As n approaches Infinity
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I hate this man
0.99999999..... = x Multiply both sides by 10
9.999999999...... = 10 x Subtract equation 2 by equation one
9.000000000.... = 9 x
1 = x
There you go
dude this is chess.com not math class
ur like the math version of blighted stop it
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#8
You’ve negated n approaching infinity. You’re not even “keeping it at bay” because of the negation.
Gregg-Turkington wrote:
You’ve negated n approaching infinity. You’re not even “keeping it at bay” because of the negation.
That's because it doesn't truly approach 1, it simply repeats but doesn't go out of the ((10^-n)-1))/(10^n)) scope.
Better than being a T*rded T*rd.
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#12
One_Zeroth wrote:
Gregg-Turkington wrote:
You’ve negated n approaching infinity. You’re not even “keeping it at bay” because of the negation.
That's because it doesn't truly approach 1, it simply repeats but doesn't go out of the ((10^-n)-1))/(10^n)) scope.
No. It’s because that is the formula you chose. Math dos not have sentience.
Gregg-Turkington wrote:
You are trying so hard.
And succeeding.
You aren't trying at all, and are failing.
Lmao why tf does this matter it’s not like your gonna be in a situation where this makes a difference
Gregg-Turkington wrote:
One_Zeroth wrote:
Gregg-Turkington wrote:
You’ve negated n approaching infinity. You’re not even “keeping it at bay” because of the negation.
That's because it doesn't truly approach 1, it simply repeats but doesn't go out of the ((10^-n)-1))/(10^n)) scope.
No. It’s because that is the formula you chose. Math dos not have sentience.
Theory presented not formula chosen.
.999 increases within the bounds of f(n)=((10^-1)-n))/(10^n))
There is no value of n for f if n that would make it arrive at 1.
I’m too dumb to understand this
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((10^n)-1))/(10^n) shall not reach 1.
As is true of 0.999 ad infinitum, which some here have argued against