I cant sleep, this shit drives me insane...
Delusions of irrational numbers again. The polarity of the fuzz could warp it out.
>>17044931You don't understand rational numbers if you struggle with "1/0", which is unrelated to your pic. It's not a difficult thing to understand from first principles if you sit down with some constructive definition of the rationals. Ultimately you can't define 1/0 meaningfully as long as you want all the other properties of the rational numbers to hold as well. For example, since all fractions have the cancellation property, you get:[math] \frac{1}{0} = \frac{c}{c0} = \frac{c}0 = c \frac{1}{0} [/math]. This alone eventually breaks the math and leads to nonsense like 1=0. You thus need to assert a lot of things that make 1/0 special. In other words there are two choices:>breaking most rules of arithmetic for exactly one element which is very easy to do consistently but means you can't do any of the things you are used to, or >omitting it and retain all propertieskeeping the structure and trying to define it leads to contradiction.Any questions?
>>17044931there is no error, you've simply proven that 1/0 is undefined in standard arithmetics
>>17044931The main error is that lines 1&3 of the “Butt”head section are false and don’t follow from anything. There are a bunch of other errors, too.
>>17044931Another freshman stumped by limits in week one.Sad.Many such cases.
>>17044937Literally this
>>17044931The error is in the last line, saying "f(0.999...) = +infinity". This is not true. It's just undefined.
>>17044931If 0.999...=1 then 1-0.999...=0.Therefore: 1/(1-x)=1/0 given x=0.999...In other words, this statement:>f(0.999...) = +infinityIs false.
>>17044994>+infinity or -infinity are not defined.Wut ? My math is full of + and - infinities.>>17044996By writing that you just said that :>If A=B>Then f(A)=/=f(B) ...which contradicts the definition of a function.>Reminder : if A=B, then f(A) is always equal to f(B)>The reciprical is not always true : f(A)=f(B) does not imply A=B>>17044992>>17044937>A function is a function, like a black box.>Some black boxes are copyrighted, dont ask.
>>17045013>My math is full of + and - infinities."Your math" is scribbles without definitions or rigor, and all your retarded questions have serious answers if you actually take the time and read definitions. You also actively ignore the answers already given that don't even try to address your sloppy nonsense
>>17045013>By writing that you just said that :>>If A=B>>Then f(A)=/=f(B)>...which contradicts the definition of a function.I said no such thing. Quite the opposite, really.
>>17044931line by line[math] a=b \Rightarrow f(a) = f(b)[/math]correct.[math]f(a) = f(b) \not\Rightarrow a=b[/math]correct.Skipping the junk>f(0.99..) = +inffalse. This kinda stuff works for continuous functions which yours is not. You would know this if you actually read a book rather than puzzling together nonsense with half forgotten high school knowledge.[eqn] \lim_{n\to\infty} f(\sum_{k=1}^n 9\cdot10^{-k}) \neq f(\lim_{n\to\infty} \sum_{k=1}^n 9\cdot10^{-k})[/eqn]
>>17045019Statement, no proof. Skipped.>>17045024Ok, but this implies that you would know the behavior of a function beforehand.>"This is always true for every function."Gives you a function which is not explicitly behaving in a singularity.>"Oh well ! Excuse me ! this function is not a function in fact."Just say that A=B <=/=> f(A)=f(B) >>17045023>(A-epsilon)=B <=> A=BThat's why it's a contradiction.
>>17045031>Ok, but this implies that you would know the behavior of a function beforehand.You do, you moron.
>>17045032>A: You know it's true.>B: Let's prove it.>A: Dont need to, it's true.>B: But the proof is not true, it's a singularity, it's not even an exception in your statement.>A: Yea I always knew it was a singularity, that's why I'm always right.Seriously ?
>>17045013>+infinity or -infinity are not defined.That's not what I said. I said that "f(0.999...) = +infinity" is not true. f(0.999...) is undefined.
