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Suppose we have a function defined like so;

$$f_n( x, x ) = f_{n+1}( x, 2 )$$

$$f_n( x, x, x ) = f_{n+1}( x, 3 )$$

$$f_n( x, x, x, x,\dots) = f_{n+1}( x, \text{number of} x\text{'s} )$$

$f_1( x, y ) = x + y$

We know that $f_2(f_1( x, y ), a ) = f_1( f_2( x, a ), f_2( y, a ) )$ because of the distributive property, but do all functions have this property?

Does $$f_{n+1}( f_n( x, y ),a ) = f_n( f_{n+1}( x, a ), f_{n+1}( y, a ) )$$ for any $n$?

mr_e_man
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Caleb Thoburn
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  • Does this answer your question? https://math.stackexchange.com/questions/1215781/example-of-algebraic-structure-that-is-non-distributive-for-both-distributive-la – CyclotomicField Apr 20 '23 at 16:16
  • No, I am inquiring about functions that are "repeated" forms of either addition, or other repeated functions (multiplaction, exponents, tetration, ect) The funciton used to answer this post mixes addition and multiplacation functions instead. – Caleb Thoburn Apr 20 '23 at 16:23
  • I'm not sure I follow. Let $f_1(x,y)=2x+3x+1$ so that $f_1(x,x)=5x+1$. Should $f_1(x,x,x)=5(5x+1)+1 =25x+6$? What is $f_2$ in this case? – CyclotomicField Apr 20 '23 at 17:21
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    @CyclotomicField - It is given that $f_1$ is addition. – mr_e_man Apr 20 '23 at 17:52
  • @CalebThoburn - So $f_1(x,y)=x+y$, and $f_2(x,y)=x\cdot y$, and $f_3(x,y)=x^y$. But then what is $f_4(x,3)$, for example? You say it's $f_3(x,x,x)$, but does that mean $x^{(x^x)}$ or $(x^x)^x$? Exponentiation is not associative. – mr_e_man Apr 20 '23 at 17:57
  • @mr_e_man f sub 4 is tetration, or repeated exponents f sub 4 (x,3) would be x^x^x, which is always read as(x^x)^x – Caleb Thoburn Apr 20 '23 at 18:03
  • Tetration would be $x^{(x^x)}$. – mr_e_man Apr 20 '23 at 18:08

1 Answers1

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I'll use infix notation for the hyperoperations:

$$a\times_0b=1+b$$ $$a\times_1b=a+b$$ $$a\times_2b=ab$$ $$a\times_3b=a^b$$ $$\qquad\qquad\qquad\qquad\quad a\times_4b=a^{(a\times_4(b-1))},\quad a\times_41=a$$

We know that multiplication distributes over addition, on both sides. Exponentiation distributes over multiplication on the right but not on the left:

$$(x\cdot y)^a=(x^a)\cdot(y^a)$$ $$\qquad\qquad a^{(x\cdot y)}\neq(a^x)\cdot(a^y)=a^{(x+y)}$$

And addition distributes over succession:

$$(x\times_0y)\times_1a=(1+y)+a$$ $$(x\times_1a)\times_0(y\times_1a)=1+(y+a)$$ $$a\times_1(x\times_0y)=a+(1+y)$$ $$(a\times_1x)\times_0(a\times_1y)=1+(a+y)$$

What about tetration?

$$(x^y)\times_4a\overset?=(x\times_4a)^{(y\times_4a)}$$ $$a\times_4(x^y)\overset?=(a\times_4x)^{(a\times_4y)}$$

It's not right-distributive, as we can see with $a=2,x=2,y=3$:

$$(2^3)\times_42\overset?=(2\times_42)^{(3\times_42)}$$ $$(2^3)^{(2^3)}\overset?=(2^2)^{(3^3)}$$ $$2^{(3\cdot2^3)}\overset?=2^{(2\cdot3^3)}$$ $$2^{24}\neq2^{54}$$

And it's not left-distributive either, as we can see with $a=2,x=1,y=1$:

$$2\times_4(1^1)\overset?=(2\times_41)^{(2\times_41)}$$ $$2\neq2^2=4$$

mr_e_man
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  • Note that we only used tetration in the form $x\times_42=x^x$, so it doesn't matter whether it associates right or left ($x\times_43=x^{(x^x)}$ or $(x^x)^x$). In either case it doesn't distribute. – mr_e_man Apr 20 '23 at 19:40