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# /sci/ - Science & Math

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 >> Anonymous Thu Feb 28 21:12:14 2019 No.10425953   >>10425928It's indeterminate.$\dfrac{0}{0} = x$$0 = 0x$x can be any number.
 >> Anonymous Thu Feb 28 21:13:38 2019 No.10425957 >>10425953But the number 1 is clearly larger than 0.999...
 >> Anonymous Thu Feb 28 21:14:45 2019 No.10425960 >>10425957It’s the same as (1-1)/(1-1)
 >> Anonymous Thu Feb 28 21:15:05 2019 No.10425961 It has to be 1. They’re going to 0 at the same rate.
 >> Anonymous Thu Feb 28 21:16:49 2019 No.10425964 >>10425960>>10425961So which is it? Two different answers here.
 >> Anonymous Thu Feb 28 21:18:12 2019 No.10425967 >>10425964real answer: divide by zero error. this doesn't indicate any sort of limit, instead 0.999... = 1 by definition so the denominator is illegal, t. NYPD
 >> Anonymous Thu Feb 28 21:24:08 2019 No.10425977 >>10425967Thank you
 >> Anonymous Thu Feb 28 21:25:41 2019 No.10425980 >>10425961They aren't going anywhere your dofus.OP's image is a png, not a gif or a webm
 >> Anonymous Thu Feb 28 21:26:44 2019 No.10425985 >>10425928Aaahhhh can you shut uppp please
 >> Anonymous Thu Feb 28 21:27:34 2019 No.10425988 let f(x) = -1 / xlimit(x->inf) f(x) = -0let g(x) = f(x) + 1limit(x->inf) g(x) = 1Also, since g(x) is an increasing function for all x > 0, this fits the conventional definition of "0.9999..."limit(x->inf) g(x) = 0.999...So:(1 - g(x) / 1 - g(x))(1 - (-1 / x + 1)) / (1 - (-1 / x + 1))(1 + 1/x - 1) / (1 + 1/x - 1)(1/x) / (1/x)x / xAnd since we established that x > 0, we haven't divided by zero at any point, so this is 1, not undefined
 >> Anonymous Thu Feb 28 21:31:56 2019 No.10425997 >>10425988yeah but your proof is invalid since g(x) is a function and not a number. l'hopital's rule doesn't exist for numerical expressions
 >> Anonymous Thu Feb 28 21:32:23 2019 No.10425998 This is dumb and you should feel bad.
 >> Anonymous Thu Feb 28 21:35:51 2019 No.10426007 >>10425997I'm didn't invoke L'opital's rule here.I showed that 1 - g(x) / 1 - g(x) is equal to 1 for every x > 0.
 >> Anonymous Thu Feb 28 21:37:00 2019 No.10426010 >>10426007fine, then that has nothing to do with dividing by 1-0.999...
 >> Anonymous Thu Feb 28 21:38:03 2019 No.10426012 >>10426010How else does 0.999... exist "in the wild", other than as a limit of a function?
 >> Anonymous Thu Feb 28 21:38:09 2019 No.10426013 >>10426007>L'opital
 >> Anonymous Thu Feb 28 21:42:39 2019 No.10426017 >>10426012you could consider "1" as just an alternate notation for 0.999... or 2-0.999... or 6-4.999..., decimal notations are om general not unique and that phenomenon exists in the wild
 >> Anonymous Thu Feb 28 21:47:00 2019 No.10426025 This whole thread is bad, it’s based off of a problem missing pieces, leading us to all assume things, not to mention that the whole 0.999... is 1 thing is also just a cheap way that our society threw out a seemingly pointless problem (we all use the logic 0.999... is 1 just so we don’t end up running in circles) without that flawed logic we could also say in reply to “real answer” that 1-0.999... is greater than zero and if you divide the same number (that is greater than zero) by the identical number (that is greater than zero) then it equals 1
 >> Anonymous Thu Feb 28 21:48:00 2019 No.10426027 >>10426025>t. mathlet
 >> Anonymous Thu Feb 28 22:01:03 2019 No.10426059 >>10426025Conventional math is different when dealing with infinites. Infinity / Infinity is undefined
 >> Anonymous Thu Feb 28 22:05:42 2019 No.10426073 >>10425928.111...--------- = 1.111...Constant divided by equivalent constant is 1, no matter the precision of the constant.
 >> Anonymous Thu Feb 28 22:07:01 2019 No.10426078 >>10426073nuh uh, 0.000.../0.000... = undefined
 >> Anonymous Thu Feb 28 22:08:43 2019 No.10426080 >>10426078Oh woops, brain fart.
