(2m+1)+k}| = |(-1) + 1 + \cdots + (-1) + 1| = 0 < \epsilon \)

now let k be any odd natural number
\( |a_{(2m+1)+1} + a_{(2m+1)+2} + \cdots + a_{(2m+1)+k}| = |(-1) + 1 + \cdots + (-1)| = |-1| = 1 \not< \epsilon \) for all \( \epsilon > 0 \)

--------

tl;dr

So I can create an absolute value of a partial sum less than all \( \epsilon > 0 \) as long as it contains an even number of terms from your series. If I use an odd number of terms, the absolute value of the partial sum will not be less than \( \epsilon \) for any \( 0 < \epsilon < 1 \) . Since you have not satisfied the Cauchy convergence criterion for all natural numbers k, then it seems to me that this series cannot converge.

Therefore, assuming that:
\( 1 + (-1) + 1 + (-1) + \cdots = x \) which implies that the series *does* converge to a number x, is a false assumption. Any conclusion reached beyond this point is a falsehood.

Can anyone find a hole in this argument? I believe I have argued correctly, but if not someone please point it out.

Chris


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## blah (Sep 19, 2010)

Yes, it _is_ false. I thought that was fairly obvious  It was meant to go along with all the other nonsense going on in this thread.


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## cmhardw (Sep 19, 2010)

blah said:


> Yes, it _is_ false. I thought that was fairly obvious  It was meant to go along with all the other nonsense going on in this thread.


 
*Takes foot and places it directly into mouth* :fp


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## Andrew Ricci (Sep 19, 2010)

cmhardw said:


> *Takes foot and places it directly into mouth* :fp


 
How does that foot taste, Chris?


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## Daniel Wu (Sep 19, 2010)

cmhardw said:


> *Takes foot and places it directly into mouth* :fp


 lololol Chris.  

<3 this thread.


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## BigSams (Sep 19, 2010)

Someone should paste this on the AoPS forums. They'd love it.


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## LewisJ (Sep 19, 2010)

blah said:


> \( \forall x_0 \in D(f), \forall \varepsilon > 0, \exists \delta \) such that \( |x-x_0|<\delta\Rightarrow|f(x)-f(x_0)|<\varepsilon \).


 
You know, colons can be used for "such that"


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## Christopher Mowla (Sep 19, 2010)

qqwref said:


> It is if you allow infinity in the domain.


Huh? I am not sure what you are talking about. The domain of \( f\left( x \right)=\frac{1}{x} \) is \( \left( -\infty ,0 \right)\cup \left( 0,\infty \right) \).



qqwref said:


> I have no clue what you were getting at with the continuity thing, since proving a function is not continuous does not in fact prove that any one of the three conditions you listed is false, merely that SOME one of them is.


The very statement \( \underset{x\to \infty }{\mathop{\lim }}\,\left[ \frac{1}{x} \right]=0 \) resembles the definition of continuity: \( \underset{x\to a}{\mathop{\lim }}\,\left[ f\left( x \right) \right]=f\left( a \right) \).

Since that's the case, we must see what conditions hold true for continuity. If the conditions do not hold true, then obviously \( \underset{x\to \infty }{\mathop{\lim }}\,\left[ \frac{1}{x} \right]=\frac{1}{\infty } \) is not a true statement because it does not fit the *definition* of continuity, that is \( \underset{x\to a}{\mathop{\lim }}\,\left[ f\left( x \right) \right]=f\left( a \right) \), because one of the three conditions for this particular case fails.

Since this particular function is not defined anywhere on the x-axis, then the conclusion is 
\( \underset{x\to \infty }{\mathop{\lim }}\,\left[ \frac{1}{x} \right]\ne \frac{1}{\infty } \)


Remember, \( \underset{x\to \infty }{\mathop{\lim }}\,\left[ \frac{1}{x} \right]=0 \) means that this function converges to (but is not equal to) zero.



qqwref said:


> Proving 1/x is continuous does not in any way imply that the limit as x approaches infinity of 1/x is NOT 1/infinity.


Where did I show the function was undefined? At the line y=0. This does show it. Why? Because \( \frac{1}{\infty }\ne 0 \).

