Given:
$\mathbb{R}\times(\mathbb{R}\setminus\{0\})$ with the following operation $\circ$
$(a_1,b_1)\circ (a_2,b_2):=(a_1\cdot b_2+a_2,b_1\cdot b_2)$
I need to prove the associative property. So:
$\bigl((a_1,b_1)\circ (a_2,b_2)\bigl)\circ (a_3,b_3)=(a_1,b_1)\circ \bigl((a_2,b_2)\circ (a_3,b_3)\bigl)$
My problem is, I am not sure how to get from $((a_1\cdot b_2+a_2)\cdot b_3+a_3,(b_1\cdot b_2)\cdot b_3)$ to $(a_1\cdot(a_2\cdot b_3+a_3),b_1\cdot(b_2\cdot b_3))$, except for the second part which is trivial $(b_1\cdot b_2)\cdot b_3=b_1\cdot(b_2\cdot b_3)$.
2026-03-26 12:42:52.1774528972
Prove associative property
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Just do it.
$((a,b)\circ (c,d))\circ(e,f) = (ad+c,bd)\circ(e,f) = ((ad+c)f+e,bdf) = (adf+cf + e, bdf)$.
$(a,b)\circ((c,d)\circ(e,f)) = (a,b)\circ(cf+e,df)=(adf+cf+e, bdf)$