>>17045040Not "you" as in "your mentally inept middle school dropout ass" but as in "a person with any mathematical ability and the intellect to use mathematical structures as defined". Let me walk your troglodytic self through it.[math] 0.999\dots = \lim_{n\to\infty} \sum_{k=1}^n 9\cdot 10^{-k}[/math]. That is the rigorous definition of "..." you are not taught as a child. If you disagree with it the conversation is over because you don't get to re-define notation, make up your own.Next, your function is not defined at 1. There exists a left handed limit and a right handed limit in the sense both grow without bound and do so unidirectionally. However, that is distinct from saying they have a limit in the reals. The limit that can be described as being +inf is [eqn]\lim_{\epsilon\to0,\,\epsilon>0} f(x-\epsilon) = +\infty, [/eqn] which is defined in the sense that [eqn] \forall r\in\mathbb{R} \exists d(r) : f(x-\epsilon) > r \forall \epsilon \in (0,d)[/eqn]Writing f(0.99..) by definition means that you take the limit before applying the function aka f(1), which is not defined. What you try to connect it to is a one sided limit because you realized that the above limit approaches from below. You are confusing things because you abuse notation, and you disagreeing with notation is not the problem of math, it is literally a skill issue on your end.
x=1 in this case, not that it matters.
>>17044931I guess you're looking for a C=1.000...1 to get a -infinity ?
>>17044942How do you solve this?[math]\frac{1}{0} = \frac{c}{c0} = \frac{c}0 = c \frac{1}{0}[/math]
>>17044931The error was in using a function that was not defined for you choice of x.Functions include their domains, in which case you would have seen that it was not defined for x=1 to begin with.Stick .9_ and 1 in some other function that is defined for them.
>>17045071The floating point standard is one approach. Let's say we include three new "numbers" which I will make the conscious decision to name strangely: [math]I,-I,*[/math]. We retain all properties of the real numbers and add the following rules of addition and subtraction (x,y is any real number), everything still commutes:[eqn]\begin{array}{c|cccc}+ & I & y & -I & * \\\hlineI & I & I & * & *\\x & I & x+y & -I&*\\-I & * & -I & -I & *\\*&*&*&*&*\end{array}\\\begin{array}{c|cccc}- & I & y & -I & * \\\hlineI & * & I & I & *\\x & -I & x-y & I&*\\-I & -I & -I & * & *\\*&*&*&*&*\end{array}[/eqn]Multiplication looks like this, r,l>0 is real[eqn]\begin{array}{c|cccccc}\cdot & I & r & 0 & -r & -I &*\\\hlineI & I & I & * & -I & -I & *\\l & I & lr & 0 & -lr & I & *\\0 & * & 0 & 0 & 0 & * & *\\-l & -I & -lr & 0 & lr & -I & *\\-I & -I & -I & * & I & I & *\\*&*&*&*&*&*&*\end{array}[/eqn]And finally, division like this:[eqn]\begin{array}{c|cccccc}\div & I & r & 0 & -r & -I &*\\\hlineI & * & I & I & -I & * & *\\l & 0 & l/r & I & -l/r & 0 & *\\0 & 0 & 0 & 0 & 0 & 0 & *\\-l & 0 & -l/r & -I & l/r & 0 & *\\-I & * & -I & -I & I & * & *\\*&*&*&*&*&*&*\end{array}[/eqn]If TeX doesn't shit itself or worse I made a massive blunder this should be it. As you can see, I=1/0 can carry a sign but absorbs anything else including itself. Because any multiple of I is still I, 0/0 does not return to 1 but becomes its own number, * (aka NaN). All these new numbers inherit the ability of 0 to "absorb" other values into itself. Where 0 just does so multiplicatively, I does additively and multiplicatively, and * must be even more extreme than either.
