...9999=-1

Steven G

Steven G

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...99999+1=...0000 so ...99999 = -1
What is wrong with this logic?
 
Steven G said:
...99999+1=...0000 so ...99999 = -1
What is wrong with this logic?
Click to expand...
It's a good example of the need to prove existence before you prove facts about things. Rules about addition apply only to numbers that are properly defined.
 
Steven G said:
...99999+1=...0000 so ...99999 = -1
What is wrong with this logic?
Click to expand...
Nothing is wrong. We know [imath] (1-x)\cdot (1+x+x^2+x^3+\ldots)=1 [/imath] and therefore [imath] \displaystyle{\sum_{k=0}^\infty q^k=\dfrac{1}{1-q}. }[/imath] Your question asks for
[math] ...999=9\cdot \sum_{k=0}^\infty 10^k =9\cdot \dfrac{1}{1-10}=-1.[/math] Everything is fine, except that the series does not converge for [imath] |q|\geq 1 [/imath] and we have [imath] q=10, [/imath] so the formula is not applicable.

The formula produces a singularity at [imath] q=1, [/imath] which in a way still makes sense since [imath] 1+1+1+\ldots =\infty [/imath] but not for [imath] |q|>1. [/imath]

Ramanujan's sum
[math] 1+2+3+4+5+\ldots = -\dfrac{1}{12} [/math]is a bit trickier but along similar lines.

Fancy things happen when dealing with infinity. The problem is that you need a definition for what you mean by [imath] \infty , [/imath] and your dots in front of the nines is an infinity. Not every definition makes sense.
 
fresh_42 said:
Everything is fine, except that the series does not converge
Click to expand...
The series \(\displaystyle ...999=9\cdot \sum_{k=0}^\infty 10^k\) \(\displaystyle \textcolor{red}{\bold{does \ in \ fact \ converge}}\)
And
\(\displaystyle \textcolor{green}{\bold{it \ exactly \ converges \ to}}\)

\(\displaystyle ...999=9\cdot \sum_{k=0}^\infty 10^k = \textcolor{blue}{\bold{-1}} \ \ \ \text{in} \ \mathbb{Q}_{10}\)

Forget the real number system for this problem. If we work in the \(\displaystyle b\)-\(\displaystyle \text{adic}\) number system, or more precisely in the \(\displaystyle 10\)-\(\displaystyle \text{adic}\) number system, we have:

\(\displaystyle v_{10}(9 \times 10^k) = k\)

and

\(\displaystyle |9 \times 10^k|_{10} = 10^{-v_{10}(9 \times 10^k)} = 10^{-k}\)

Since the norm of \(\displaystyle 9*10^k\) is \(\displaystyle 10^{-k}\) (Or more precisely \(\displaystyle |10|_{10} = 10^{-1} < 1\)) the geometric series converges in the \(\displaystyle 10\)-\(\displaystyle \text{adic}\) norm.

I know that this idea is beyond Steven's knowledge. But there's no harm that he leaves the corner for awhile and learns something new. It will take him a lot of time but I'm sure that eventually he will grasp it.

The OP question has shown everyone why Steven is ranked the worst in my List. And I don't understand how he is a math teacher!

🙄🤔 1756498553915.png1756498565231.png
 
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There is no [imath] \mathbb{Q}_{10}. [/imath]

[imath] p [/imath]-adic requires a prime, [imath] b [/imath]-adic is just notational nonsense.
 
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fresh_42 said:
There is no [imath] \mathbb{Q}_{10}. [/imath]
Click to expand...
Of course. This ambiguous notation is a useful shorthand for the set of rational numbers completed with respect to the \(\displaystyle 10\)-adic norm, but you are right as formally, this object doesn't exist as a field because \(\displaystyle 10\) is not a prime.
 
[imath]...999 = 9 \cdot 10^0 + 9 \cdot 10^1 + 9 \cdot 10^2 + \dots = \infty [/imath]

[imath]...999 + 1 = 1 + 9 + 9 \cdot 10^1 + \dots[/imath]

[imath]= 100 + 9 \cdot10^2 + 9 \cdot 10^3 + \dots = \infty [/imath]
 
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