 >> Anonymous Thu Feb 28 22:09:34 2019 No.10426083 >>10426073>1 - 0.99 = .11
 >> Anonymous Thu Feb 28 22:25:24 2019 No.10426113 >>10426083That's what I meant, but it's still an irrational constant. Analyzing it's indeterminate nature is useless because of this.The numerator and denominator are infinitesimally small, but more importantly, they're infinitesimally equivalent. Their ratio remains the same because of this, at 1.
 >> Anonymous Thu Feb 28 22:34:48 2019 No.10426130 >>10426013el hospital
 >> Anonymous Thu Feb 28 22:44:11 2019 No.10426146 >>10426078>>10426078Lol wut 1-0.999... will be indefinitely small but ultimately greater than zero still (ie. 0.0000............01) so that over the same thing = 1
 >> Anonymous Thu Feb 28 23:05:12 2019 No.10426175 >>10426146>1-0.999... will be indefinitely small but ultimately greater than zero still (ie. 0.0000............01)Bzzt. Wrong.
 >> Anonymous Thu Feb 28 23:11:59 2019 No.10426184 >>10426146>(ie. 0.0000............01)That's not how it works, if you know the end of the sequence it's not infinite, which means it is not 0.999....
 >> Anonymous Fri Mar 1 00:02:27 2019 No.10426250 >>10425928its one stupidthe top is numerator and denominator are equal
 >> Anonymous Fri Mar 1 00:09:23 2019 No.10426258 >>10426184>it's not infiniteyes it is, if it has '...' it's infinite
 >> Anonymous Fri Mar 1 00:12:43 2019 No.10426264 >>10426012> I heard smart people say "in the wild">>10426025Take your meds
 >> Anonymous Fri Mar 1 00:13:44 2019 No.10426265 >>10425988>>10426007The correct answer is undefined. Real numbers are defined as the limit of a cauchy sequence of rationals. The number one could generally be defined more succintly (with ordinal sets), but the cauchy sequence {.9, .99, .999, . . . } is just an obstuse manner of describing the number 1.If this is unclear, we have:a = (.9999. . .)(10) = 9.9999. . .whileb = (1)(10) = 10So that (9/10)a = 9.999. . . - .999. . . = 9and (9/10)b = 9 Hence (9)(0.999. . .) = 9 = (9)(1) so that by cancellation 0.999. . . = 1
 >> Anonymous Fri Mar 1 00:29:56 2019 No.10426281 1 in math, divide by zero error in inferior computation.
 >> Anonymous Fri Mar 1 00:30:01 2019 No.10426282 >>10425961They are not "going to 0 at the same rate". The numerator and denominator are identically zero. By convention, 0/0 is left undefined.
 >> Anonymous Fri Mar 1 01:17:41 2019 No.10426350 >>10425961>numbers are little diesel engines chugging alongtopkek
 >> Anonymous Fri Mar 1 01:40:26 2019 No.10426367 $\displaystyle\boxed{0 < p < 1} \\p^n-1 = (p-1)(p^{n-1}+p^{n-2}+ \dots +p+1) \\\dfrac{p^n-1}{p-1} = \sum \limits_{j=0}^{n-1}p^j \\\displaystyle\lim_{n \to \infty} \dfrac{p^n-1}{p-1} = \lim_{n \to \infty} \sum \limits_{j=0}^{n-1}p^j \\\displaystyle\dfrac{0-1}{p-1} = \sum \limits_{j=0}^{\infty}p^j \implies \dfrac{1}{1-p} = \sum \limits_{j=0}^{\infty}p^j$
 >> Anonymous Fri Mar 1 01:58:43 2019 No.10426395 >>10425928$\frac{1-\sum_{k=1}^{\infty}\frac{9}{10^k}}{1-\sum_{k=1}^{\infty}\frac{9}{10^k}}$Then see where it converges. Don't have time to do it correctly now and calculating in $\LaTeX$is hard.
 >> Anonymous Fri Mar 1 02:42:18 2019 No.10426449 File: 112 KB, 953x613, C7473FA965194B0B8FE526924E00C01D.jpg [View same] [iqdb] [saucenao] [google] [report]
 >> Anonymous Fri Mar 1 04:21:43 2019 No.10426573 P = (1-0.999...)(1-0.999...)/(1-0.999...) = P/P=1:^]
 >> Anonymous Fri Mar 1 04:25:50 2019 No.10426579 >>10426449So what's the trick in each one of those?
