# Argument from Motion

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Once again, you are making assertions. You assert that an intellect created time and space. Now prove it. You assert that time is not eternal. And that is another assertion. Please prove these assertions or true, or troll away

The [ Math Processing Error ] is gone now....

Yes, something like that...

"I + 2 = I". I'm guessing your goal is to manipulate this equation into an explicit contraction, so if you aren't using the idea I suggested above (where I reversed your substitution, then manipulated the equation), could you please explicitly post a different manipulation?

/e The problem with the manipulation I posted above is that it is incorrect. I sneakily inserted a mistake in the simplification of the right hand side between lines 2 and 3. Infinity - infinity is indeterminate (you can't safely say it equals 0).

We start from the idea:

0) I + 2 = I

Your manipulation was:

a) 2 = I - I

b) 2=0

If 2=0 is a logical contradiction, then by modus tollens, I+2=I is also a contradiction, and is false.

Though my idea is that I + 2 = I is directly a contradiction, and hence it needs not to be manipulated to another form...

Just stating the axiom: [ A+B is never equal to A unless if B = 0 ] would be sufficient...

What are your thoughts on this? Can I + 2 = I? Or that ( I + 2 ) is greater than I ?

It is.

NO! I just told you that I reached 2=0 from I+2=I|I = ∞ by intentionally making a mistake! Therefore the truth value of 2=0 infers NOTHING about the truth value of I+2=I.

I + 2 = I | I = ∞, doesn't seem inconsistent, it doesn't even violate your axiom. If you think it is inconsistent you are going to need to demonstrate it explicitly (which I'd bet can't be done, but we'll see).

skd6348:

Again, you are confusing infinity with a number. There is no contradiction because infinity doesn't obey all the rules for numbers.

Ummm... What was the mistake? Both steps in your manipulation are valid for all numbers.

The axiom is, and I quote again:

[ A+B is never equal to A unless if B = 0 ]

In the specific case: I + 2 = I, A represents I, and B represents 2.

In other words, I + 2 = I is the same as saying:

[ I+2 IS equal to I where where 2 != 0 ]

A contradiction to the axiom.

No, I already explained this:

skd6348:

In case you missed the other two posts, infinity is not a number and does not obey all the rules for numbers.

I-I is not 0 (or any other number). As Nyarlathotep said, it is indeterminate.

skd6348:

Addition and subtraction are defined for numbers, not for infinity.

You have to understand when it comes to operations with infinity...

The reason I-I is considered indeterminate by some is because Infinity is not considered to be actual...

And it's not just I - I.

Even " I + 2 " is considered indeterminate.

Why? Because THEY LEAD TO CONTRADICTIONS.

What I am showing you is that, IF infinity is treated with the "same axioms" as all numbers are, it leads to a contradiction.

A-A = 0. This is just an implication of the Law of Identity.

So I-I=0. Why wouldn't it be? How can I - I = 3 or anything other than 0? You're subtracting the exact same thing from itself...

Unless one is committing the fallacy of equivocation, and the two "I"s in I - I are not the same, I - I = 0.

In other words, the reason I - I is considered indeterminate because it leads to situations like 2 = 0...

So saying I + 2 = I is just as much as a mistake as saying I - I = 0.

In fact, it is much worse...

At least I - I = 0 does not violate any well defined mathematical axioms...

I + 2 = I DOES.

Great another arm chair guy trying to tell us that infinities don't exist, despite the fact that we use them every day... They are coming out of the woodwork now!

Why not just jump to: negative numbers don't exists because I've never see a negative number of apples?

skd6348:

Infinity is a limit concept, not a number. If you insist on treating it as a number you will certainly get the contradictions that you got. As a limit concept, it is certainly as "real" as numbers are, except that it isn't a number.

skd6348:

"What I am showing you is that, IF infinity is treated with the "same axioms" as all numbers are, it leads to a contradiction."

Mathematicians already know that. What's the point?

skd6348:

Consider: I - I =0

Consider: (x times x) - x = 0 (letting x-->infinity gives the above)

Consider that: (x times x) - x = x(x-1)

Now if we let x-->infinity we wind up with: (I times I) instead of 0.

Therefore: I-I = I squared!

Conclusion: Infinity doesn't obey the arithmetic operations for numbers.

