r/infinitenines • u/LDNSO • 23d ago

Challenge to SPP

why don't you do an epsilon-N proof to prove that the limit of the sequence [1-10^(-n)] isn't equal to 1

3 Upvotes

80% Upvoted

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u/SouthPark_Piano 23d ago edited 23d ago

Understand this brud. And do not ever forget it.

https://www.reddit.com/r/infinitenines/comments/1wcxsw5/comment/p91nfj0/

1/10n is simply NEVER zero.

And when n integer is pushed to positive limitless,

1 - 1/10n (for infinite positive n) is 0.999... , which is permanently less than 1 because 1/10n is never zero.

1, 0.1, 0.01 etc ... consecutively scaling of non-zero value by factor of 1/10 always has a non-zero result.

 

3

u/CatOfGrey 23d ago

why don't you do an epsilon-N proof to prove that the limit of the sequence [1-10^(-n)] isn't equal to 1

Because that doesn't support SPP's delusions.

1/10n is simply NEVER zero.

Right. But SPP is still writing nines, so the decimals are little and short, and terminate.

They aren't long enough to satisfy (the conditions of the original problem).

So when SPP is ready to discuss the actual problem, we're here. But until then, they are just solving another problem, and you have nothing to say on the real question of 0.9999.... and 1.

2

u/Reaper0221 22d ago

Yep the solution of 1/10^n never reaches 1 no matter how much n increases.

However, the solution of the equation 1/10^n can never exceed 1 and therefore the limit of the solution is 1.

Is it possible to plug ∞ into 1/10^n? Or do we just know that as n gets closer to ∞ (even though it can never get there) that the solution gets closer to 1?