(∞)+1 is larger than (∞) Some infinities are larger than others.
there are infinite numbers between 1 and 1.1. There are infinite numbers between 1.1 and 1.2 but the sum of the first set is lower than the sum of the second set.
edit: reworded to make more sense.

If you take
1 + S(∀x) (where 1 > x > 0)=
1 + S(∀x) (where 2 > x > 1)
You are comparing 2 infinites both of which have 1 added to them and of which the second is larger and find you have a false statement.

As stated, I was talking about a sequential (x > 0) with no upper limit, so that doesn't apply. I don't know enough about the actual names to say the proper writing for it.

Your argument requires a specific value of infinity (x>0) but it is still incorrect in that form.
You are saying that
S(∀x) (where x >0) =
1+ S(∀x) (where x >0)
If we show S(∀x) as "m" we get
m = m+1 which is obviously false.

I don't think you understand infinity very well. You can't have a number larger than a sequential infinity simply because that number would still be a result of that infinity. These infinity is simply infinitely big and therefore cannot have a larger number because it is defined by being the largest "number".

It is not possible mathematically to have a largest number. The quote "there is always a bigger fish" is a basic mathematic law. You don't seem to understand how mathematical proofs work. My argument proves your's wrong by contradiction. Unless you can show mathematical proof of your argument just saying "its infinity so its the biggest" your argument has no ground to stand on.

infinity by it's very nature can't be counted. Infinity is a really weird concept that produces ridiculous theoretical paradoxes when you toy with it. It's like a reverse singularity.

You were arguing that adding 1 to
S(∀x) (x>0) does nothing. I used the most basic mathematical proof format to prove that wrong and then you change what you are saying you were arguing. If anything is a sign of not having a leg to stand on that most definitely is.

If we’re taking one as a constant and not a higher order of infinity. Then his reply makes no difference. But since we don’t know what f(x) is we can’t say for certain what the limit is approaching.

bethorien· 9 weeks ago · FIRSTthere are infinite numbers between 1 and 1.1. There are infinite numbers between 1.1 and 1.2 but the sum of the first set is lower than the sum of the second set.

edit: reworded to make more sense.

timebender25· 9 weeks agoAleph one is a larger infinity.

spiderwoman· 9 weeks agoYou can't even add numbers to infinity. It's like "FISH + 1". Infinity isn't a number.

timebender25· 9 weeks agobethorien· 9 weeks ago1 + S(∀x) (where 1 > x > 0)=

1 + S(∀x) (where 2 > x > 1)

You are comparing 2 infinites both of which have 1 added to them and of which the second is larger and find you have a false statement.

spiderwoman· 9 weeks agobethorien· 9 weeks agoYou are saying that

S(∀x) (where x >0) =

1+ S(∀x) (where x >0)

If we show S(∀x) as "m" we get

m = m+1 which is obviously false.

spiderwoman· 9 weeks agobethorien· 9 weeks agoimrichbitch· 9 weeks agospiderwoman· 9 weeks agocryoenthusiast· 9 weeks agofunkmasterrex· 9 weeks agocryoenthusiast· 9 weeks agobethorien· 9 weeks agoS(∀x) (x>0) does nothing. I used the most basic mathematical proof format to prove that wrong and then you change what you are saying you were arguing. If anything is a sign of not having a leg to stand on that most definitely is.

cryoenthusiast· 9 weeks ago