Pi is exactly three
Point
You can round that up to 5
>>16801277One
>>16801271it's exactly 1 in base pi, checkmate irrationalists
>>16801513Pi is not a valid numerical base, for a number to be able to serve as a valid base, you have to be able to count from 1 to 10 in that number of steps, so unless you can figure out how to count from 1 to 10 in an irrational number of steps, irrational numbers will never be a valid base.
>>16801513Also, 1 is 1/base in every base, pi would be exactly 10 if pi were a valid base.
>>16801575No, you can use [math]\beta[/math]-expansion. Let's consider a [math]\pi[/math] based numeral system. It will use a digit set [math]D = \{ 0, 1, 2, 3 \}[/math], where the digits have according values (zero, one, two, three). To convert a [math]n_{10}[/math] number, written in base 10, to [math]\pi[/math]-expansion form is a bit tricky. Let's explore one case [math]4_{10}[/math]: obviously you can't just write [math]10_{\pi}[/math] because it will be equal to [math]\pi_{10}[/math], and can't write [math]3_{\pi}[/math] because it equals to [math]3_{10}[/math]. So the only option is somehow to use powers of reciprocals of pi to represent one, so we will have something like [math]3.d_1 d_2 d_3..._{\pi}[/math]. How to get [math]d_i \in \{ 0, 1, 2, 3 \}[/math]?We can use the following algorithm: 1. Find out [math] [\frac{1}{\pi^{-1}}] [/math], if it is less than or equal to three, it will be our [math]d_1[/math], otherwise we will choose 3 for [math]d_1[/math].2. Substract [math] 1 - \frac{d_1}{\pi} = r_1[/math].3. Find out [math][\frac{r_1}{\pi^{-2}}][/math], if it is less than or equal to three, it will be our [math]d_2[/math], otherwise we will choose 3 for [math]d_2[/math].3. Substract [math] r_1 - \frac{d_2}{\pi^2} = r_2[/math]....Why this algorithm works? First of all, the series [eqn] \frac{d_1}{\pi} + \frac{d_2}{\pi^2}} + \frac{d_3}{\pi^3} + ... [/eqn] converges, because [math]d_i \leq 3[/math] and [eqn] 1 + \frac{1}{\pi} + \frac{1}{\pi^2}} + \frac{1}{\pi^3} + ... [/eqn] converges ([math]\frac{1}{\pi} < 1 [/math]). And our [math]r_i[/math] are just remainders from substracting partial sum [math]S_i[/math]. We have that [math]r_n \rightarrow 0[/math] as [math]n[/math] goes to inf, because we always are subtracting on even steps. On odd steps we are just finding out how many [math]\pi^{-i}[/math] are fitting in the number we are considering.At the end we will get [math]4_{10} = 3.22012202112111030100001011001001023..._{\pi}[/math].
>>16801611I don't know why sums are broken. Anyway, generally speaking you can literally have any number as a base: imaginary, complex, non-integer, negative integer etc. And you can even negative integers as digits. Although I don't know if you can use non-integer numbers as digits. I think you can't.
>>16801611Also in [math]\pi[/math]-expansion you have multiple variants for encoding a number:[math] 1_{10} = 1_{\pi}[/math], and[math] 1_{10} = 0.22012202112111030100001011001001023..._{\pi} [/math]
>>16801611>a digit set [math]D = \{ 0, 1, 2, 3 \ where the digits have according values (zero, one, two, three). That isn't pi base, that is base 3.>>16801612>Anyway, generally speaking you can literally have any number as a baseNot for pi or any other irrational until you can count from 1 to 10 in pi or any other irrational number of steps.
>>16801621It's not even base 3 if you are counting numbers of digits, but base 4.In [math]\beta[/math]-expansion we use digits that are non-negative integers less than [math]\beta[/math], so 0,1,2,3 in the case of [math]\pi[/math].
>>16801636>>16801636Either way, base 3 or 4, its definitely not base pi and there is no way to create pi number of symbols since, as you said, that set needs to be a nonnegative integer and pi does not fit the description.
>>16801271picrel means 4 though. he has his thumb bent and all his other fingers stretched, and that means 4
>>16801575 >>16801621 >>16801646your denial seems familiar, have you by any chance autistically spouted your nonsense when shown that any number can be used as a base before?
>>16801575>for a number to be able to serve as a valid base, you have to be able to count from 1 to 10 in that number of steps,which emperor made that a law?
>>16801921Emperor Wildberger
>>16801790Did you completely fail to actually demonstrate you can count from 1 to 10 in pi steps and devolve into retarded ad hominem attacks in another thread too or something?
