Is 1 = 0.999... ? Really?

1/infinity is not a number. It’s hypothetically, a procedure. It is most definitely not a number.

It’s not a procedure.

(\infty) represents a number greater than every number of the form (n, n \in N).

Similarly, (\frac{1}{\infty}) represents a number greater than (0) but less than every number of the form (\frac{1}{n}, n \in N).

(0 < \frac{1}{\infty} < \cdots < \frac{1}{3} < \frac{1}{2} < \frac{1}{1} < 2 < 3 < \cdots < \infty)

Also:
(0.\dot9 + 0.\dot01 = 1)

Note that (0.\dot01) or (\frac{1}{10^\infty}) is actually smaller than (\frac{1}{\infty}).

Ok, fine. I don’t buy this, but let’s say that what you’re saying is absolutely true.

Per the argument I leveled. That means every whole number is EXACTLY equal to the lowest possible “number” (your argument, not mine) that’s not zero.

My argument still stands. It’s absurd.

How did you arrive at the conclusion that every whole number is exactly equal to the lowest number?

That’s not even true for (\frac{1}{\infty}) let alone for whole numbers.

Really Magnus ?!

I’ll use your own post for it!

viewtopic.php?p=2758485#p2758485

You’re right. 1=0 is a constradiction.

My argument proves that when numbers converge at infinity (and in saying this, infinity is NOT A NUMBER!)

That 1=0.

Thus, infinities do not converge.

All you did was change infinity to “lowest possible ‘number’ that’s not equal to zero, which by my argument, makes every whole number equal to “the lowest possible number not equal to zero” which is still a contradiction.

That means that 1=2!! Contradiction !

What argument, Ecmandu? Where is it?

You replied to the post yourself ! Honestly! This is getting absurd!

viewtopic.php?p=2758484#p2758484

What exactly does that prove?

Just what I said it does. Infinite series don’t converge.

To answer your question.

This started out with you asking:

“You’d have to explain why you’re limiting yourself to integers.”

To which I said: “Try it without integers. It doesn’t work.”

I was showing you it doesn’t work. Of course, what you really meant was the non-integer (\infty).

I don’t get the distinction between these two cases. Sounds like the exact same case just worded differently. In the one case you’re saying x can equal (\infty), in the other that it can’t.

This logic:

(x = 0.\dot9)
(10x = 9.\dot9)
(10x = 9 + 0.\dot9)
(10x = 9 + x)
(9x = 9)
(x = 1)

A sum cannot stop after an infinite number of terms because if it could it would be finite so the concept of completed infinity is entirely fallacious
And so your first sentence and third sentence contradict each other because if (1) increases by (1) and does not end then logically the sum cannot stop

I0x = 9.999…
I0x = 9 + .999…
I0x = 9 + x
I0x = 9 + I
I0x = I0
x = I

Not really.

I addressed this “proof” around 20 pages ago and I can restate what’s wrong with it but I think it’s pointless since you don’t agree that we can do arithmetic with infinite quantities.

Basically, you don’t agree that adding a green apple to an infinite line of red apples increases the number of apples in the line. Instead, you prefer to contradict yourself by saying that the number of apples remains the same.

It has been claimed that it’s a contradiction in terms to say that an infinite sequence has a beginning and an end.

An infinite sequence has no end, so you cannot say that it has an end.

Well, you actually can, provided that the first occurrence of the word “end” and the second occurrence of the word “end” mean two different things (i.e. provided that they refer to two different ends.)

Don’t be fooled by homonyms.

(S = (e_1, e_2, e_3, \dotso, e_L)) is one such sequence. It’s an infinite sequence with a beginning and an end.

Note that a sequence with no repetitions is no more than a relation between the set of positions and the set of elements.

(S = (e_1, e_2, e_3, \dotso, e_L)) is a relation between the set of positions (P = {1, 2, 3, \dotso, \infty}) and the set of elements (E = {e_1, e_2, e_3, \dotso, e_L}).

Note that sets have no order. This means that, when visually representing a set, you can place its elements anywhere you want. This means that (P = {1, 2, 3, \dotso, \infty} = {\infty, 1, 2, 3, \dotso}). The same applies to (E). By moving the last element of the set to the beginning of the set, there are no longer any elements after the ellipsis, so there is less to complain about (:

Let’s represent the sequence as a set of pairs ((\text{position}, \text{element})). (S = (e_1, e_2, e_3, \dotso, e_L) = {(\infty, e_L), (1, e_1), (2, e_2), (3, e_3), \dotso}). And voila! There is nothing beyond the ellipsis anymore, so absolutely nothing to complain about (((:

“The last position in the sequence” refers to the largest number in the set of positions (P). Either there is such a number or there is not. In the case of our sequence, there is such a number and it is (\infty).

This is not the same as “The number of elements in the set of positions (P)”. This is an entirely different thing. In the case of our sequence, the number of positions is infinite (i.e. there is no end to the number of elements.) It’s also not the same as “The last element in the set of positions (P)”. No such thing exists, not because the set is infinite, but because sets have no order.

Aw, what a sad way to go out. We were so close to a break through. I don’t know why you didn’t want to get into hyperreals. I think that’s where you had your best shot and where I think you might have had a point. But I guess frustration got the better of you. Sayonara chico.

Talking about hyperreals is both unnecessary and pointless. How can you accept hyperreals if “you don’t agree that adding a green apple to an infinite line of red apples increases the number of apples in the line”?

I don’t accept hyperreals. But I’m willing to entertain them conditionally. Under the condition that you can have numbers greater than infinity (or numbers that are infinitely small), then hyperreals become not only a possibility but a necessity. We could then go on to debate the logic of hyperreals, argue about what can and can’t be said about them.

You need to address this argument:

Infinites can be measured?