Well, infinity isn't "close" to numbers since most mathematical statements you're used to are in the form of limits. What we're talking about goes beyond numbers into set theory. A set's size can be infinite, not just a number that approaches infinity (like a limit). I'm not really sure where the relation to number is here.

If you add X+A, you get a real number which is not in the set of integers. I'm not really sure why you're adding them together. Our goal was to create a bijection between the the two sets and we showed the act of trying to do that creates a contradiction. Then by our definition, the reals is a much larger set than the integers. However, our definition can actually be applied to find greater or less than for any set, finite or infinite, so it's a much better definition.

x=0

observe:
1x=x (this is still valid)

now lets use some algebra and divide by X on both sides

1=(x/x)

x/x = 1? where x is zero? i think not. the fact is, x divided by itself is equal to all real numbers. "infinity" or the idea of everything can be represented as 0/0


There are more infinities than one.

There are an infinite amount of numbers that are between one and zero, an infinity inside of one so to speak.
There is an infinite amount of irrational numbers (numbers that cannot be generated by dividing two integers)

lrn2divide
You cannot do that if x = 0.

I find these discussions rather boring. Infinity contradicts itself, which places it not only beyond our minds but beyond mathematics. It's damn annoying.

Your inability to understand a topic does not mean you can just talk out of your ass. Try again k thnx.

Infinity is not self-contradictory... it only contradicts the idea that "infinity" is a quantity. Infinity is used in limits, and is essential to understanding many different parts of calculus.

Here is Taylor's Expression:

e^x=(1+x/1!)+(x^2/2!)+(x^3/3!) . . . (Cont'd),-∞<x<∞

Notice it uses positive and negative infinity to describe that X can be all real numbers, positive or negative using mathematical terms.

"lrn2divide
You cannot do that if x = 0."

You most likely say that i cannot because you dont understand the idea of 0 divided by itself, that is okay... i just recommend that you explore the concept.
(the calculator does not accept it, because it is programmed not to accept divisions by 0. Heres a thought experiment: If you have 0 cookies shared amongst 0 people... how many cookies get shared?)

Midget man. Let me go through what you originally said:

Ok you've defined x to be 0.

Right, this is valid. 1x=x. This is also equal to 1(0) = 0 since x = 0 as you've defined.

No. You cannot do this if you've already defined x to be 0.

In short, you're hiding the divide by zero error with a variable that is already zero.

It's not programmed in. It's the definition of division. Rather than asking 0 cookies, how about 5 cookies. You have 5 cookies divided amongst 0 people. How many cookies are shared? From here, we can immediately see that the question itself is flawed. Your wording is an attempt to place emphasis on the 0 cookies; since we are so programmed to think zero multiplied by any value is zero, we would assume the answer is zero.

One final tip, you probably shouldn't tell clokr he doesn't understand math since he's a math major.

I thought he was a biology major.. But whatever, Clokr_ and Cheeze are right.

View below, your own shit:


If x=0
then 1x=x == 0
then 1 != (x/x) which is in turn (0/0)

Zero means zilch. Nada. Null.

You cannot divide something that is not already there.

Therefore, 1=(0/0) != True

Thx, bubye.

To be most precise, when you divide by x on both sides, what you actually have is 1(x/x) = (x/x). This is true. You may then replace (x/x) with 1 only if you assert that x ≠ 0.

Graph the function [1/x=y] in the window
Xmin=-1
Xmax=1
Ymin=-20
Ymax=20

Here's where it comes to understanding infinity:
The function completely dodges the value 0, the smaller the number for x values the greater the absolute value of the Y. In order to get a VERY small number you need a VERY large number under the division sign. No matter what X equals, Y cannot equal zero. You can cut up a cookie as many times as you want, but you will still have crumbs.

"x/x = 1? where x is zero? i think not." - i know basic algebra people, i was just going beyond it. i was trying to get a concept out, notice my dissonance at the end.

>.>

Never, and I mean never, confuse the concept of infinity in limits (something that never stops growing) with the concept of infinite sets (infinity as a quantity). They are by no way related and in fact an infinite set is not represented by the simbol ∞, but by the symbol aleph:
http://en.wikipedia.org/wiki/Aleph_number

Again here you're refering to the limits infinite. The expresion -∞<x<∞ is just notation, and will only be well-defined if you define the operator < for the extended real numbers: http://en.wikipedia.org/wiki/Extended_real_number
However extended real numbers don't behave as well as the standard real numbers since indeterminations cannot be operated.

"lrn2divide
You cannot do that if x = 0."


I recomend you to explore the concept:
6/3 = 2 because 2 * 3 = 6
21/7 = 3 because 3 * 7 = 21
But 0/0 = 0 because 0 * 0 = 0
and 0/0 = 3 because 0 * 3 = 0
and 0/0 = a because 0 * a = 0
It's not well-defined at all. The result of an operation cannot be "any real number works". Even if you define a to be not a real number, but a new number which behaves differently (similar to how the imaginary unit was created) you'll get tons of contradictions when you work with that new number. So sorry but 0/0 is not defined.


