$(\mathbb R, \oplus)$ is a group. Define a multiplication with which we get a field. Where $a \oplus b = a + b +1$

Question

I have this problem from the book Real Analysis of Miklos Laczkovich (p. $35$):

Let $\oplus : \mathbb R \times \mathbb R\to \mathbb R $, where $(a,b) \mapsto a + b +1$. Then, $(\mathbb R, \oplus)$ is a group. Define a multiplication with which we get a field.

But in the exercise there are no hints. Could anyone help me?

Answer 1

If you consider $a\otimes b= ab+a+b$. Is clear that is a commutative operation because of the structure. Notice that:

$$(a\otimes b) \oplus (a\otimes c)= (ab+a+b)\oplus (ac+a+c)=(ab+a+b)+(ac+a+c)+1$$ $$=a(b+c)+2a+b+c+1$$ That is equal to $$a\otimes(b\oplus c)=a\otimes(b+c+1)=a(b+c+1)+a+(b+c+1)$$ $$=a(b+c)+2a+b+c+1$$

Finally look that $$a\otimes(b\otimes c)=a\otimes (bc+b+c)=a(bc+b+c)+a+(bc+b+c)$$ $$=abc+ab+ac+bc+a+b+c$$ And $$(a\otimes b)\otimes c=(ab+a+b)\otimes c=(ab+a+b)c+(ab+a+b)+c$$ $$=abc+ab+ac+bc+a+b+c$$

Answer 2

The most straightforward approach to solving this problem is to prove $(\mathbb{R}, \oplus) \cong (\mathbb{R}, +)$. Once you've done that, you can use the isomorphism to define $\otimes$ in terms of $\times$.

Answer 3

Let us start by noticing that $$(a-1)\oplus(b-1)=(a-1)+(b-1)+1=(a+b)-1.$$ This means that $f\colon\mathbb R\to\mathbb R$ defined by $f(x)=x-1$ is a group isomorphism between $(\mathbb R,+)$ and $(\mathbb R,\oplus)$. (The map $f$ is bijection and fulfills $f(a+b)=f(a)\oplus f(b)$.) And the inverse $g(x)=x+1$ is an isomorphism $(\mathbb R,\oplus)$ to $(\mathbb R,+)$.

I had a brief glance at the book mentioned in the question - it seems that the notion of of group and the notion of isomorphism is not introduced in the book at this point. (Of course, I might have missed it.) But the basic intuition can be explained in a few words. We are basically saying that after "renaming" the objects the two structures $(\mathbb R,+)$ and $(\mathbb R,\oplus)$ are the same.

Just notice this part of Cayley table showing results of operation. (Of course, since we are working with an infinite set, we cannot put the whole set $\mathbb R$ into a finite table.)

$$ \begin{array}{cc} \begin{array}{c|ccccc} + &-2 &-1 & 0 & 1 & 2 \\\hline -2 &-4 &-3 &-2 &-1 & 0 \\ -1 &-3 &-2 &-1 & 0 & 1 \\ 0 &-2 &-1 & 0 & 1 & 2 \\ 1 &-1 & 0 & 1 & 2 & 3 \\ 2 & 0 & 1 & 2 & 3 & 4 \end{array} & \begin{array}{c|ccccc} \oplus &-3 &-2 &-1 & 0 & 1 \\\hline -3 &-5 &-4 &-3 &-2 &-1 \\ -2 &-4 &-3 &-2 &-1 & 0 \\ -1 &-3 &-2 &-1 & 0 & 1 \\ 0 &-2 &-1 & 0 & 1 & 2 \\ 1 &-1 & 0 & 1 & 2 & 3 \end{array} \end{array} $$ You can see that if we replace in the first table each occurrence of $-4$ by $-5$ and similarly $-3\mapsto-4$, $-2\mapsto-3$, $-1\mapsto-2$, etc., then we get the second table.

In this sense, the two operations are "the same", we just "renamed" the elements. (And $x\mapsto x-1$ is the dictionary.)

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Now we know that $(\mathbb R,+,\cdot)$ is a field. We also know that using "renaming" $f\colon x\mapsto x-1$ creates $\oplus$ from $+$. What happens if we apply the same thing to the usual multiplication $\cdot$? $$a\odot b = (a+1)\cdot(b+1)-1=ab+a+b$$ or $$a\odot b +1 = (a+1)\cdot(b+1) +1.$$ Therefore $a\odot b=ab+a+b$ seems like a natural candidate for the multiplication. (Notice that I have used the inverse mapping $x\mapsto x+1$.)

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This can be formulated a bit simpler in case you are already familiar with the notion of group isomorphism and field isomorphism. I have posted a bit more lengthy answer on purpose. (I was not sure whether the OP is familiar with these notions. And for the readers who are already familiar with them, there already is a shorter answer. Perhaps this longer answer might be useful for some people reading this question, too.)

I will add the link to this post, which might also help to gain some insight: Intuition on group homomorphisms.

And I will also add another link to a post where at least some of the field axioms for $(\mathbb R,\oplus,\odot)$ are verified: Prove that R is a ring under 'special' definitions of multiplication and addition.

Answer 4

Hint $ $ Any set bijection $\,h\,:\,R'\to R\,$ serves to transport the (ring) structure of $\,(R,+,*,0,1),$ to $\,(R',\oplus,\otimes,0',1') \,$ by simply defining the operations in $\,R'\,$ to make $\,h\,$ be a (ring) isomorphism

$$\begin{align} &a \oplus b\, =\ h^{-1}(h(a) + h(b)),\quad 0' = h^{-1}(0)\\ &a \otimes b\, =\, h^{-1}(h(a)\, * h(b)),\quad\, 1' = h^{-1}(1)\end{align}\qquad $$

Yours is the special case $\ h(x) = x+1.$

As an example let's explicitly examine how the associative law is transported from $R$ to $R'$.

$$ \begin{align} h(a\oplus(b\oplus c)) \,&=\, h(a)+h(b\oplus c)\\ &=\, h(a)+(h(b) + h(c))\\ &=\, (h(a) + h(b))+h(c)\\ &=\, h(a\oplus b)+h(c)\\ &=\, h((a\oplus b)\oplus c)\end{align}\qquad $$

Therefore we conclude that $\ a\oplus (b\oplus c) = (a\oplus b)\oplus c\ $ by applying $h^{-1}$ to above. The other (universal equational) group axioms transport in exactly the same way.