 >> Anonymous Fri Mar 1 04:28:53 2019 No.10426581 >>10426579the trick is that 0.999... and 1 are different ways to write the same thing
 >> Anonymous Fri Mar 1 04:29:22 2019 No.10426584 looks like you've discovered a situation that causes two axioms to come into conflicteither 0/0 is undefined, therefore x/x is not always 1, or 0/0 is 1, therefore 0/0 is not undefined
 >> Anonymous Fri Mar 1 04:30:33 2019 No.10426585 >>10426573Nigger
 >> Anonymous Fri Mar 1 04:34:32 2019 No.10426588 >>10426581Brainlet
 >> Anonymous Fri Mar 1 04:36:09 2019 No.10426589 >>10426588retarded brainless fungus
 >> Anonymous Fri Mar 1 04:38:39 2019 No.10426590 >>10426588>>10426589>typical /sci/ discussion
 >> Anonymous Fri Mar 1 04:39:01 2019 No.10426592 >>10425928surely anything divided by itself is 1?
 >> Anonymous Fri Mar 1 04:41:27 2019 No.10426594 >>10426592huurrr
 >> Anonymous Fri Mar 1 04:44:21 2019 No.10426598 >>10426592>b-but muh undefined
 >> Anonymous Fri Mar 1 04:44:24 2019 No.10426599 >>10426594It isn't zero it's a number greater than zero, if you want to debate zero divided by zero that's a different bag of worms, some number greater than zero divided by itself is one
 >> Anonymous Fri Mar 1 04:46:42 2019 No.10426600 >>10426579>>10426581when you divide by 1 by 3 there is a remainder. 10/3 = 3+R1100/3= 33+R1they just choose to ignore it
 >> Anonymous Fri Mar 1 04:58:10 2019 No.10426606 $x= \frac{1}{10} \\0. \overline{9}=9x+9x^2+9x^3+9x^4+ \cdots \\0. \overline{9}=9x \left (1+x+x^2+x^3+ \cdots \right ) \\0. \overline{9}=(1-x) \left (1+\mathbf{x}+x^2+\mathbf{x^3}+x^4+ \cdots \right ) \\0. \overline{9}=1-x+ \mathbf{x-x^2}+x^2-x^3+ \mathbf{x^3-x^4}+x^4-x^5+ \cdots \\0. \overline{9}=1$
 >> Anonymous Fri Mar 1 04:59:20 2019 No.10426607 >>10426599>1-1 greater than zerobravo retard
 >> Anonymous Fri Mar 1 05:14:00 2019 No.10426627 >>10426600it's exactly $0.1_3$
 >> Anonymous Fri Mar 1 05:16:27 2019 No.10426631 >>10426449>hyperreals dont existbrainlet detected
 >> Anonymous Fri Mar 1 05:22:29 2019 No.10426641 Wow this thread really reveals how mathematically inept the people on this board are. Anyone who does not immediately know the answer is 1 should forever leave.
 >> Anonymous Fri Mar 1 05:26:58 2019 No.10426652 >>10426606Wrong. Wow you are so stupid. Your calculation has a blatant mistake. Not just the first but also the last element does not cancel out. Therefore the jump from line 5 to line 6 is wrong.
 >> Anonymous Fri Mar 1 05:31:26 2019 No.10426655 >>10426652> the last element no such thing
 >> Anonymous Fri Mar 1 05:32:29 2019 No.10426657 >>10426641so leave
 >> Anonymous Fri Mar 1 05:35:23 2019 No.10426662 >>10426655Yes there is. Your mind is too small to see it.
 >> Anonymous Fri Mar 1 05:42:31 2019 No.10426670 File: 1005 KB, 220x260, 1545869393864.gif [View same] [iqdb] [saucenao] [google] [report] >>10426631>hyperreals are a subset of reals
 >> Anonymous Fri Mar 1 05:50:35 2019 No.10426687 File: 8 KB, 169x220, x.jpg [View same] [iqdb] [saucenao] [google] [report] >>10426662sure bud, you have the large mind
 >> Anonymous Fri Mar 1 07:58:01 2019 No.10426827 >>10425928A 1-0.999.../1-0.999...=0/0=errorB 1-0.999.../1-0.999...= 1-1-1=-2C 1-0.999.../1-0.999...=0.00...01/0.00...01=1I know that A is correct but couldnt B be right?Or that is only if you dont have it set up like this
 >> Anonymous Fri Mar 1 08:44:16 2019 No.10426922 >>10426827For B:1 - (0.999.../1) - (0.999...) = ?1 - 0.999... - 0.999... = -0.999...8
 >> Anonymous Fri Mar 1 08:45:23 2019 No.10426925 File: 502 KB, 1080x1451, x7wg0u7xu8j21.jpg [View same] [iqdb] [saucenao] [google] [report] >>10426662VERY big brained my guy! Good job!
