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I am reading "Higher Algebra" by A. Kurosh.

The following sentence is in this book:

Analogously, the associative law of addition leads to the concept of a multiple, $na$, of the element $a$ by a positive integral coefficient $n$

If the associative law doesn't hold, then we cannot define $a^n$.
Is it really true?

Original image: https://i.sstatic.net/YbaAy.png

cineel
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tchappy ha
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    Rings are usually defined to be associative ... if you want an example of a non-associative operation where this does not hold, consider the positive integers ${1, 2, 3, \ldots}$ and the exponentiation operator. $3^{3^3} = 3^{27}$, but $(3^3)^3 = 3^9$, and $3^{27} \neq 3^9$. – Joppy Jan 14 '22 at 01:14
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    @Joppy exponentiation is not distributive – Kenta S Jan 14 '22 at 01:20
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    Perhaps unsatisfying, the free nonassociative algebra on 2 generators (over your favorite field) should work – Chris Grossack Jan 14 '22 at 01:38
  • @Joppy Thank you very much. – tchappy ha Jan 14 '22 at 01:45
  • @HallaSurvivor Thank you very much for your comment. – tchappy ha Jan 14 '22 at 01:46
  • Somos, Thank you very much for your edit. – tchappy ha Jan 14 '22 at 01:48
  • @KentaS Thank you very much for your comment. – tchappy ha Jan 14 '22 at 01:55
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    The cross product of vectors in $\mathbb R^3$ is not associative. The NOR operator in logic is communative but not associative. – bof Jan 14 '22 at 02:23
  • @bof Thank you very much for your comment. – tchappy ha Jan 14 '22 at 02:26
  • If the operation is nonassociative (see dupes for examples) then there can be more than one way to associate $3\cdot a = a+a+a,,$ viz. $,(a+a)+a,$ or $,a+(a+a),$ and ditto for $,n\cdot a,$ for larger $n$. So any bracketless notation is not well-defined when the operation is not associative absent any convention, e.g. define it to be left-associated e.g. $(((a+a)+a)+a)\ \ $ – Bill Dubuque Jan 14 '22 at 03:07

1 Answers1

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Suppose we define a non-associative binary operation $\circ$ on $\Bbb R$.
$x\circ y:=x-y$
trying to figure out the third (or higher) power causes problems.
$1\circ (1\circ 1)=1-(1-1)=1$
$(1\circ 1)\circ 1=(1-1)-1=-1$
$(\Bbb R,\circ)$ is a non-associative magma, so it is not a semigroup. it is not an example of a non-associative ring or algebra though.

cineel
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