Again, remember, \( \underset{x\to \infty }{\mathop{\lim }}\,\left[ \frac{1}{x} \right]=0 \) means that this function converges to (but is not equal to) zero. This limit does not converge to \( \frac{1}{\infty } \) (which is undefined), but to zero.



qqwref said:


> Even if you believe that, since infinity is not a real number, it is not possible to plug it into the function 1/x,


Treating infinity as a finite number is obviously wrong, however that does not affect substitution into a function such as \( f\left( x \right)=\frac{1}{x} \) where it is very obvious what the result will be. No indeterminate forms exist here.

At the same time, we need to be careful to substitute infinity into the function keeping the concept of infinity in mind. (And there is a result with this substitution only in context of limits. Merely substituting infinity into this function alone is just undefined).



qqwref said:


> I'd say that in some contexts this is completely possible and gives a well-defined result (that is, 0).


 Yes, of course, if the very small decimal value is negligible for a certain application. However, \( \frac{1}{\infty } \) is still never \( 0.\bar{0} \).



qqwref said:


> Unless there was a mistake in your proof, of course. If your proof is convincing and true, why would this disclaimer be necessary?


I am asking myself the same question. You are proving nothing here, only adding unnecessary confusion.


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## BigSams (Sep 19, 2010)

LewisJ said:


> You know, colons can be used for "such that"


 
Or an epsilon that's flipped across a vertical line


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## blah (Sep 20, 2010)

BigSams said:


> Or an epsilon that's flipped across a vertical line


Neither of which is standard notation.

Besides, I prefer the topological definition of continuity, which really has nothing to do with anything in this thread.


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## KboyForeverB (Sep 20, 2010)

I unstood the first few sentences, then I lost track, but still an interesting point


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## qqwref (Sep 20, 2010)

cmowla said:


> Since that's the case, we must see what conditions hold true for continuity. If the conditions do not hold true, then obviously \( \underset{x\to \infty }{\mathop{\lim }}\,\left[ \frac{1}{x} \right]=\frac{1}{\infty } \) is not a true statement because it does not fit the *definition* of continuity, that is \( \underset{x\to a}{\mathop{\lim }}\,\left[ f\left( x \right) \right]=f\left( a \right) \), because one of the three conditions for this particular case fails.


Hang on, if you have proved 1/x is not continuous - which means there is a discontinuity somewhere - it doesn't mean you proved that it is discontinuous at -infinty. A discontinuous function does not have to have lim_a f(x) != f(a) at ALL points a - just at least one of them. In fact, you didn't even put infinity in your domain, so this continuity argument you're making contradicts your own terms - the function cannot be either continuous or discontinuous at a point that is not in its domain. (I suppose this would apply to 0 too...?)



cmowla said:


> Where did I show the function was undefined? At the line y=0. This does show it. Why? Because \( \frac{1}{\infty }\ne 0 \).
> 
> Again, remember, \( \underset{x\to \infty }{\mathop{\lim }}\,\left[ \frac{1}{x} \right]=0 \) means that this function converges to (but is not equal to) zero. This limit does not converge to \( \frac{1}{\infty } \) (which is undefined), but to zero.


A function of x cannot be "undefined" on a horizontal line. That does not make any sense. Did you see the little links in my last post? I didn't see anyone specify "we are working with the standard real numbers here ONLY"; there are formally defined - and useful - mathematical setups where 1/infinity is very much defined and also equal to 0.

And you don't have to tell me what a limit is; I obviously know since that's what we are talking about. It seems like you don't know, since you just said that the limit being equal to zero MEANS that the function is not equal to zero. A limit of a function at a point says absolutely nothing about the function's value at that particular point.



cmowla said:


> Treating infinity as a finite number is obviously wrong


I never did this - infinity is not finite. But it does exist and is a number.



cmowla said:


> At the same time, we need to be careful to substitute infinity into the function keeping the concept of infinity in mind. (And there is a result with this substitution only in context of limits. Merely substituting infinity into this function alone is just undefined).
> 
> Yes, of course, if the very small decimal value is negligible for a certain application. However, \( \frac{1}{\infty } \) is still never \( 0.\bar{0} \).


There is no such thing as a "very small decimal value" that is smaller than any other decimal value but not zero, at least in the number spaces suggested so far. There are no infinitesimals in the reals or the two similar systems I linked to. Any two numbers which are infinitely close are the same; if 1/infinity exists, it cannot be anything but zero.



cmowla said:


> I am asking myself the same question. You are proving nothing here, only adding unnecessary confusion.