That's not true. The assumption that x ≠ 0 has to be stated BEFORE dividing by x. That's the point of most of the 1 = 2 proofs around internet. Get in mind that x is not variable, it is a constant since you have not defined any function, and thus it is just a number. 1x = x thus 1(x/x) = (x/x) is not a valid step if x = 0.

The function does not dodge the value 0. It has a hole right there. It is small yeah, smaller than anything you will ever be able to plot, but the hole is there, and 0/0 is not defined. Almost0/almost0 however is, and the result sure is 1, but as stated above if you try to assert that 0/0 = 1 it'll result in lots of contradictions, including 1 = 2.

In fact, here is that proof:

gl hf

1-] a = b
2-] ab = b²
3-] -(ab) = -(b²)
4-] a²-ab = a²-b²
5-] a(a-b) = (a+b)(a-b)
6-] a = a+b
7-] a = 2a
8-] 1 = 2

From line 5 to line six is the big scandal, correct? there is a division of zero by zero. a=b, and they are dividing by (a-b).. right? i suppose then, you are correct. 0 dividing anything is just silly.

thanks clokr_ i have gained knowledge in this process

Well true, but these inequality proofs usually are meant to read "assume x is variable, and then plug in zero at the end," so the formal contradiction is when trying to say x = 0 contradicts the assertion that made the division legal. It's sort of silly to use it as a constant for such a small proof. Btw, my favorite inequality proof is the one that works by saying that since (5 - 9/2)² = (4 - 9/2)², then 5 - 9/2 = 4 - 9/2.

You know you should've just understood that just by looking at the sentences above.
"There are 5 cookies to be divided upon 0 people." So where are the people? There was simple logic in just simply not defining x/0. It's not knowledge. It's pure logic.

Lets say x is bigger number than 0, and t is smaller number than 0. Then thats why when something is x² = t² then you go x = |t|.

My logic was distorted, so in the end, i was left with realization. knowledge was gained, the knowledge that 0/0 is irrelevant.

Here is a true statement:

(5-9/2)²=(4-(9/2))²
When we square root, we musn't forget ±

so, ±√( (5-(9/2)² )=∓√( (4−(92))² )

Bedazed, i like seeing things head on. that "proof" just made more sense.

Post has been edited 2 time(s), last time on Jan 14 2009, 5:02 pm by midget_man_66.

If you had x = 0
1x=x,
divide both sides by x, (supposing we are allowed to divide by 0 and it will act like a real number)
1(x/x) = (x/x)
1*1 = 1
1 = 1,
How does that say anything about infinity? 1x=x is just x=x.

I'm just curious, where was your thought process going with that? Sorry if you guys think this was exhausted already.

No you? It's accepted that all infinities are equal. Yet one set of infinite length with all integers that are divisible by four is shorter then the set of all integers for pretty much any segment (with exception to for example 1-7, which only includes 4 as a member of the other set, but the concept of unequal lengths is supported). The same basic concept is shown by comparing points along a horizontal and diagonal line.

Your inability to understand a topic does not mean you can blabber mindlessly in it, jackass. Infinity as a length is contradictory.

Oh, and I'm with Vrael in complete incomprehension of what Midgetman was trying to show there.

Wrong. Go read up on countably infinite (set of integers) and uncountably infinite (set of reals). A brief argument for the existence of a larger infinite set was presented above by both me and clokr.

Did you just contradict yourself? ... Yes. Yes you did.

That's what talking out of your ass means. Which is what you shouldn't be doing. If you want to learn, I'd be more than happy to explain my arguments; but until then, once again, don't talk out of your ass.

Fine. Is that all? It doesn't break my arguments, since both integers and multiples of four are subsets of natural numbers.

When one's proving contradiction, yes, one usually has to mention a contradiction...

When you actually show me wrong, your opinion will be worth something. Addressing small points while blabbering about me being wrong in the big picture doesn't impress.

Post has been edited 1 time(s), last time on Jan 15 2009, 12:35 am by Centreri.

Fine, let me prove that the set of integers and the set of integers that are multiples of four is the same size.

Proof:
Define Z to be the set of integers.
Define F to be the set of integers that are multiples of four.

I define a bijection between Z and F:
f(x) = 4x.
This is a bijection because it is both surjective and injective:
Surjective:
y = 4x. Hence, x = y/4. Thus f^-1(y) = x exists.
Therefore, f(x) is surjective.

Injective:
Define a and b to be in domain. Let f(a) = f(b). Then we see that 4a=4b. Thus, a = b.
Therefore, f(x) is injective.

Since f(x) is both surjective and injective, it is bijective. Since I have found a bijective function that relates the set of integers to the set of integers of multiples of four, the two sets are equal in size.
Therefore, |Z| = |F|. (|Z| is the size of Z and |F| is the size of F)