Calculation order[]
There is a contradiction in the article:
A chain of multiplications should be solved from the right to the left. Example:
\(3 \cdot 3 \cdot 3 = (3+3+3)\cdot 3 = 333 \cdot 3 = \underbrace{3+3+3...3+3+3}_{333}\)
In the example we are solving the expression from the left to the right. What's wrong? Ikosarakt1 (talk ^ contribs) 17:21, October 26, 2013 (UTC)
I reversed right and left, but it is fixed now. Wythagoras (talk) 17:36, October 26, 2013 (UTC)
Shorten[]
let me show you how to shorten number using hypermathematics and copy notation define addition to be equal to concatenation 1000000000 = 1+0[9] 1222222222 = 1+2[9] 1111111111= 1[10] 65.26.80.144 11:30, March 24, 2014 (UTC)
copy notation[]
Hypermathematics has some pretty neat connections to SpongeTechX's copy notation:
a*b = a[b]
a^b = a[[b]]
a^^b = a[[[b]]]
a^^^b = a[[[[b]]]]
etc.
Cookiefonster (talk) 00:52, September 24, 2014 (UTC)
I can easily see how this form of mathematics lacks any familiar commutativity. No big deal, but this means that ordering is going to need to be more explicitly defined. A noted example on the page was \[(4•3)•3=444•3=\underbrace{333...333}_{444}\] The first step of which suggests that the operator (right) concatenates the operand (left), whereas the second part shows the opposite; the right number being concatenated by the left. I don't know if I'm missing anything here, but I just wanted to point that out. This Hypermath does seem really interesting and promising though. QuasarBooster (talk) 01:13, December 12, 2015 (UTC)
Discrepancy between the article and the source[]
There is a discrepancy between how hypermathematical multiplication is treated in the article versus how it's done in the source, namely which number in \(a\cdot b\) represents what. In the \(333\cdot 3\) example, the source interprets this as \(333333333\), so that we take \(b\) repetitions of \(a\). On the other hand, the article writes that it is equal to \(333\)-fold concatenation of \(3\), meaning it's \(a\) repetitions of \(b\). Because of that it seems that all examples in the article are wrong. LittlePeng9 (talk) 13:09, January 21, 2017 (UTC)
- Also (as I've just noticed), QuasarBooster above mentions that the article isn't even being consistent about which of the two conventions is being used. LittlePeng9 (talk) 13:11, January 21, 2017 (UTC)
hmmm interesting[]
Some of my classmates would let me solve a few addition problems involving hypermathematics. So that means 1+0 is 10? and 1*(0*100,000) is 10^100,000?? —Preceding unsigned comment added by Hypertetrakulus44 (talk • contribs)
It is unclear but I would treat zero as a "zero-digit number" and say 1+0 = 1 and 1+(0*100,000) = 1. Also, always sign your posts with four tildes (~~~~). A Hippopotatomus (talk) 01:42, 20 April 2021 (UTC)
Contenciation[]
Dosen't it mean that it means gluing 2 numbers? so 1+0 is equal to 10! ( not 10 factorial ok?) Hypertetrakulus44 (talk) 10:27, 20 April 2021 (UTC)
About a more formal way, I am not sure if you can define \(a\,\vert\vert\,b\ :=\ a\times 10^{\lfloor\textrm{log}(b)\rfloor+1}+b\) for \(a,b\in\mathbb N\), since then (1\vert\vert001=11\)This handles b=0 wrong
Math tags[]
It looks like another user replaces MathJax with math tags, can someone warn the user? ARsygo (talk) 09:58, 15 August 2021 (UTC)
Question[]
A user in Japanese Googology Wiki pointed out that the example in the article
3*3*3 = 333*3 = 3+3+…+3+3 (333 3s) = 33…33 (333 3s)
is inconsistent with the example in the first source
3*3*3 = 333*3 = 333+333+333 = 333333333 (9 3s).
Since the example is explicitly written in the source, I wonder why the inconsistency occurred. One possibility is that the source has been changed after the creation of the article. Another possibility is that people who wrote this article referred to another source (such as a blog post by the creator) rather than the first source. Does anyone have a clue?
p-adic 12:10, 8 March 2022 (UTC)
- It seems that there is a conversation about it in the top of this page. 🐟 Fish fish fish ... 🐠 14:10, 8 March 2022 (UTC)
- It seems that it is about this edit. Somehow Wythagoras believed that this order is correct. I don't know why... 🐟 Fish fish fish ... 🐠 14:15, 8 March 2022 (UTC)
- I first thought the same thing as you thought when I saw the first section of this talk page, but it is a little different issue. I think that they are taking about whether a * b * c should be solved as (a * b) * c or a * (b * c), i.e. whether 3-ary * is left associative or right associative. But the current issue is whether a * b is solved as a + … + a (b a's) or b + … + b (a b's), i.e. the choice of the recursive formula (a+1) * b = b + (a-1)*b or a * (b+1) = a + a * b. (Since * is non-commutative as a result unlike the usual multiplication, the choise actually matters).
- p-adic 14:32, 8 March 2022 (UTC)
- Oh... Why is this article so inconsistent with the source and itself...?
- p-adic 15:13, 8 March 2022 (UTC)
- Maybe Whytagoras made a mistake and no one realized it until now. Calculations of the numbers in Category:Hypermathematics may also require reevaluation. 🐟 Fish fish fish ... 🐠 15:18, 8 March 2022 (UTC)
- I see. Then is it ok to first write something like WARNING: This article is based on descriptions which conflicts the first source. We are now trying to fix them. (partially because I am too lazy to fix the issues right now).
- p-adic 00:41, 9 March 2022 (UTC)
- It is perhaps correct, but... the source seems to be deleted these days. Therefore I am not certain about the consistency now.
- By the way, although the multiplication is not commutative, it is interesting to have
- abc = a(usual multiplication of b and c) = a(usual multiplication of c and b) = acb
- for any positive integers a, b, and c.
- p-adic 22:55, 9 March 2022 (UTC)
- Oh, I did not know the moval to a new URL. Thank you.
- p-adic 09:02, 10 March 2022 (UTC)