What is the solution to this week's Problem of the Week?

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    2016
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SUMMARY

This week's Problem of the Week (POTW) involves a set \( S \) with two binary operations, \( \# \) and \( * \), governed by specific axioms. The axioms establish the existence of an element \( z \) such that \( z \# s = s \) for all \( s \in S \), and provide conditions for uniqueness and operation interaction. The conclusion drawn from the axioms is that the only element in set \( S \) is \( z \), thus proving \( S = \{ z \} \). Notable contributors to the discussion include Opalg, kiwi, and Deveno, who provided correct solutions.

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Mathematicians, students of abstract algebra, and anyone interested in the foundations of mathematical structures will benefit from this discussion.

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Here is this week's POTW:

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A set $S$ has two binary operations $\#$ and $*$ on it, and the following axioms hold:
  1. There is an element $z$ in $S$ such that $z\# s=s$ for all $s\in S$.
  2. For all $s,t,u\in S$ if $s\# u=t\#u$ then $s=t$.
  3. For all $s,t\in S$ if $z*s=z*t$ then $s=t$.
  4. For all $s,t,u\in S$, $(s\# t)*u=(s*u)\#(t*u)$.
Prove that $S=\{z\}$.

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Congratulations to Opalg, kiwi, and Deveno for their correct solutions! Opalg's solution follows:

1. There is an element $z$ in $S$ such that $z\# s=s$ for all $s\in S$.2. For all $s,t,u\in S$ if $s\# u=t\#u$ then $s=t$.3. For all $s,t\in S$ if $z*s=z*t$ then $s=t$.4. For all $s,t,u\in S$, $(s\# t)*u=(s*u)\#(t*u)$.5. $z\#z = z$ (from 1.).6. $z*t = (z\#z)*t$ for all $t\in S$ (from 5.).7. $z*t = (z*t)\#(z*t)$ for all $t\in S$ (from 6. and 4.).8. $z*t = z\#(z*t)$ for all $t\in S$ (from 1.).9. $z = z*t$ for all $t\in S$ (from 7., 8. and 2.).10. $z = z*z$ (from 9.).11. $z*t = z*z$ for all $t\in S$ (from 9. and 10.).12. $t = z$ for all $t\in S$ (from 11. and 3.).From 12., $z$ is the only element of $S$, in other words $S = \{z\}$.
 

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