I'm not trying to prove anything, but to disprove. If you're making unqualified and incorrect statements that pretend to be rigorous mathematical proofs, I think arguing against them removes confusion.


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## Olivér Perge (Sep 21, 2010)

I just love how the thread is called "People don't know how to math".


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## Lucas Garron (Sep 21, 2010)

blah said:


> \( \aleph_0<\aleph_1 \)
> 
> kthxbai


Yeah, but what's between those?


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## xFear of Napalm (Sep 21, 2010)

If the number line truly is a loop, then it can't ever be proved, because 0 and infinity aren't technically numbers in my mind. 0, to me, is just a placeholder and representation of nothing. Also, infinity can never be reached, because it is only a representation of the lack of a limit. I didn't check the other conversation on this thread because I wouldn't be able to understand it, but this is just my idea. It's a difficult task in itself to define infinity.


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## hawkmp4 (Sep 22, 2010)

xFear of Napalm said:


> If the number line truly is a loop, then it can't ever be proved, because 0 and infinity aren't technically numbers in my mind. 0, to me, is just a placeholder and representation of nothing. Also, infinity can never be reached, because it is only a representation of the lack of a limit. I didn't check the other conversation on this thread because I wouldn't be able to understand it, but this is just my idea. It's a difficult task in itself to define infinity.


Just because you don't think zero and \( \infty \) are numbers doesn't mean that you can't prove things with them. Your personal beliefs don't affect the truth of mathematical statements, sorry.


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## qqwref (Sep 22, 2010)

Zero... isn't a number? So should we say that -1 + 1 is undefined? Or do you also not accept negative numbers?


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## blah (Sep 22, 2010)

Lucas Garron said:


> Yeah, but what's between those?


What are you talking about? \( \aleph_\frac{1}{2} \) of course!


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## imaghost (Sep 22, 2010)

qqwref said:


> Zero... isn't a number? So should we say that -1 + 1 is undefined? Or do you also not accept negative numbers?



Actually zero is not a number. At least it never was. Not long ago have people actually considered it a number. It used to never exist. He is right by saying that it means nothing. -1 + 1 is nothing, not undefined. There is a lot more to it but it is not my thing. 

this also reminds me of robotics club. Last year we came up, somehow, with the equation 0/0=the last digit of pi to the power of a rocket ship. I actually solved it by saying that i is the last digit of pi, i is imaginary. I took the power of the Saturn V rocket, since it was one of the only ones to carry people into space, making it a ship. so then i took i and raised it to the power of the number I found and it equals 1. 
Obviously this is not true, but still, we all have our moments.


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## ~Adam~ (Sep 22, 2010)

AvidCuber said:


> From one simple statement, we have proved that ∞=-∞=0, that the number line is actually a number circle, and that ∏=2.



Is it just me or is he not really talking about a circle but an oblong with zero width.


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## shelley (Sep 22, 2010)

Olivér Perge said:


> I just love how the thread is called "People don't know how to math".



It seemed quite appropriate. Especially considering it contains posts like these:



imaghost said:


> this also reminds me of robotics club. Last year we came up, somehow, with the equation 0/0=the last digit of pi to the power of a rocket ship. I actually solved it by saying that i is the last digit of pi, i is imaginary. I took the power of the Saturn V rocket, since it was one of the only ones to carry people into space, making it a ship. so then i took i and raised it to the power of the number I found and it equals 1.
> Obviously this is not true, but still, we all have our moments.


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## imaghost (Sep 22, 2010)

Hey, it was off-season and we had a lot of time. I am good at math, actually very good, but we were just having fun.


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## Hiero (Sep 23, 2010)

You must realize that something is wrong with your original argument, right? What kind of math class was this, what did the teacher say? I teach 6th grade and don't think I'd let this fuzzy math fly. The main flaw in your argument is that infinity equals infinity. A basic proof of first semester theoretical math is that infinity doesn't equal infinity. I don't remember the whole proof but, for example, compare integers with real numbers. Over any given range of numbers, the real numbers will outnumber integers, so it makes sense that over a larger amount, such as infinity, the real numbers will still outnumber the integers. Some infinities are larger than others, so you can't just perform operations on infinity as you would on other numbers. 