You might be confusing potential infinities with actual infinities...

Potential infinities are just conceptual....

Examples:

1) recurring numbers, like 0.333...

2) Transcendental numbers, like pi and Euler's constant.

3) Various conceptual sets, like the set of natural numbers, set of multiples of 5, ....

4) Summation of various series to infinity converge to certain finite values...

These entities are different then actual infinities...

WHAT actual infinities do you use in everyday life?

Has a super computer ever performed " An infinite number of operations? ".

skd6348:

What's this business about "actual infinities"? I thought that I had already point out (and you already acknowledged) that in an "infinite" past there is no actual moment of time that isn't a finite number of years away. Therefore, there seems to be no danger of having to cross an "actual infinity."

I apologize for not noticing your replies earlier...

You are saying that infinity isn't a number, and therefore the axiom:

[ A+B != A unless B=0 ]

Does not apply to it.

Your way of viewing time as endless is interesting. It is true that if we take a certain moment from the past, the duration to the present would be finite.

But I'm not referring to a length of moment between the present and a specific moment in the past, but the past as a whole considered as a set.

If the present time is t, then set P:

P = { x: t(x) is less than t }

where x refers to an event in time, and t(x) represents the corresponding value of time.

My question is, what happens to the cardinality of P as we add more moments to P while moving ahead in future?

x, t ∈ Z

A = P(t=2) = {..., -1, 0, 1}

B = P(t=3) = {..., 0, 1, 2}

Finding a bijective function from A to B is easy: f(n)= n + 1; so they have the same cardinality (the cardinality does not change). And clearly we can find it for any other value of t with something like f(n) = n + Δt

But a bijective function won't change the fact that:

B-A = {2}

So how can their cardinality be the same? I'm aware of some of the conventions used when dealing with infinity, but I would be interested in a rational behind them.

When we express sets like the multiples of 2 ( N-2) and 6 ( N-6):

N-2 = { 2,4,6,8,10,12,..}

N-6 = { 6,12,.. }

It is clear that f(n) = 3n; is a function that can be used to get any element of N-6 from N-2, but still N-2 does have more members than N-6. [ It has all members of N-6 as well, making N-6 a proper subset of N-2 ]

These sets represent conceptual iterators. P is related to an actual sequence(s) of events that have occurred.

No, that is a mistake. There exists a 1:1 mapping from every element in N2 to a unique element in N6. There also exists a mapping from every element in N6 to a unique element in N2. This demands that they have the same exact number of elements.

[handwaving]Perhaps it would help to think about it this way: You have a small child, and she isn't very good at counting yet. In fact the only numbers she understands is 0 and 1, and even this is only a visceral understanding; she knows the very real difference between having 0 and having 1 candy bar. Now we present her with two stacks of candy bars (A and B) and ask her which stack has more candy bars.

It seems like a hopeless task but she comes up with an idea: She takes one candy bar from A and one from B then puts them in a two new stacks C and D. She then repeats this process, each time matching one candy bar from A and B and adding them to the corresponding new stacks (C and D). If at any time she is unable to find a match from B for the candy bar she removed from A, she knows that stack A + C has more candy bars. If at any time she is unable to find a match from A for the candy bar she removed from B, she knows that stack B + D has more candy bars. If she successfully matches all the candy bars she knows both C and D have the same amount of candy bars. It is interesting that this primitive yet successful strategy doesn't actually require her to count either stack, and that is a good thing because she doesn't know how to count!

This is essentially what we are doing with a bijection, because we don't want to actually try to count the elements in those infinite sets. Since we have exactly matched every element in N2 with a unique element in N6 they must have the same number of elements, despite the fact that your common sense might trick you into guessing otherwise.[/handwaving]

You said:

"No, that is a mistake. There exists a 1:1 mapping from every element in N2 to a unique element in N6. There also exists a mapping from every element in N6 to a unique element in N2. This demands that they have the same exact number of elements. "

N-2 - N-6 = {2,4,8,10,14,16,20,22,...}

This isn't an empty set. One could go the other way and say that since (N-6) - (N-2) != {} , this demands that they must not have the same number of elements. What I mean is that all members of N-6 are clearly in N-2, and N-2 still has more members.