 >> Anonymous Fri Mar 1 08:50:38 2019 No.10426930 >>10425928You would get 0/0.0.999999 = 1 has been shown to be true for a long time.
 >> Anonymous Fri Mar 1 08:54:47 2019 No.10426934 >>10426930>0.999999 = 1 has been shown to be true for a long time.nope, it's 0.9... = 1
 >> Anonymous Fri Mar 1 09:05:17 2019 No.10426945 >>10426258nopean infinite number wont have an end
 >> Anonymous Fri Mar 1 09:14:41 2019 No.10426956 >>10426945>an infinite number wont have an endcorrect, that's why writing anything after '...' is meaningless
 >> Anonymous Fri Mar 1 13:54:40 2019 No.10427550 >>10425928so waht is the answer?it's 1 isn't it?
 >> Anonymous Fri Mar 1 14:02:31 2019 No.10427562 1 - 0.999... = 1 - (1 - 1/inf) = 1/inf(1/inf) / (1/inf) = inf/inf = undefinedcheckmate matheists
 >> Anonymous Fri Mar 1 14:18:41 2019 No.10427592 Why can't we just say .999... is a poorly defined expression (leaves ambiguity in value when contested with 1) and that the question itself is flawed.I don't think the term .999... itself should really come into question here. This shit is a funny allegory for the kind of thinking that's been fucking up quantum mechanics for years now.
 >> Anonymous Fri Mar 1 14:23:44 2019 No.10427602 $\lim_{n \to \infty}1^n = 1$. But is $\lim_{n \to \infty}0.999...^n = 1$?
 >> Anonymous Fri Mar 1 14:24:32 2019 No.10427606 >>10427592because it is a well defined expression
 >> Anonymous Fri Mar 1 14:28:26 2019 No.10427616 >>10427592>.999... is a poorly defined expression It's perfectly well defined, brainlet
 >> Anonymous Fri Mar 1 14:29:43 2019 No.10427617 >>10427592Because it is perfectly well-defined. The fact that mathlets don't understand it is immaterial.
 >> Anonymous Fri Mar 1 14:30:00 2019 No.10427618 >>10427606How so?
 >> Anonymous Fri Mar 1 14:30:40 2019 No.10427621   >>10427602$\displaystyle\lim_{n \to \infty}1^n = 1 \\ 1^ \infty \textsf{ is undefined} \\\lim_{n \to \infty}0.9...^n = 1$
 >> Anonymous Fri Mar 1 14:30:45 2019 No.10427622 >>10427602Yes.
 >> Anonymous Fri Mar 1 14:32:54 2019 No.10427628 >>10427602$\displaystyle\lim_{n \to \infty}1^n = 1 \\ 1^ \infty \textsf{ is undefined} \\\displaystyle\lim_{n \to \infty}0.9...^n = 1$
 >> Anonymous Fri Mar 1 14:33:58 2019 No.10427631 >>10427618$1 = \dfrac{3}{3} = 3 \cdot \dfrac{1}{3} = 3 \cdot 0.\bar{3} = 0.\bar{9}$
 >> Anonymous Fri Mar 1 14:38:10 2019 No.10427641 So... if you guys are gonna insist on this... then would .999...99 = .999...?Also, isn't the fraction a better expression for what you're trying to express? Why bother switching to decimal?
 >> Anonymous Fri Mar 1 14:39:42 2019 No.10427642 >>10427641>.999...99 = .999...?yes >>10426956
 >> Anonymous Fri Mar 1 14:40:32 2019 No.10427645 >>10426657Baited
 >> Anonymous Fri Mar 1 14:43:14 2019 No.10427650 >>10427645look what the cat brought in
 >> Anonymous Fri Mar 1 14:43:27 2019 No.10427651 I'm thinking the guys here don't really care about context... I would think it's common sense that .999... in the current setting could result in ambiguity of an infinitely small number divided by itself or 0/0. And if 1-.999... isn't a good way to describe an infinitely small number... then what is?
