Magnus! Honestly! sigh
This is what I was trying to explain to you before.
If you halve 1, it’s 1/2 + 1/2
If you halve THAT!
It’s 1/4+1/4+1/4+1/4. Etc…
If you keeps doing this:
At infinity (convergence)
EVERY!!! Whole number solves as zero.
I still think you are confusing all your arguments. You switch back and forth when it suits you without understanding (or caring) about the implications.
Well, you didn’t say much. What you said is basically:
(1 = \frac{1}{2} + \frac{1}{2} = 2 \times \frac{1}{2})
(1 = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = 4 \times \frac{1}{4})
(\dotso)
(1 = 0 + 0 + 0 + \cdots = \infty \times 0)
How can that be the case when zero times anything is zero?
Properly speaking, (1 = \frac{1}{\infty} + \frac{1}{\infty} + \frac{1}{\infty} + \cdots = \infty \times \frac{1}{\infty}).
I said a lot! I said that it’s impossible for convergence theory to be true (standard calculus and other mathematics)
Ultimately it equals (according to convergence theory)
0+0+0+0 (etc…) at infinity! (Convergence)
That means that every whole number equals zero. (Contradiction)
How’s that relevant to anything of what I’m saying?
How’s what you said relevant to what I said?!?!
1/infinity is a meaningless mathematical phrase.
You quoted me. I guess it’s somehow relevant to what I said in that quote.
(\frac{1}{\infty}) is a number greater than (0) but smaller than every number of the form (\frac{1}{n}, n \in N). That’s what it means (so it’s not really meaningless.)
(0 < \frac{1}{\infty} < \cdots < \frac{1}{4} < \frac{1}{3} < \frac{1}{2} < \frac{1}{1})
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!
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.