Infinity also doesn't equal 0. In Calculus we used infinity quite a bit and never was it considered the same thing as 0. We also couldn't perform operations on infinity as you are doing. Positive infinity is similar to negative infinity, but they are not one and the same. Take the limit of an expression as you get closer to infinity and negative infinity and you will not get the same answer. You also can't take a number line and start performing geometric calculations on it or give it a shape like a circle. It would make just as much sense to bend it into a triangle or quadrilateral and start finding the area and perimeter of it. Infinity is a theoretical concept, you can't just change it into a closed shape.

Hopefully your teacher lead you back on the right direction and away from this psuedo-math.


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## maggot (Sep 23, 2010)

Siraj A. said:


> Infinity is equal to zero in the sense that 1/∞ = 0 because 1/∞ is infinitesimally small. (like when doing limits)


correction: when doing limits
lim x=>inf 1/x = 0. because 1/x as x becomes larger the value approaches 0. 

0 does not equal infinity. infinity is greater than any real number. i can go on and on about infinity, but ... proof that inf/inf is not =1

1= inf/inf
inf + inf = inf
---------
1= (inf +inf) / inf 
1 = (inf/inf) + (inf/inf)
1 = 1 + 1 
1 = 2

i definately agree with pochmann. i think you posted a bunch of rubbish. 
i was going to post more on how infinity is handled in many other types of situations and prove it cannot be handled as a real. however, most of this is to complex for your brain im sure.


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## AvidCuber (Sep 23, 2010)

hiero said:


> You must realize that something is wrong with your original argument, right? What kind of math class was this, what did the teacher say? I teach 6th grade and don't think I'd let this fuzzy math fly. The main flaw in your argument is that infinity equals infinity. A basic proof of first semester theoretical math is that infinity doesn't equal infinity. I don't remember the whole proof but, for example, compare integers with real numbers. Over any given range of numbers, the real numbers will outnumber integers, so it makes sense that over a larger amount, such as infinity, the real numbers will still outnumber the integers. Some infinities are larger than others, so you can't just perform operations on infinity as you would on other numbers.
> 
> Infinity also doesn't equal 0. In Calculus we used infinity quite a bit and never was it considered the same thing as 0. We also couldn't perform operations on infinity as you are doing. Positive infinity is similar to negative infinity, but they are not one and the same. Take the limit of an expression as you get closer to infinity and negative infinity and you will not get the same answer. You also can't take a number line and start performing geometric calculations on it or give it a shape like a circle. It would make just as much sense to bend it into a triangle or quadrilateral and start finding the area and perimeter of it. Infinity is a theoretical concept, you can't just change it into a closed shape.
> 
> Hopefully your teacher lead you back on the right direction and away from this psuedo-math.


Well, we got the zero=infinity=negative infinity stuff from a book, which, as Stefan pointed out, ended up not being what the author had meant.

This was pre-calculus/trig. We showed our work the next day to a college math professor and he thought, like you, that zero didn't equal infinity nor negative infinity, but assuming this was true, the rest of the arguments were solid.

Also, the number line being a circle ended up not being some crazy idea of ours, but it is called the projective line.


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## Christopher Mowla (Sep 23, 2010)

@maggot,

Siraj A. has already mentioned in post #60 that he was wrong, okay?


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## blah (Sep 23, 2010)

Hiero said:


> I don't remember the whole proof but, for example, compare integers with real numbers. Over any given range of numbers, the real numbers will outnumber integers, so it makes sense that over a larger amount, such as infinity, the real numbers will still outnumber the integers. Some infinities are larger than others, so you can't just perform operations on infinity as you would on other numbers.





blah said:


> \( \aleph_0<\aleph_1 \)


This.

It's not a proof, but here's an intuitive explanation: You can count the natural numbers, or all the integers, even all the rational numbers. But you can't count all the real numbers. Thus, there are more reals than naturals/integers/rationals (you can even put all of them together and there would still be more reals than naturals, integers, and rationals combined).



Hiero said:


> You also can't take a number line and start performing geometric calculations on it or give it a shape like a circle. It would make just as much sense to bend it into a triangle or quadrilateral and start finding the area and perimeter of it. Infinity is a theoretical concept, you can't just change it into a closed shape.


