Chatgpt and Claude cant resolve this.Is your iq 80+ enough to solve this ?
>>16943882I don't like the way you wrote it It sucks
>>16943882f(0.999...) is also undefined.
>>16943885That's not the point, they are both undefined ok, but not in the same way."undefined positive infinity" is totally different from "strictly undefined".
>in order to input repeating decimals to Desmos write the number as a fraction>google how to write .999... as a fraction>.999... as a fraction is equal to 1/1>1/(1-(1/1)) = undefined
>>16943882Your first mistake was asking AI.Anyways...>>>/sci/ai+containment+thread
>>16943896Hey Anon, we are just here to make fun of "AGI is NIGH! Repent before your new Overlord!".
>>169438820 is unsignedt. 81 IQ math chad>>16943883>>16943885>>16943891you will never be an alumna
>>16943891>undefined positive infinityNo. It's just undefined for the exact same set of reasons 1/(1-1) is undefined.You can say either *approach* positive infinity as x *approaches* 1 or 0.999... but that, again, makes them identical.
>>16943911>undefined for the exact same set of reasons 1/(1-1) is undefined.True>*approach*Doesn't matter, nothing needs to approach.0.999... just means there's a 9 ouskon veut
>>16943911f(x)=1/(1-x)Questions :(1 + 000...1) = 1 ?(1 - 0.000...1) = 1 ?f(1 + 000...1) = -infinityf(1 - 0.000...1) = +infinitySo +infinity = -infinity ????
>>16943915Nigga what>(1 + 000...1) = 1 ?(1 + 000...1) = 2Rewrite yo shit
>>16943915I'm intrigued
>>16943915I think the fault has to be in the assumption that 1^-inf holds the same properties as 0.999...
>>16943915>000...1No such thing.
>>16943924>>16943917Trolls
>>16943917>>16943924>You math stalinists.f(x)=1/(1-x)Questions :(1 + 0.000...1) = 1 ?(1 - 0.000...1) = 1 ?f(1 + 0.000...1) = -infinityf(1 - 0.000...1) = +infinityBut +infinity >>> -infinity
>>16943882the "but" part is wrong. it is not +inf it too is undefined. all repeating decimals have a representation as a fraction. what fraction equals 0.9...?
>>16943924Sure there is. It's a 1 after some arbitrary string of 0>>16943920About?
>>16943930Graphs an shit
>>16943927>(1 + 0.000...1) = 1No, 1 + 0.000...1 = 1.000...1>(1 - 0.000...1) = 1 ?No, 1 - 0.000...1 = 0.000...999...
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>>16943939Get that gay twerking alien out of my face degenerate
>>16943932>No, 1 - 0.000...1 = 0.000...999...Wut ?>1 - 0.000...1 = 0.000...999...>1 = 0.000...999... + 0.000...1>1 = 0.000...1???
>>16943947>You math stalinists.No, 1 - 0.000...1 = 0.999...900...
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>>16943958It is virtuous for a man to proclaim that which he finds worthy of contempt
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>>16943882>f(0.999...) = +infIncorrect.>f(1) = undefinedAlso technically incorrect. You cannot have something equal to undefined. You can only say that f(1) *is* undefined. But that's just nitpicking really.>f(1 + 000...1) = -infinityNo.>f(1 - 0.000...1) = +infinityNo.They are both undefined. They don't equal anything. They are not numbers. You can't give f an argument of 1 (or anything equivalent).
>>16944030>They are both undefined. They don't equal anything. They are not numbers. You can't give f an argument of 1 (or anything equivalent).Ok.Can it be written like this then :Let's say A = BThen f(A) = f(B)If :f(x) = 1/(1-x)A = (1 + epsilon)B = (1 - epsilon)"epsilon" is an infinity small number.So A = (1 + epsilon) = B = (1 - epsilon)Then :f(1 + epsilon) = f(1 - epsilon) ...but it's false.The truth is :f(B) >>> f(A)So A = (1 + epsilon) and B = (1 - epsilon) are at least different.
>>16944079If epsilon is too small to be a number, 1+e = 1-e = 1
>>16944086f(x) = 1/(1-x)You said >1+e = 1-e = 1But :f(1+e) , f(1-e) and f(1) give 3 different results.So1+e =/= 1-e =/= 1
>>16944097You said epsilon wasn't a number, so nothing changes with respect to f(1+e) , f(1-e), f(1)
>>16944079>"epsilon" is an infinity small numberNo such thing. Learn about limits aka real analysis if you want to do any of this properly. You're wasting your time.