 >> Anonymous Fri Mar 1 14:46:05 2019 No.10427656
 >> Anonymous Fri Mar 1 14:48:45 2019 No.10427661 >>10427656That's a cool wiki page i agree... Is there any context you could give me where there's utility in using .999... instead of 1 though?
 >> Anonymous Fri Mar 1 15:05:12 2019 No.10427689 File: 99 KB, 414x446, smug perturabo.jpg [View same] [iqdb] [saucenao] [google] [report] Imagining believing that one branch of maths acts consistently in all situations
 >> Anonymous Fri Mar 1 15:07:11 2019 No.10427692 Can anyone here give me a context where using .999... instead of 1 would provide more utility?If not... I'm gonna go ahead and keep spouting that .999... is ill defined when used with 1.
 >> Anonymous Fri Mar 1 15:13:54 2019 No.10427705 >>10427617Yup. It simply becomes a very small number divided by itself. Infinitely small, but not zero, and because it's divided by itself the answer is 1.
 >> Anonymous Fri Mar 1 15:14:32 2019 No.10427709 >>10425961good bait bro
 >> Anonymous Fri Mar 1 15:59:52 2019 No.10427804 >>104276611/3 = 0.333...3 * (1/3) = 0.999...
 >> Anonymous Fri Mar 1 16:25:54 2019 No.10427893 >>10427705>Such a brainlet that they didn't even know that the guy they responded to was talking about the well-definedness of the expression 0.999..., and not the well-definedness of OP's expression.>Such a brainlet that, not only do they make the above mistake, but they conclude that a fraction equivalent to an indeterminate form (0/0) is a well-defined expression.Time to stop posting.
 >> Anonymous Fri Mar 1 16:29:51 2019 No.10427907   >>10427893The answer is still 1 no matter how hard you many people you call brainlet
 >> Anonymous Fri Mar 1 17:16:19 2019 No.10428051 >>10427705>Infinitely small, but not zeroFalseYou don't understand what infinite means
 >> Anonymous Fri Mar 1 17:17:51 2019 No.10428054 >>10428051You don't know what zero means
 >> Anonymous Fri Mar 1 17:24:16 2019 No.10428072 >>10427705>Infinitely small, but not zeroThere are no infinitesimals in R you retarded ape, R is an Archimedean field, this is not validYou could use any formulation of the completeness axiom you wanted to show infinitesimals aren't valid in R.
 >> Anonymous Fri Mar 1 18:00:15 2019 No.10428176 >>10428072>it is wrong because I arbitrarily define it to be so
 >> Anonymous Fri Mar 1 18:04:28 2019 No.10428199 >>10427804But you forgot the remainder.Consider a)10/3 = 3+R1b)100/3 = 33+R1c)1000/3 = 333+R1d)10000/3 = 3333+R1e)100000/3 = 33333+R1Oh wait there's that pesky remainder you keep forgetting about. It's negligible in many everyday computations, that's why the math works so well above and beyond. Not to mention random noise outright discards this noise. But you know darn right that remainder will be there even in an infinite sequence. Maybe it is clear now, or maybe you are still skeptical but 1/3 will yield a remainder now matter how far you go because 10/3 will never be equal to 3 evenly. That is a mathematical fact so basic that higher levels can only forget about it when it becomes infinitely small(not 0). So as you can probably see now, 0333... != R1+0.3333.....; these are two different numbers not a way to write >>bbbbut, since 0.333... is infinite you will never get to add back in the remainder. That is true. But what is also true, is that this can be rewritten as a finitely long number: 0.333(10/3) and likewise this number can be written as 0.3333(10/3) = 0.33333(10/3)=0.333333333333333333(10/3).It should not be hard to see that if I multiply this number by 3, the result is identically 1.
 >> Anonymous Fri Mar 1 18:05:37 2019 No.10428202 >>10425928Are you serious?
 >> Anonymous Fri Mar 1 23:17:27 2019 No.10428986 >>10428199>1/3 currently does not have an exactly equal (aka valid) translation to decimalIn base-3 it is exactly 0.1 = 1*3^(-1).There is nothing 'clean' about base-10.1/10 looks nice in base-10 because monkeys have 10 fingers and we created the symbols accordingly.In hexadecimal (base-16) it's a fucking mess:0.1999...https://www.rapidtables.com/convert/number/decimal-to-hex.html
 >> Anonymous Fri Mar 1 23:20:59 2019 No.10428994 >>10428176>arbitrarily definelook who's talking
 >> Anonymous Fri Mar 1 23:26:15 2019 No.10429008 Well, 1-1 equals zero. So zero divided by zero is DNE. Why is this a fucking question?