>>16801921Its just the definition of a numerical base. See how this dipshit >>16801636 actually has to try to pass off base 4 as base pi to have a coherent numerical system?
How do I represent 5 in the pi number system?
>>16801271Thank you for the cool hat professor frink
>>16803328Its not a real number system, it can't be used for counting, its just a conversion factor that arbitrarily uses pi and converting the decimal 5 results in 11.220122021121..._ฯ.
>>16803328We can try to go 2 ways:1. Trying to represent two with the sum of powers of reciprocals if pi, then we will get [math]5_{10} = 3.d_1 d_2 d_3 ..._{\pi}[/math], where [math]2_{10} = 0.d_1 d_2 d_3 ..._{\pi}[/math]2. Trying to use the algorithm from >>16801611 but with changes so we can get the integer part. In the end we will have something like [math]5_{10} = k_n k_{n-1} ... k_1 . d_1 d_2 d_3 ..._{\pi}[/math]Let's try the first way. What will equal the sum [math]1 + \frac{1}{\pi} + \frac{1}{\pi^2} + \frac{1}{\pi^3} + ...[/math]? Using formula for geometric series we obtain [eqn]\frac{1}{1 - \frac{1}{\pi}} = \frac{1}{\frac{\pi}{\pi} - \frac{1}{\pi}} = 1 + \frac{1}{\pi - 1}[/eqn] And we know that [math]\pi - 1 > 1[/math], therefore we can't write two as sum of powers of reciprocals of pi. Then we must use the other method.Okay, how the freak we will get the integer part? Well, we have that [math]2 > 5 - \pi > 1[/math] (all in base ten). From this we can try to use the integer part [math]11_{\pi} = \pi_{10} + 1_{10}[/math]. Now we just need to find [math]0.d_1 d_2 d_3 ..._{\pi}[/math] that will equal to [math]5_{10} - \pi_{10} - 1_{10} < 1[/math]. And we can just use the original algorithm, only the first and second step will change (the following is in base ten):1. Find out [math][\frac{5 - \pi - 1}{\pi^{-1}}][/math], if it is less thatn or equal to three, it will be our [math]d_1[/math], otherwise we will choose 3 for [math]d_1[/math].2. Substract [math]5 - \pi - 1 - \frac{d_1}{\pi} = r_1[/math]....The answer will be [math] 5_{10} = 11.220122021121110301..._{\pi} [/math]P.S: in the post >>16801611 there is a mistake. The right answer will be [math]1_{10} = 0.301102..._{\pi}[/math], [math]4_{10} = 3.301102..._{\pi}[/math]. And yes, we can also write [math] 4_{10} = 10.220122021121110301..._{\pi} [/math]P.S.S: You can use wolframalpha.com to find the [math]\pi[/math]-expansion form for any number. Just write "[math]n[/math] in base pi".
>>16803783>>16803788Yet its still just a semi-arbitrary conversion factor and not part of an actual functional number system since you can't even use it to count.
>>16803788For those of you who don't understand why don't the number of digits in the digit set equal to the radix of the (positional) numeral system: it's a rule only if the radix is a positive integer. In other cases this rule doesn't apply (e.g. negabinary system has two digits, but the radix is equal to -2, not 2).Generally speaking the radix only matters when we try to find the value of the number. For example, if ten is [math](n_3 n_2 n_1)_b[/math] in a number system with radix [math]b[/math], then [math]10_{10} = (n_3)_{10} \times b_{10}^{2} + (n_2)_{10} \times b_{10} + n_1[/math].Essentially, a positional numeral system is a tripple [math](b,D,f_D)[/math], where [math]b[/math] is the radix, [math]D[/math] is the digit set, and [math]f_D[/math] gives value for every digit in the digit set. Some tripples may not describe every value for every number that exists (e.g. they describe only even numbers, or only positive numbers, or only a specific set of numbers), it must be proved separately that they describe every number.
>Pi is exactly 3:|>Pi is exactly 3.14:O
>>16803872Don't forget that according to the base pi guy, pi is actually exactly 4.>>16801611>>16801636
>>16801271sneed
>>16803911And where exactly it is stated?
>>16803950>>16801636>but base 4He said base pi and base 4 are the same thing which is only possible if pi is exactly 4.
>>16803371lol
>>16801575>numericalanything is valid if you make a rule for it. computers have -0. irrational is z dimension, so surely a valid base
>>16801271Thatโs what the Bible says.1 Kings 7:23And he made a molten sea, ten cubits from the one brim to the other: it was round all about, and his height was five cubits: and a line of thirty cubits did compass it round about.
>the circumference of the circle = 3>checkmate pitards