>>16944128Even if you allow "epsilon" to be infinitely small, it defines some number 0.000...000... ... ...1 with a last digit somewhere, in which case you can write 1-e as 0.999...999... ... ...9 which also has a last digit somewhere. So 1-e is always less than 0.999... because 0.999... doesn't have a last digit.
>>16943882Saying that 1-0.999... has a different outcome than 1-1 is implicitly stating that 0.999... =/=1. This doesn't prove anything because depending on what side of this you land you get two answers
>>16944164There aren't two sides or two answers. In decimal notation, 1 - 0.999... can only ever = 0. Doesn't matter whether you're using real numbers, hyperreal numbers, or the most jihadi of finitist computers that explodes whenever its storage capacity is exceeded.
>>16944155>with a last digit somewhereIt's not infinitely small if it has a last digit somewhere. There's no such thing as "infinitely small" using regular real numbers, unless you count zero, which you shouldn't because "infinitely small" is bad terminology.
>>16944191Some well-meaning people are convinced that you can graft two or more infinite strings of digits together so as to define an infinitely small number of the form e = 0 + 0/∞ + ... + n/∞^x + m/∞^(x+1) with n > 0 and m = 00.999... has no m = 0, making it greater than any of that.
>>16943882>Chatgpt and Claude cant resolve this.Prove it
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>>16943882based as fuck ai. fuck miggers.
>>16943882I'm sure what I'll say in some form has already been said, but I'm not gonna read all that.Just take this as further confirmation to that take (which is the correct one, I mean I'm an unc with a PhD):The real numbers u=0.999... and v=1 are equal. Take any framework, say ZFC, and ramp up the formalism to define them, and the theory will prove u=v. Your f also is not defined v=1. The value is not even in the domain if f is a function to R. So if f(v) is not defined, then f(u), which is equal to f(v) is not defined either.It's not "undefined in another way". Since u and v are provably equal, you don't have "another way" there.PS people in academic math don't commonly reason too much in terms of some base 10 digit expansion anyway, the "0.999..." thingy barely doesn't come up.But if you'd actually formalize this, then see above.
>>169438821/(1-0.999…)=1/01/(1-1)=1/0Both are undefined
>>169438822 divided by -1 is not 2This function does not prove identity
>>16943882
>>16944944>1/(1-0.999…)=1/0Nope, its :1/(1-0.999…)=+inf>1/(1-1)=1/0Yep, it's undefined.The point is "+inf" and "undefined" are perhaps both undefined, but not in the same manner."+inf" undefined in the sense that "it's a big positive number that cant be defined, but still a big positive number".1/0 is strictly undefined : It's all numbers and none.>>16945428I've a counter argument with upper bound :Limits between different bases.0.999...(base 10) = 0.111... (base 2) ~ 1-But the "speed" at which both [0.999...(base 10)] and [0.111... (base 2)] have there digits go to 1 are different.The digits of [0.999...(base 10)] goes to 1 way faster than [0.111... (base 2)] :0.111 (base 2) = (0.5 + 0.25 + 0.125) (base 10) = 0.875 (base 10)Etc. with 0.1111 and 0.11111 etc.0.999(base 10) > 0.111(base 2)0.9999(base 10) > 0.1111(base 2)Etc.In this way, you can always find an under bound between the limit "1" and 0.999...(base 10) :0.FFF... (base 16) goes faster to "1" than 0.999... (base 10)And same with 0.GGG... (base 17) that goes faster than 0.FFF... (base 16)Etc. untill 0.(+inf -1)(+inf -1)(+inf -1)... (base +inf)>>16944960>2 divided by -1 is not 2Wut ? Where it's written 2/(-1)=2 in OP ?
>>169438820.999... is not 1 yeswhat are we resolving here
>>169438821.4+1.4=2.8Round your figures. 1+1=3 MOM'S GONNA FREAK
>>16943924[math] \displaystyle 0. \bar{0}1 = \lim_{n \to \infty} 0. \underbrace{0 \dots 0}_{n ~ \text{times}}1 = \lim_{n \to \infty} \left [ \left ( \sum_{k=1}^n \dfrac{0}{10^k} \right ) + \dfrac{1}{10^{n+1}}\right ]=0[/math]