Relatively, for every one member in N-6, N-2 has 3 members.

You said:

" This is essentially what we are doing with a bijection, because we don't want to actually try to count the elements in those infinite sets. Since we have exactly matched every element in N2 with a unique element in N6 they must have the same number of elements, despite the fact that your common sense might trick you into guessing otherwise."

But one can put your example another way.

Suppose I start removing the members from N-2 and N-6.

After 100 removals,

N-2 ( modified ) = { 202,.. }

N-6 ( modified ) = { 606,... }

Once could argue that N-2 is more dense, so it has more members. [ Also note about N-2 - N-6 ]

---------------------------------------------

OR suppose the girl is told that no two candy bars in the same box have the same color for its wrapper.

Now the girl removes all the candy bars from A, and removes a bar from B with the same color if it exists.

She finds that all the colors in A are also in B.

She also finds that once she has removed all bars from A, and their corresponding ones from B, B still has candies left. Would this mean B has more candy bars?

Remember you asked about cardinality, the question is which pile has more bars. It doesn't matter if they are different colors, or even if the bars aren't the same size. If pile A has 10 tiny red bars, and pile B has 10 huge bars each of a different color, they have the same cardinality. If every element in A can be paired with a unique element in B (and vice-versa) they MUST have the same number of elements and therefore the same cardinality. By adding coloir you are complicating something you already don't understand. But I will address it any way:

No, because we can not ensure a 1:1 mapping because of the bolded part. A trivial counter-example: A = {red}, B= {blue}. She remove the red from A and since there is no red in B she does nothing to B. Now she finds 1 element remaining in B. So then falsely concludes that B had more elements than A (originally). Clearly this method is flawed.

[handwaving]In my original example, every-time she removed one from one pile, she removed one from the other, this is what allowed her to get away with saying which pile has more bars without actually counting. In your suggestion sometimes she removes them in pairs and sometimes she does not. If you do it that way you will have to count how many times you didn't remove a pair if you want to get the correct answer. And remember she doesn't know how to count (we don't want to have to count infinite stuff).[/handwaving]

Again, the question you asked was about cardinality, not density. As a side note you might notice that comparing densities of objects does not indicate which object has more components, so this dog don't hunt.

You said:

" Remember you asked about cardinality, the question is which pile has more bars. It doesn't matter if they are different colors, or even if the bars aren't the same size. If pile A has 10 tiny red bars, and pile B has 10 huge bars each of a different color, they have the same cardinality. If every element in A can be paired with a unique element in B (and vice-versa) they MUST have the same number of elements and therefore the same cardinality. By adding coloir you are complicating something you already don't understand. But I will address it any way: "

The reason it had different colors was just to make it easier for the girl to match members between A and B.

You said :

"//skd6348 - no two candy bars in the same box have the same color...Now the girl removes all the candy bars from A, and removes a bar from B with the same color if it exists....She also finds that once she has removed all bars from A, and their corresponding ones from B, B still has candies left. Would this mean B has more candy bars?//

No, because we can not ensure a 1:1 mapping because of the bolded part. A trivial counter-example: A = {red}, B= {blue}. She remove the red from A and since there is no red in B she does nothing to B. Now she finds 1 element remaining in B. So then falsely concludes that B had more elements than A (originally). Clearly this method is flawed."

-------------------------------

OK.......

Why are you cutting out this important part from my words:

" She finds that all the colors in A are also in B. "

Really? It's true that without this, the method is flawed... but you seem to quote my entire example except this. It was already verified by the girl that all colors in A were in B too.

I make it real simple.

The girl removes a red, green, blue and yellow bar from A. She also removes one of each of these from B as well. If there are still bars left in B, but no bars in A, B has more bars than A.

Likewise, if:

(N-2) - (N-6) != {},

AND...

All members of N-6 are in N-2.

Then wouldn't N-2 have more members?

Oh sorry I missed that part. Yes then she has in fact removed them in pairs, so if there are any left in B then it has more elements.

No problem. :)

Again you are concentrating on the value of the elements and not the number of elements.

This is not a concentration of values. It is that N-2 has all the members of N-6, and yet more members.

It sure seems that way don't it! But it's wrong! Isn't that interesting? I think its fascinating, personally.

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