Yes, you can give it a shape. While we're still at shapes, there is no topological difference between a circle, a triangle, and a quadrilateral. So yes, it would indeed make just as much sense to bend it into a triangle or a quadrilateral. This "bending" stuff is called the Alexandroff compactification of \( \mathbb{R} \): turning a non-compact space into a compact space by adding a point to it, which is possible in this case because \( \mathbb{R} \) is a locally compact Hausdorff space. That said, \( \mathbb{R}^* \) (the compactified version of \( \mathbb{R} \)) is not the same as \( \mathbb{R} \). As mentioned earlier, \( \mathbb{R}^*\setminus\mathbb{R} \) is only one point (conventionally labeled as \( \infty \)), which is why the Alexandroff compactification is sometimes known as the Alexandroff one-point compactification.

No, you can't start performing geometric calculations on it for obvious reasons.



AvidCuber said:


> We showed our work the next day to a college math professor and he *thought*, like you, that zero didn't equal infinity nor negative infinity


Thought?



AvidCuber said:


> but assuming this was true, the rest of the arguments were solid.


Here's a logically true statement for you: "Assuming this was true", bananas eat monkeys and humans sh*t on pigeons.

A true implication with a false hypothesis proves nothing about the conclusion.


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## riffz (Sep 23, 2010)

blah said:


> AvidCuber said:
> 
> 
> > This was pre-calculus/trig. We showed our work the next day to a *college* math professor and he thought, like you, that zero didn't equal infinity nor negative infinity, but assuming this was true, the rest of the arguments were solid.
> ...



Explanation bolded.


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## BigSams (Sep 23, 2010)

Would you guys stop bumping this thread and just let it die?! You've poked his eyes out, run him over with a truck, dumped a load on him. He gets it; the overall argument is wrong. He'll fully understand it if he takes analysis, topology, etc later on (not that I understood much of it ).


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## Siraj A. (Sep 23, 2010)

@maggot


cmowla said:


> @maggot,
> 
> Siraj A. has already mentioned in post #60 that he was wrong, okay?


---


maggot said:


> i was going to post more on how infinity is handled in many other types of situations and prove it cannot be handled as a real. however, most of this is to complex for your brain im sure.



You should really read the entire thread before posting something. I never actually _thought_ infinity = 0 in the first place >_> http://www.speedsolving.com/forum/s...ow-how-to-math&p=457108&viewfull=1#post457108


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## Christopher Mowla (Sep 23, 2010)

blah said:


> That said, \( \mathbb{R}^* \) (the compactified version of \( \mathbb{R} \)) is not the same as \( \mathbb{R} \)


@blah and qqwref,

I think this is why there is so much debate on this (the quote above). According to my understanding of this concept, the compactified version doesn't treat infinity like the real number system (which is why I think blah stated that fact above). I have just read somewhere that there are different interpretations of "infinity", depending on the context given (this could explain why you guys are speaking of concepts outside of basic math).

So why (blah and qqwref) are you guys referring to another type of infinity when Siraj A. mentioned basic calculus? You guys are correct for the "restriction" you are setting here, but how are these two infinities the same? What's the problem in saying that, in the real number system (which follows the axiom of completeness), 
\( \frac{1}{\infty }\ne 0 \)?

I am just trying to see why you guys are mentioning this, when it is not at all what most people are thinking of when they hear "infinity".

_Note: I don't want to fight. If either of you post a flame post, I will ignore you. I am simply asking a question in a calm and honest tone. Don't take it negatively, please?_ _It's just a question.

_EDIT:
And please don't ask me to read a page in wikipedia. Just give the the facts on that page you want me to understand and please explain them to me in English. You guys are the "experts", and I am the "student", so *teach* me. I am sure others who don't understand what you guys are talking about would appreciate a clear explanation as well.


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## Stefan (Sep 23, 2010)

BigSams said:


> You've poked *his *eyes out, run *him *over with a truck, dumped a load on *him*. *He *gets it; the overall argument is wrong. *He*'ll fully understand it if ...


 
Who?


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## blah (Sep 23, 2010)

But the real number system *doesn't* contain infinity. No one's talking about two different things. That one thing everyone's talking about doesn't even exist in the context that everyone's talking about in the first place.

Okay, so maybe I _was_ talking about "different infinities" (whatever that means) with the alephs, but that was all a joke and no one but Lucas responded to that, so I'm assuming that's not what you're talking about. Besides, I really know nothing about alephs.

I'm a student as well. If math is real, then I am epsilon - that's how much I know about math, to be really honest 

Edit: Where *did* you get the idea that we're talking about two different infinities? Quote? I don't think anyone mentioned anything about different "kinds" of infinities at any point.


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