 >> sage goes in all fields Fri Mar 1 23:28:19 2019 No.10429011 File: 417 KB, 1846x867, the absolute state of sci.png [View same] [iqdb] [saucenao] [google] [report] ITT
 >> Anonymous Fri Mar 1 23:40:38 2019 No.10429038 >>10429011this, but also for literally every thread on sci
 >> Anonymous Sat Mar 2 00:32:55 2019 No.10429152 >>10425928you are asking what is the limit of x/x as x approaches 0 from the positive sidethis is not mind blowing mathematics guys the answer is 1. you can graph it on desmos or something its first semester college maths.
 >> Anonymous Sat Mar 2 00:40:39 2019 No.10429164 >>10429152>you are asking what is the limit of x/x as x approaches 0 from the positive sideno he's not
 >> Anonymous Sat Mar 2 00:51:45 2019 No.10429185 >>10429164this
 >> Anonymous Sat Mar 2 00:54:37 2019 No.10429190 >>10428986Changingbase also requires a change in definitio of division so....
 >> Anonymous Sat Mar 2 00:57:09 2019 No.10429197 >>10429190[citation needed]
 >> Anonymous Sat Mar 2 01:20:12 2019 No.10429239 >>10429197Umm how about binary? even additionand subtractionare not the same hence anstract algebra
 >> Anonymous Sat Mar 2 01:21:31 2019 No.10429242 >>10429239different bases arent different algebraic structuresyoure thinking of modular numbers
 >> Anonymous Sat Mar 2 01:45:18 2019 No.10429311 >>104292391+1=10 --- 1+1=210+1=11 --- 2+1=311+1=100 --- 3+1=4it's exactly the same, wtf are you talking about.the numbers look different, but the math is the same
 >> Anonymous Sat Mar 2 02:04:40 2019 No.10429379 >>10428054Mathlet please leave
 >> Anonymous Sat Mar 2 02:05:41 2019 No.10429387 >>10429379So a number greater than zero is zero?
 >> Anonymous Sat Mar 2 02:07:22 2019 No.10429398 >>104293871/inf =0
 >> Anonymous Sat Mar 2 02:52:02 2019 No.10429652 >>10425928if 0.999... is a function approaching 1 as it's argument approaches inf, then the answer is 1.If 0.999... is a number then it's 0/0 so indeterminate.
 >> Anonymous Sat Mar 2 03:08:51 2019 No.10429791 >>10429652>approaching'...' means it's already at infinity, not approaching
 >> Anonymous Sat Mar 2 03:19:23 2019 No.10429848 >>10429652>if 0.999... is a functionIt is not.>If 0.999... is a number then it's 0/0Indeed.>so indeterminate.No. That word doesn't mean what you think it means.
 >> Anonymous Sat Mar 2 05:17:10 2019 No.10430227 >>10425928zero fuck you if u wanna read all these real analysis bullshit
 >> Anonymous Sat Mar 2 05:39:48 2019 No.10430265 >>10429848that what does indeterminate means good sir?
 >> Anonymous Sat Mar 2 06:03:50 2019 No.10430290 >>10430265something exists but is unknownundefined = doesn't exist>puts feeding spoon back in drawer
 >> Anonymous Sat Mar 2 06:06:44 2019 No.10430293 >>10429791when do we arrive at infinity ?
 >> Anonymous Sat Mar 2 06:20:56 2019 No.10430310 >>10426579Part of it is psychological. The bottom left one where it has "9x = 9" is incorrect, its actually "9x = 8.999...1". The repeating number 9 in the ones above makes you more susceptible to accepting the 9 as correct.
 >> Anonymous Sat Mar 2 06:28:49 2019 No.10430330 File: 3 KB, 635x223, r8.png [View same] [iqdb] [saucenao] [google] [report] >>10430293>no approachingdid I fucking stutter?
 >> Anonymous Sat Mar 2 06:30:34 2019 No.10430333 >>104303108.999...1 = 8.9... = 9
 >> Anonymous Sat Mar 2 10:49:58 2019 No.10430907 >>10425928If 0.999... = 1, does 0.0999 = 0.1?
 >> Anonymous Sat Mar 2 11:02:31 2019 No.10430935 >>10430907yes
 >> Anonymous Sat Mar 2 11:04:13 2019 No.10430942   >>10430907you mean 0.09... = 1yes
 >> Anonymous Sat Mar 2 11:05:17 2019 No.10430945 >>10430907you mean 0.09... = 0.1yes
>>