>>17045082There is one typo![math]0/0 = *[/math]
In other words:[eqn]\begin{array}{c|cccccc} \div & I & r & 0 & -r & -I &*\\\hline I & * & I & I & -I & * & *\\ l & 0 & l/r & I & -l/r & 0 & *\\ 0 & 0 & 0 & * & 0 & 0 & *\\ -l & 0 & -l/r & -I & l/r & 0 & *\\ -I & * & -I & -I & I & * & *\\ *&*&*&*&*&*&* \end{array}[/eqn]fml
>>17045048What if I can prove that your pic is wrong ?And if your 1st definition is wrong, then everything else falls appart...So who is the trollgolist now ?0.999... , 0.999...8 or 0.999...x are all equals to 1-epsilon0.999...8 = 0.8 + 0.1999...80.999...8 = 0.8 + 0.18 + 0.018 + 0.0018 + 0.00018 + ...0.999...8 = 0.8 + 2*(0.09 + 0.009 + 0.0009 + 0.00009 + ...)0.999...8 = 0.8 + 2*(0.0999...)0.999...8 = 0.8 + 2*(0.1)0.999...8 = 1It works with whatever the 0.999...x ends :0.999...3 = 0.3 + 0.6999...30.999...3 = 0.3 + (0.63 + 0.063 + 0.0063 + 0.00063 + ...)0.999...3 = 0.3 + 7*(0.09 + 0.009 + 0.0009 + 0.00009 + ...)0.999...3 = 0.3 + 7*(0.0999...)0.999...3 = 0.3 + 7*(0.1)0.999...3 = 1
>>17045139>What if I can prove [a rigorous definition] wrong [without providing a rigorous definition of my own terms]>le ebin 0.0...x memeBro we are in our mid 30s can we let those stupid ass memes go already and approach things with a modicum of sincerity?
>>17045145>modicum of sincerity?
>>17044931Whats driving you insane is trying to get logic behind cantors definition of existance of actual infinity. It doesnt exist and it killed math. Only potential infinity exists, but a "set with infinite many numbers" doesnt exist.But within this framework>>17044990is correct. You need to take the limit first, you cant just take that out of f like its nothing.
>>17045411finitism solves a lot of problems of contemporary applied math, but I think a lot of it is just popsci bullshit artistry. Take for example P=NP, there could hardly be a less practically relevant notion in complexity theory, nobody gives a fuck if an algo scales with x^101 or x!^x!. That said I would say infinity has its place, especially things like ordinals are very useful.
>>17044942if sqrt -1 can be 'i' imaginary can we not manipulate 1/0 as undefined numbers>>17044931
>>17044931Just use wheel theory bro.
>>17045919We can and I even provided an explicit example thereof ITT. See >>17045082and >>17045087 and note that there are two sign errors in the multiplication table that are easily fixed.The special thing about complex numbers is, by the way, that you can derive them without ever asserting the existence of "i". You only need to make 3 assumptions:1.) The "number set" [math]\mathbb{M}\supset\mathbb{R}[/math] outght to contain the real numbers, but contains at least one additional element [math] \mathbb{R}\not\ni x \in\mathbb{M} [/math]2.) The arithmetic of multiplication and addition is preserved meaning [math](\mathbb{M},+,\cdot)[/math] is a field. This is what people mean by an "extension" of the real numbers.These two already imply that every integer power of x is in the new set, and [math]\mathbb{C}[/math] is uniquely determined if you assert that3.) Not all integer powers are distinct, meaning there is an eventual finite N such that [math]x^N = \sum_{k=0}^{N-1} c_k x^k [/math] for some choice of real numbers [math]c_k[/math]. It turns out you can show that if a solution exists for some N, it also exists for N=2, and that therefor every number in that set can be written using two real numbers (a,b), yielding the form a+bx. From there you can show that there must necessarily be a combination of real numbers c,d such that (c+dx)^2 = -1. It's a tedious way of doing it but much, much more educational. In short:Any extension of the reals is either the complex numbers, or has infinitely many "terms" (like a power series). Any other construction that is neither is necessarily not a field. Hyperreals violate 3.), Quaternions violate 2.), so does the above example.
>>17044931>Wtf is 1/0if you divide 1 zero times what are you left with? if you give $100 zero times to losers, how much money are you left with?
