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 Zeration marraco Fellow Posts: 93 Threads: 11 Joined: Apr 2011 03/20/2015, 10:41 PM (This post was last modified: 03/21/2015, 04:41 AM by marraco.) Now, let's suppose that somebody only knows a number n, and the product operation. Then he can operate n.n and get n² he would get a "natural" sequence of numbers n, n², n³.... nᵐ then, if he wants to solve any equation, he will discover the neutral 1 and the inverses of product, which he may name the "integer" sequence: nˉᵐ, nˉ³, nˉ², nˉ¹, 1, n, n², n³.... nᵐ Then, he may ask what operation lies before the product, what is addition. He will have a hard time, because he does not know what we call integer numbers. He only knows powers of n, so he will try to assign a=1.a a+a=n.a a+a+a=n².a a+a+a+a=n³.a and of course, he will go nowhere, because we know that this is the answer: 0=nᶫᶰ°.a a=nᶫᶰ¹.a a+a=nᶫᶰ².a a+a+a=nᶫᶰ³.a ^ LNx means ln(x)/ln(n), but I'm using "our" representation for x, not "his" x We have a similar problem. If we define zeration to satisfy ln(a+b)= ln(a)°ln(b) we find that a₁°a₂°a₃°...°aₓ=a+ln(x) but because I'm using "our" x instead of the x element of zeration, it should be -∞=a+ln(-∞) a=a+ln(f) a°a=a+ln(f°f) a°a°a=a+ln(f°f°f) a°a°a°a=a+ln(f°f°f°f) where f is the first number known by somebody who his first operation is zeration, and maybe "that" logarithm is not what we know as "our" logarithm. marraco Fellow Posts: 93 Threads: 11 Joined: Apr 2011 03/21/2015, 12:35 AM (03/20/2015, 10:41 PM)marraco Wrote: He will have a hard time, because he does not know what we call integer numbers. He only knows powers of n, so he will try to assign a=1.a a+a=n.a a+a+a=n².a a+a+a+a=n³.a If n=2 then [only we know that] the correct addition would be a=1.a a+a=2.a (a+a) + (a+a)=n².a ((a+a) + (a+a)) + ((a+a) + (a+a)) =n³.a I terms of zeration, it should be f=f+0 f°f=f+1 (f°f) ° (f°f)=f+2 ((f°f) ° (f°f)) ° ((f°f) ° (f°f)) =f+3 (f+n)°(f+n)=f+(n+1) and the natural choice for f, should be the neutral element of addition, so 0 = 0 0°0 = 1 (0°0) ° (0°0) = 1°1 = 2 ((0°0) ° (0°0)) ° ((0°0) ° (0°0)) = 2°2 =3 (0+n)°(0+n) = n°n = (n+1) marraco Fellow Posts: 93 Threads: 11 Joined: Apr 2011 03/21/2015, 01:44 AM (This post was last modified: 03/21/2015, 06:47 AM by marraco.) (03/21/2015, 12:35 AM)marraco Wrote: f=f+0 f°f=f+1 (f°f) ° (f°f)=f+2 ((f°f) ° (f°f)) ° ((f°f) ° (f°f)) =f+3 (f+n)°(f+n)=f+(n+1) and the natural choice for f, should be the neutral element of addition, so 0 = 0 0°0 = 1 (0°0) ° (0°0) = 1°1 = 2 ((0°0) ° (0°0)) ° ((0°0) ° (0°0)) = 2°2 =3 (0+n)°(0+n) = n°n = (n+1) let's call $\circ{n} \,=\, (\cdots (((-\infty \,\circ \, 0_1) \,\circ \, 0_2) \,\circ \, 0_3) \, \cdots \,\circ \, 0_n)$ then we have: ○0=-∞ ○1=0 ○2=1 ○3=1○0 ○4=2 ○5=○4○0 ○6=○5○0 ○7=○6○0 ○8=○7○0 ○16=4 ⁞ ○2 ͫ =m (this may only be valid for m ∈ ℤ) ¿can this zeration be associative? If it is associative, then this is a zeration table for °0 to °16 The blue numbers are the natural numbers. I don't know if in general ○n ∈ ℝ, so I wrote $\circ{n} \,=\, (\cdots (((-\infty \,\circ \, 0_1) \,\circ \, 0_2) \,\circ \, 0_3) \, \cdots \,\circ \, 0_n)$ If "our" logarithm is the same that "zeration" logarithm (which is dubious), then a°b=ln₂(2ᵃ+2ᵇ) ○a°○b=○(a+b) A question arises: can the symbol ○ ,used as a sign, be identified with a number or an operation? the symbol "-" can be identified with -1, and "+" with +1 and 0+ -n = -1.n +n = 0+n We can also use (/n = 1/n) or (÷n = 1/n) or (±in = ±i1.n) or Lm=ln(m) we can combine and operate those symbols -i = /i = ÷i We can also define the operation inverse of zeration this way (☺n=InverseOf(n) ⇔ ☺n○n = -∞) or (☺n = -∞☺n) (Sorry, I didn't found the combination of - and ○) Does this mean anything: -☺/i n ? Is (-☺/i n = ☺i/- n) ? From the table, ○-∞=-∞ ○0=-∞ ○1 = 0 ○2 = 1 ○4 = 2 ○2ⁿ=n but this is associated to a capricious choice of f and 2. Maybe is more "natural" to use (○eⁿ = n) or (○n = ln(n)) That would make ○e = 1 + i2.n.$^{\pi}$ ○-1 = i$^{\pi}$(1+2n) ○i = i$^{\pi}$(½+2n) e°⁻¹+1 = ○1 ^ There I'm "ignoring" that "our" logarithm may not be the same thing as "zeration" logarithm. maybe it should be necessary to specify what f and n numbers are used $a \,\circ_f^n\, b$ where f is the first number known by "zerationists", and n the one of "additionists", from "zerationists" viewpoint. the natural zeration would be $a \,\circ_{\small{0}}^{\small{e}}\, b$ or $a \,\circ_{\small{0}}^{\small{2}}\, b$ This shows that zeration may be as complex as exponentiation, so maybe addition is the real "zero" operation (the simplest one), and zeration is the "negation", or "real zeration" lies between addition and product. $^{ \circ_{-1} \,\leftarrow\, +_0 \,\rightarrow\, \times _1 \,\rightarrow\, \wedge\,_2 \,\rightarrow\, \wedge\wedge\,_3}$ or $^{ \circ_{-2} \,\leftarrow\, +_{-1} \,\leftarrow\, \small{''real\,zeration''}_0 \,\rightarrow\, \times _1 \,\rightarrow\, \wedge\,_2 \,\rightarrow\, \wedge\wedge\,_3}$ marraco Fellow Posts: 93 Threads: 11 Joined: Apr 2011 03/21/2015, 04:10 AM (This post was last modified: 03/21/2015, 04:18 AM by marraco.) (03/21/2015, 01:44 AM)marraco Wrote: the natural zeration would be $a \,\circ_{\small{0}}^{\small{e}}\, b$ or $a \,\circ_{\small{0}}^{\small{2}}\, b$ If we take this definition for zeration, then this may be a link between zeration and tetration. If we represent iterated zeration symbol this way: $(\circ_{\small{0}}^{\small{a}})^n1 \,=\, \circ_{\small{n}}\circ_{\small{n-1}}...\circ_{\small{1}} 1$ then $(\circ_{\small{0}}^{\small{a}})^n1 \,=\,\circ^n\,1 \,=\, ^{-n}a$ which is valid if tetration to negative integer exponents is taken as $^{-n}a \,=\, ln_{a_n}(ln_{a_{n-1}}(ln_{a_{n-2}}(...ln_{a_0}(1=e^{i2m\pi})...))))$ as explained here http://math.eretrandre.org/tetrationforu...66#pid7666 MphLee Fellow Posts: 95 Threads: 7 Joined: May 2013 03/21/2015, 11:53 AM (This post was last modified: 03/21/2015, 11:56 AM by MphLee.) wow, I had this idea at least 3 years ago but then I worked on other stuff and completely forgot it! I happy that this idea come up again. Btw Marraco, have you checked this related threads at Tetrationforum? Distibutive property (on Bennet Hyperoperations, homomorphic defined operations and Tommy's distributive property) http://math.eretrandre.org/tetrationforu...hp?tid=520 Zeration=Inconsistent (on Zeration from Godstein classical successor to Rubtsov-Romerio's preadditive paradigm) http://math.eretrandre.org/tetrationforu...hp?tid=926 PS: you can use "\ominus"$a \ominus b=...