u/Ill_Contract_5878

1/10^(n) is the limit we are evaluating. 1/10^(n) = 0. Define n —> ∞ and the limit is equivalent to zero. In my opinion, ”but not for any integer/finite n” is overstated and it’s unlikely to reach someone who does possess the misconception (although on this subreddit, there’s the people who deny the limit altogether). Clarifying something should be integrated into a response gracefully and you should address the rest of the content of what you’re responding to instead of only stating a clarification (i.e avoid responding with a clarification only to nitpick or change the subject). I’ll conclude by saying that you shouldn’t start a dispute which isn’t original and reflective of value.

Consider x as the number of trials performed, with a limit of x —> ∞. Each trial involves flipping a coin, and the question posed is the probability of landing at least one heads in x trials. We can construct this limit as 1 - (1/2)^(n)- 0.5, 0.75, 0.875, 0.9375. At n = 10 for instance, the solution is 0.9990234375.

Thinking with standard probability theory, only real numbers can be probability values. 1 - (1/2)^(n) as constructed has increasingly numerous nines as n —> ∞. 0.999…7 for instance is not a real number and it would not matter if it was standardized in a different system either because we need a real number for a probability value. The limit of 1 - (1/2)^(n) is equivalent to 1- any proposed value in the manner of 0.999…7 isn’t a real number (nor well-defined automatically) and would imply a terminating decimal such as 0.9997. If 1 - (1/2)^(n) = 0.999… (an infinite decimal expansion of nines), and 1 - (1/2)^(n) = 1, then 0.999… = 1.

A case to the contrary would necessitate:

1. Rigorously demonstrating 1 - (1/2)^(n) ≠ 1. 2. Demonstrating 0.999…7 (The last digits can be any values you please) is well-defined and a real number. 3. Demonstrating 0.999…7 ≠ 0.999… 4. Demonstrating all other possible proofs in the real numbers that 0.999… = 1 are false (without redefining what the real numbers are)

The probability of meeting those standards is vanishingly small.

What values are greater than 0.(9) but less than 1? There aren’t any consistent answers to that, so it follows that 0.(9) = 1 as real numbers which are distinct from each other have infinitely many other real numbers between them.

SPP claims to agree with the definition of 0.(9) as 0.9 + 0.09 + 0.009…, although he calls limits “snake oil.” Someone else’s inconsistent definition that appeared here is that: 1 > 0.999… > x for any x < 1 and ≠ 0.(9). Immediately, this definition defeats itself by leading to 0.(9) = 1 (because no real numbers exist between them here). In the intended spirit, if we hold it to be true and acknowledge some contradictions that arise in the real numbers, we could address another question. Under this framework, define n = (0.(9) + 1) / 2. To understand this n under the framework, consider: The decimal expansion of n, if n > 0.(9), and if n < 1. If you haven’t noticed, n would be “0.999…5”. Let’s move on and allow that framework some space.

Despite the contradictions that result, SPP has asserted infinitely many values exist between his 0.(9) and 1. Past examples include 0.999…1, 0.999…5, and even 0.999…9. A value cannot both terminate and not terminate like this, thus meaning these values terminate at some point and discrediting that SPP even knows what 0.(9) is. He abuses these notations as well by asserting they exist independently in some discussions but then conflating them with 0.(9) in other discussions (notably, “0.999…9 = 0.999…”). SPP also has to contend with values approaching 1 from the right (that is they’re > 1 in SPP’s system but whose limit should equal 1). In the past, he’s asserted 1 = 0.999… + 0.000…1 (that is ill-defined itself), which raises the question of how to handle x where x < 1 and x > 0.(9) (for instance the notation 0.999…9) in the case where x + y where y approaches zero from the right. When values for x aren’t conflated with 0.(9), we could resolve 0.999…1 + 0.000…1 within the framework of Real Deal Math. 0.999…1 + 0.000…1 = 1.000…1, which is contradictory as either 1.000…1 should terminate (such as with 1.0001) or equal 1 (due to no point where the value terminates with a final digit of 1). SPP’s 0.(9) isn’t 0.(9) in any case because his 0.(9) isn’t subject to any proof that 0.(9) = 1, implying that 0.(9) terminates at some point (such as 0.999), invalidating his 0.(9)‘s utility as a limit (His Real Deal Math isn’t very useful by rejecting foundations of calculus).

What else has been spewed around here? Let’s examine SPP’s ”golden equation”, that is 1 - 1/10^(n)- usually this would be a limit equivalent to 1 (and a demonstration of 0.999 = 1). A claimed motive for alignment with SPP has been “finitism”, but the golden equation is a demonstration of why finitism isn’t a counter to 0.999… = 1 in the reals. If you’re a finitist and claim 0.999… < 1, you accept that the infinite decimal expansion is valid (while other finitists often don’t accept the validity of infinite decimal expansions). Recall that your system lacks much utility if you reject the premise of 0.999... altogether. If you’re a finitist who rejects that 0.999… is complete (rejecting limits involving it as a result), you still cannot disprove that 0.(9) = 1 in other frameworks. Finitists, any positive nonzero gap between 0.(9) and 1 you propose (such as 0.0001) won’t be sufficient, demonstrable via the golden equation (such as the result of 1/10^(5)). You cannot claim n will approach infinity (or is infinite) without accepting you are no longer arguing from a strictly finitist perspective.

Finally, I’m just going to say that cardinality ultimately is not relevant to the relation of 0(9) and 1. (This has also been spewed here recently). Just remember everyone, that rejecting limits doesn’t get you anywhere. It just harms any credibility your “framework” has.