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Say we have the set of real numbers. Can we construct a different ring than the usual ring of real numbers?

I am trying to wrap my head around the idea of rings and I couldn't find two other operations that will hold distributive property other than usual addition and multiplication.

If we can find another ring with the set of real numbers, is there a set with only one ring structure? If we can't find another ring with the set of real numbers, is there a set with different ring structures?

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    $\begingroup$ Definitely. Consider that $\Bbb Z$ and $\Bbb Q$, for example, can be considered the same set since they are in bijection. Their usual ring structures are clearly not-equivalent since $\Bbb Q$ is a field, for example. $\endgroup$
    – pancini
    Jun 1, 2021 at 23:53
  • $\begingroup$ Any set with two elements has precisely one ring structure: $\mathbb{Z}/2\mathbb{Z}$ $\endgroup$
    – E G
    Jun 1, 2021 at 23:57

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Here is a more explicit example if my comment about $\Bbb Z$ and $\Bbb Q$ was unconvincing.

Let $S=\{a,b,c,d\}$. Let $R_1$ be the ring structure on $S$ inherited from $\Bbb Z/(2)\times\Bbb Z/(2)$ under the bijection $a\mapsto (0,0)$, $b\mapsto (1,0)$, $c\mapsto (0,1)$, and $d\mapsto (1,1)$. For example, $ab=a$, $b+d=c$, and $cd=c$.

Now let $R_2$ be the ring on $S$ inherited from $\Bbb Z/(4)$ with the bijection $a\mapsto 0$, $b\mapsto 1$, $c\mapsto 2$, and $d\mapsto 3$.

You can verify that $R_1$ and $R_2$ are not isomorphic since, for example, $R_1$ has no element of (additive) order $4$.

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The other examples given are nice, but to give an example where the set is $\mathbb{R}$ as you were originally wondering about you take $\mathbb{R}$ with the normal multiplication but where we define the multiplication operation to always evaluate to $0$. Under this definition we have a different ring structure and it's certainly not isomorphic since this structure is not a field. This isn't a particularly nice ring since it doesn't have unity.

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  • $\begingroup$ I don’t think that will be a unital ring since $(2)(2)=(1+1)(1+1)=(1)(1)+\cdots+(1)(1)=4$ $\endgroup$
    – pancini
    Jun 2, 2021 at 0:09
  • $\begingroup$ @ElliotG correct, but OP didn't ask for that $\endgroup$ Jun 2, 2021 at 0:18
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Given any bijection $\,f:\mathbb{R}\to\mathbb{R},\,$ define new addition and multiplication operations by $\,x\oplus y:=f^{-1}(f(x)+f(y))\,$ and $\,x\otimes y:=f^{-1}(f(x)\cdot f(y)).\,$ These two operations form $\,\mathbb{R}\,$ into a ring isomorphic the the usual ring of reals. As a concrete example, try $\,f(x):=x+1.\,$ This example is given in an answer to MSE question 1911294 "Operations on the vector set $\mathbb{R}$ that will provide a vector space".

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  • $\begingroup$ Being isomorphic it is not a "different ring structure". This is known as transport of structure $\endgroup$ Jun 2, 2021 at 10:33
  • $\begingroup$ @BillDubuque That depends on exactly what the definition of "different ring structure" is. The question was not precise on that point. I did mention that the new ring is isomorphic to the original ring. Finding a non trivial non isomorphic ring on $\mathbb{R}$ is a much more difficult question. $\endgroup$
    – Somos
    Jun 2, 2021 at 10:46

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