0

In general, if $(R, +, *)$ is a ring, then for all $i, j \in \mathbb{Z}$ and $r \in R$, $$(i \cdot r) * (j \cdot r) = (ij)*r \text,$$ where $i \cdot r$ denotes the $i$th power of $r$ as an element of $(R, +)$. A product of sums is a sum of products.

If the ring is $\mathbb{R}$, then there is an additional property that $$i \cdot r = i * r \text.$$ This makes sense: multiplication is iterated addition. $4 \cdot 3 = 3 + 3 + 3 + 3 = 4 * 3$. What is a rigorous proof of this property?

I expect this proof will require some construction of the real numbers; I don't have a preference for any particular construction. Or maybe this is true for any field.

jskattt797
  • 1,731

1 Answers1

1

I think this is an interesting question, and the Q&A here goes a long way to understanding it: In a ring, result of multiple (of "addition" operation) is not the same as result of multiplication, correct?

Essentially, for any ring with unity $(R, +_R, *)$ we can say that $$n\cdot x = n=\underbrace{x+_Rx+_R\cdots+_Rx}_{n\text{ times}}= n*x$$ so long as we identify that $n$ on the RHS is defined as $n \cdot 1$. The reason this is necessary is because the expression $n*x$ is only meaningful if $n$ is an element of the (arbitrary) ring (and the natural number $n$ is not necessarily an element of said ring).

So certainly for the field $\mathbb{R}$, this will hold. To prove this then, we have to prove that a natural number $n$, is equal to $n \cdot 1$, where we are now working in the field $\mathbb{R}$.

But in some systems, by its very definition $2 = 1+1$. Particularly, using Peano axioms you can show using successor functions (and then assigning the symbols $1,2,..$ etc to these objects) that $n = n \cdot 1$ will hold for any natural number $n$. Thus the formula holds for any natural number $n$.

masiewpao
  • 2,217