Hyperoperation is a field of mathematics which studies indexed families of binary operations, Hyperoperations families, that generalize and extend the standard sequence of the basic arithmetic operations of addition, multiplication and exponentiation.
Questions tagged [hyperoperation]
145 questions
10
votes
1 answer
Has this phenomenon been discovered and named?
If $$x-\frac{x}{2}=\frac{x}{2},$$ and $$\frac{x}{\sqrt{x}}=\sqrt{x},$$
and $$x-\uparrow(x-\uparrow^22)=x-\uparrow^22$$
when $(x\uparrow^n-[A])\uparrow^nA=x$, where $A$ is some constant, and one uses standard Knuth up-arrow notation[where…
9
votes
3 answers
Does anything precede incrementation in the operator "hierarchy?"
I here define the hierarchy of basic mathematical operators and their respective "inverse" operation (see hyperoperation).
$$
\begin{array}{c|c|c|}
& \text{Operator} & \text{"Inverse"} \\ \hline
\text{Incrementation} & a+1 & a-1 \\…
Argon
- 25,303
5
votes
2 answers
Could we define negative hyper operators?
So, I've researched a little bit about this topic, and hyper operators are written $H_n(a, b)$ with :
$H_0(a, b) = b+1$ (succession)
$H_1(a, b) = a+b$ (addition)
$H_2(a, b) = ab$ (multiplication)
$H_3(a, b) = a^b$ (exponentiation)
$H_4(a, b) =…
Pierre Carlier
- 305
4
votes
2 answers
We know that $2+2 = 2 \times 2 = 2^2$. Is this true for all successive hyperoperations?
Given that multiplication is repeated addition and exponentiation is repeated multiplication, if we were to continue onwards, what would the result be for successive hyperoperations? If we take the next step after exponentiation which is tetration,…
Lawrence Yan
- 41
2
votes
1 answer
Hyperoperation: Why does $H_n(0,b) = 0$ for $n\ge4$, $b$ odd ($\ge -1$)
On the Wikipedia for hyperoperations is says that $H_n(0,b) = 0$ if $n\ge4$, $b$ odd ($\ge -1$).
From what I gether, the basic jist of what this is saying is that $0^{0^0}=0$ and from that we can gather that when $b$ is odd and $\ge1$, $H_n(0,b) =…
Jacob Claassen
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1
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Laws of Hyperoperations
I was curious about this equation involving hyperoperations:
$$H_a(H_b(x,y),n)=H_b(H_a(x,n),H_a(y,n)), a,b \in \mathbb{N}$$
where $H_1(x,y)=x+y, H_2(x,y)=xy$ and so on. I found that this leads both to some actual laws such as $(x+y)*n=(x*n)+(y*n)$…
James Smith
- 21
1
vote
0 answers
Is there a way to have non-integer hyperoperations?
I'm pretty interested into hyper operators, but they are only defined for integers :
$H_0(a,b)=b+1$ succession
$H_1(a,b)=a+b$ addition
$H_2(a,b)=ab$ multiplication
$H_3(a,b)=a^b$ exponentiation
...
I already proposed a extension to relative numbers…
Pierre Carlier
- 305
1
vote
0 answers
Inverse hyperoperation code library
I hope you're having a nice day,
I am an Electrical Engineer and I am trying to do a math paper, but I need a coding library in any coding language that can do inverse hyperoperations, do you know of any? and if you don't, do you have an generalized…
rogegar
- 11
- 1
0
votes
1 answer
Why does 2 result in the same value regardless of whether it is added to itself, multiplied by itself, or put to the power of itself?
I'm inferring that any hyperoperation you could apply here using two for every value would result in four. Why is this?
Isaac Philo
- 81
0
votes
1 answer
Compare the growth rate of $f(x)$ and $g(x)$ and show how much quicker the quicker of the two grow
Which of the 2 grows quicker? $f(x)$, $g(x)$, or both grow equally as quick?
Definitions
Notation
$a [x] b = a ↑^{x-2} b$
Functions
$f(x) :=
\begin{cases}
1, & \text{if } x=0\\
f(x-1) [f(x-1)] f(x-1), & \text{if } x<0
…
Aphrontos
- 113
0
votes
1 answer
Are there more identity numbers past 2?
I apologize in advance if this question is too ill-posed, but here we go.
As an additive identity, $x+0=x$.
As a multiplicative identity, $x\times 1=x$.
$2$ feels similar in a way I can't define as well, but $2+2=2\times 2=2^2={^2}{2}=H_{a}(2,2)$,…
Trevor
- 6,022
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0
votes
0 answers
Can there be fractional hyper operators and if so how to they function?
If we call addition (and subtraction as they are the same) the first hyperoperator, multiplication (and division) the second, exponentiation (radicals, indices and logarithms) the third, then tetration the fourth and so on, we then have a definition…
Logan M
- 477
0
votes
2 answers
Are hyperoperations < 3 to a reciprocal of a positive integer equivalent to the 'root' inverse to that integer?
Can the logic of $\sqrt[n]x = x^{1/n}$ be applied to tetration and other natural numbered hyperoperations greater than exponentiation, or, do reciprocals of positive integers as the second argument of a hyperoperation greater than exponentiation…
user7778287
- 268
0
votes
0 answers
hyperoperation sequence with non-integer values of n
This probably has a very simple answer of some sort, but I'm not a mathematician. For the hyperoperation sequence:
$$G(n,a,b)$$
...there are obvious defined values for positive integer values of $n$
$$G(1,a,b)=a+b$$
$$G(2,a,b)=a \cdot…
marcus erronius
- 101
-4
votes
1 answer
What does $i\uparrow\uparrow i$ equal to?(or, it has no result?)
What does $i\uparrow\uparrow i$ (or, i↑↑i) equal to?
$i+i=2i$
$ii=-1$
$i\uparrow i=i^i≈0.20787957635076193$
Chen-Y0y0
- 13