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I am curious whether in real numbers,

the multiplication as we know it is the only operation that

is distributive over addition and associative and commutative?

I heard it is. But I am not sure and is there any way to prove it?

Since the comments below shows that x*y=0 for all x, y and

x*y=r x y also satisfies all 3 laws, it is false that the multiplication is

the only operation that satisfies all 3 laws.

What about when the multiplication has an identity element which is 1,

which means that for all x, 1*x=x*1=x.

Sorry to add, but I would appreciate the proof!

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3  
What about defining all products to be $0$? –  Tobias Kildetoft Sep 10 '13 at 17:27
1  
Also $x \ast y = r x y$, for any constant $r$. –  David Speyer Sep 10 '13 at 17:29
    
Hmm. You got a point. Let's say a*1=a, which means it has an identity element, and it is 1. –  KH Kim Sep 10 '13 at 17:29
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3 Answers

up vote 7 down vote accepted

Let us call a multiplication (on the real numbers) a map $\beta : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ which is bilinear, symmetric, associative, and satisfies $\beta(a,1)=a$. Note that we have $\beta(a,b)=ab$ for $a,b \in \mathbb{Q}$, i.e. on rational numbers such a multiplication has to be the usual one. In particular: Every continuous multiplication equals the usual one on all real numbers. However, it is easy to "twist" the usual multiplication with non-continuous maps:

Choose some transcendental number, for example Euler's number $e$. Since $1,e,e^{-1}$ as well as $1,e,e^2$ are linearly independent over $\mathbb{Q}$, we can extend them to bases (Zorn's Lemma) and then find a bijective $\mathbb{Q}$-linear map $f : \mathbb{R} \to \mathbb{R}$ such that $f(1)=1$, $f(e)=e$ and $f(e^{-1})=e^2$. Then $\beta(a,b):=f^{-1}(f(a) f(b))$ is a multiplication. Since $f$ is not multiplicative (we have $f(1)=1 \neq e^3 = f(e) f(e^{-1})$), it doesn't coincide with the usual multiplication.

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Can you prove it that with the assumption of continuity, the multiplication is the only operation? And sorry for my ignorance, I want to ask if it is in rational numbers, the multiplication is the only operation without the assumption of continuity. –  KH Kim Sep 10 '13 at 17:49
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You can do the following. Let $f:\Bbb{R}\to\Bbb{R}$ be any $\Bbb{Q}$-linear bijection that satisfies $f(1)=1$, but $f$ is not the identity. Finding non-trivial examples of these is difficult, but if you believe in axiom of choice, then it follows that they exist. Then you can just pull back the usual multiplication and define a new one, call it $*$, by the recipe $$ a*b=f(f^{-1}(a)f^{-1}(b)), $$ where the multiplication inside the argument of $f$ is the usual one. We have $$ a*b=f(f^{-1}(a)f^{-1}(b))=f(f^{-1}(b)f^{-1}(a))=b*a, $$ because the usual multiplication is commutative. Associativity is "inherited" in the same way. As $f(1)=1$, the number $1$ is a neutral element for the new operation as well. Distributivity follows from the distributivity of the usual operations and the fact that $f(x+y)=f(x)+f(y)$ for all $x,y$. The multiplication will be different, if $f$ is not the identity mappig. For example, we can prove the existence of such a mapping $f$ that interchanges $\sqrt2$ and $\pi$. With that $f$ we then get $\pi*\pi=2$, as $\pi$ takes the role of $\sqrt2$.

None of the non-identity mappings $f$ is continuous (w.r.t the usual topology) though.

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"The multiplication will be different, if f is not the identity mappig." should be "The multiplication will be different, if f is not multiplicative." I had made the same mistake, and have now polished my answer. –  Martin Brandenburg Sep 10 '13 at 20:22
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After all the added assumptions by the OP, I think the problem is easier now.

Associativity, commutativity, distributivity and unity uniquely define multiplication on $\mathbb Z$.

$\mathbb Q$ is the ring of fraction of $\mathbb Z$, so it derives a unique multiplication from $\mathbb Z$.

Assuming continuity, the multiplication on $\mathbb R$ is uniquely determined by the one on $\mathbb Q$.

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You mean "Assuming continuity". –  Martin Brandenburg Sep 10 '13 at 20:35
    
@MartinBrandenburg Ok I corrected that. I am under the impression that the OP wants that. –  Tunococ Sep 10 '13 at 20:38
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