$ MathStackExchange account:MphLee tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 03/21/2015, 11:11 PM Not that I have become a " believer " but apart from a[0]b = max(a,b) + 1 reference : {max,+} algebra , linear algebra. a[0]b = ln( exp(a) + exp(b) ) reference : Bennet Hyperoperations, homomorphic defined operations and Tommy's distributive property. Another thing makes sense , and will probably lead to the above 2 ... but if not that would be intresting ( I doubt it ! ) : Rethinking : what is required for an operator [q] such that a[q]b = b[q]a and this is somehow related to iterations ? Probably this equation : Let N be a " neutral element " as a function of q. Call this N_q. then a[q]b = b[q]a implies : f and g are some function : f^[a - N_q](N_q[q]b) = g^[b - N_q](N_q[q]a) that makes sense ! example [q] = + a+b = b+a "add+1"^[a - 0](0+b) = "add+1"^[b - 0](0+a) example [q] = * a*b = b*a "add+b"^[a-1](1*b) = "add+a"^[b-1](1*a) However notice that an expression like f^[a - oo](...) does not make sense as a nonconstant function. Also combining " all nice properties " seems impossible for zeration. For instance max(a,b)+1 is not an analytic function. Or a^^b =/= b^^a ... regards tommy1729 marraco Fellow Posts: 93 Threads: 11 Joined: Apr 2011 03/23/2015, 07:58 AM (03/21/2015, 11:53 AM)MphLee Wrote: wow, I had this idea at least 3 years ago but then I worked on other stuff and completely forgot it! I happy that this idea come up again. Btw Marraco, have you checked this related threads at Tetrationforum?I'm just starting. I didn't even read this entire thread. marraco Fellow Posts: 93 Threads: 11 Joined: Apr 2011 03/23/2015, 08:05 AM (This post was last modified: 03/23/2015, 12:30 PM by marraco.) (03/21/2015, 11:11 PM)tommy1729 Wrote: f^[a - N_q](N_q[q]b) = g^[b - N_q](N_q[q]a) I have a hard time following that. ¿Is it latex? Can you show me the thread were that notation is explained? MphLee Fellow Posts: 95 Threads: 7 Joined: May 2013 03/23/2015, 09:00 AM He means $f^{a - N_q}(N_q[q]b) = g^{b - N_q}(N_q[q]a)$ where $f^x$ is the xth iteration of the function f and [q] is the q-rank hyperoperation. MathStackExchange account:MphLee marraco Fellow Posts: 93 Threads: 11 Joined: Apr 2011 03/23/2015, 01:39 PM (This post was last modified: 03/23/2015, 01:56 PM by marraco.) (03/23/2015, 09:00 AM)MphLee Wrote: He means $f^{a - N_q}(N_q[q]b) = g^{b - N_q}(N_q[q]a)$ where $f^x$ is the xth iteration of the function f and [q] is the q-rank hyperoperation. Oh. Thanks. If the minus symbol do ever means subtraction, and not the inverse operation of [q] (which one?, some [q] have many inverses and even multivalued inverses), then it does not make sense, as tommy1729 says: (03/21/2015, 11:11 PM)tommy1729 Wrote: a*b = b*a "add+b"^[a-1](1*b) = "add+a"^[b-1](1*a)That would mean b*(a-1)=a*(b-1) but if "-" is an inverse operator of [q], such that x-y is defined as x-y=x[q]-y, and -y is defined as y[q]-y=N(q,y) then for [q]=product, "-" would be division, so (03/21/2015, 11:11 PM)tommy1729 Wrote: "add+b"^[a-1](1*b) = "add+a"^[b-1](1*a) would be "add+b"^[a/1](1*b) = "add+a"^[b/1](1*a) That would mean b*(a/1)=a*(b/1) (03/21/2015, 11:11 PM)tommy1729 Wrote: However notice that an expression like f^[a - oo](...) does not make sense as a nonconstant function.then it would make sense if "-" is the inverse operator of zeration. -∞ is used on all the basic operations: the neutral of addition is ÷∞=1/∞=0 the neutral of product is ∞√=∞√n=n^÷∞=1 the neutral of exponentiation is $^{-\infty}n=n^{\frac{1}{n}}$ from product viewpoint, all numbers smaller than 0 are transfinite. from exponentiation of n viewpoint, all numbers smaller than 1 are transfinite. from tetration of n viewpoint, all numbers smaller than $n^{\frac{1}{n}}$ are transfinite. « Next Oldest | Next Newest »

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