Q: Basic ring-based (set) algebra. Does this work? I am trying to figure out a basic algebraic concept in ring-based sets. I have a set $X$ with a ring operation $(X, \otimes)$. The operations are defined as follows: For all $x,y,z \in X$, $$x \otimes (y \otimes z) = (x \otimes y) \otimes z$$ $$x \otimes 1 = x$$ $$1 \otimes x = x \otimes 1 = x$$ I am having a difficult time showing associativity. Here is my attempt: Associativity: \begin{align} (x \otimes y) \otimes z &= x \otimes (y \otimes z) \\ &= (x \otimes y) \otimes z \\ &= x \otimes (y \otimes z) \\ \end{align} This is where I am stuck. Does this work? Or am I off in a totally different direction? Any help would be appreciated. A: This kind of proof does work. There is nothing wrong with it. Find us on Asexuality denied due to social bias Amanda Tan (left), an asexual student who has never kissed or held hands with a romantic partner, is joined by fellow asexual student Rayen Peiris, who joined their first social group after deciding that she wanted to date asexual people. Tan said that although she has asexual peers in her university, she is still excluded from activities due to her asexual identity. “I think that the reason I feel very uncomfortable with asexuality is due to the fact that asexuality is such a minority issue,” Tan, a 21-year-old undergraduate student at a private university in Singapore, told BuzzFeed. “As it is a smaller community, I feel like there’s not a lot of people who relate to it, or understand it.” “Asexuality is defined as not having sexual attraction to