this post was submitted on 22 Oct 2023
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Lemmy Shitpost

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[–] kicksystem 18 points 1 year ago (6 children)

wait till she finds out that 0.99999... 9's to infinity is the same as 1

[–] KeisukeTakatou 7 points 1 year ago (1 children)

Lmao how about ...99999 = -1?

[–] ledtasso 16 points 1 year ago* (last edited 1 year ago) (2 children)

This one has always bothered me a bit because ....999999 is the same as infinity, so when you're "proving" this, you're doing math using infinity as a real number which we all know it's not.

[–] [email protected] 2 points 1 year ago

Yes, you're right this doesn't work for real numbers.

It does however work for 10-adic numbers which are not real numbers. They're part of a different number system where this is allowed.

[–] Snazz 1 points 1 year ago* (last edited 1 year ago) (1 children)

You can also prove it a different way if you allow the use of the formula for finding the limit of the sum of a geometric series on a non-convergent series.

Sum(ar^n, n=0, inf) = a/(1-r)

So,

…999999

= 9 + 90 + 900 + 9000…

= 9x10^0 + 9x10^1 + 9x10^2 + 9x10^3…

= Sum(9x10^n, n=0, inf)

= 9/(1-10)

= -1

[–] [email protected] 2 points 1 year ago (1 children)
[–] Snazz 1 points 1 year ago* (last edited 1 year ago)

Because you could argue that the series converges to …999999 in some sense

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