In the set of real numbers, I wonder whether the distributive law uniquely determines multiplication. Suppose that for a function $f$: $\Bbb{R}\times\Bbb{R}$ $\to$ $\Bbb{R}$ the following hold for every $x,y,z$, where $+$ is the usual addition (as defined via Cauchy sequences of rationals), and $1$ is the known natural number:

  1. $f(x+y,z) = f(x,z) + f(y,z)$
  2. $f(x, y+z) = f(x,y) + f(x,z)$
  3. $f(1,x) = f(x,1) = x $

From the above does it follow that $f(x,y) = xy$, the usual multiplication?

In this post: Are the addition and multiplication of real numbers, as we know them, unique?
a somewhat related "dual" question is asked concerning addition, and a simple solution is given in the form of $(x^3+y^3)^{1/3}$. So am I missing something obvious here?

  • $\begingroup$ We can do that for natural numbers then for rationals hence for real numbers $\endgroup$
    – IrbidMath
    Commented Aug 8, 2019 at 19:36
  • 2
    $\begingroup$ @Ameryr: no, we can only do it if we know that $f$ is continuous. See the answer from Gae.S. $\endgroup$
    – TonyK
    Commented Aug 9, 2019 at 20:34

3 Answers 3


For one thing $\Bbb R$ is a $\Bbb Q$-vector space, therefore by choosing a Hamel basis it is possible to define uncountably many symmetric $\Bbb Q$-bilinear maps $\Bbb R\times\Bbb R\to\Bbb R$, even with the restriction $\phi(1,\bullet)=\phi(\bullet,1)=id$. The only continuous one among these is the usual product, though.

  • 2
    $\begingroup$ Good answer! Would be good to detail the way to define a $\mathbb Q$-bilinear map from a Hamel basis and what a Hamel basis is, as the OP is probably not familiar with it. $\endgroup$ Commented Aug 8, 2019 at 19:21
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    $\begingroup$ @exp8j: A Hamel basis is trivially obtained by transfinite recursion or Zorn's lemma. You of course have to learn them first, and then such facts would become trivial to you. Naturally, before you can learn about them you would need to work in a system that actually supports them, such as ZFC. So unfortunately you cannot avoid learning basic set theory as well. If you want a quick route, I suggest first learning precisely what ZFC is (the axiom schemas plus a deductive system for FOL), and then carefully going through the proofs of transfinite recursion and Zorn's lemma. $\endgroup$
    – user21820
    Commented Aug 9, 2019 at 4:27
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    $\begingroup$ Is it possible to explicitly construct such a bilinear map? Or is it one of those things we know must exist via the Axiom of Choice, but that gives you a headache if you think about it too hard? $\endgroup$ Commented Aug 9, 2019 at 13:15
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    $\begingroup$ I don't think it's necessary to learn foundations of set theory to understand what a Hamel basis is. You just need to take as an axiom that every $\mathbb{Q}$-vector space has a basis, and then in particular $\mathbb{R}$ has one. $\endgroup$
    – hunter
    Commented Aug 9, 2019 at 16:09
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    $\begingroup$ (Not to bash on set theory, which is great, just, if one is only trying to prove things about vector spaces, the best formulation of the axiom of choice is "every vector space has a basis." There's no need to understand why this is equivalent to all the other formulations.) $\endgroup$
    – hunter
    Commented Aug 9, 2019 at 16:10

First for a natural $n$ $f(x,n)= \underbrace{f(x,1)+...+f(x,1)}_{n \text{. times}}=nx$

Then for a rational $1/n$ $x=f(x,1)=f(x,n/n) = n f(x,1/n) $ so $f(x,1/n) =x/n$

Now for a $y$ which is a Cauchy sequence of rational numbers $y=\lim r_n$ $f(x,y) = f(x,\lim r_n) = \lim f(x,r_n)= \lim x r_n = xy$ pulling limit outside require the continuity of $f$

  • $\begingroup$ Thank you, this argument is pretty clear. I would accept your answer too, but it seems that I can accept only one! $\endgroup$
    – exp8j
    Commented Aug 9, 2019 at 4:36
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    $\begingroup$ This is good and provides more detail of the continuous case. However, it says nothing about the non-continuous case which Gae does. $\endgroup$
    – badjohn
    Commented Aug 9, 2019 at 9:07

It's a bilinear map with an additional constraint

This defines a unique function when at least one argument is rational, i.e. over $(\mathbb{Q} \times \mathbb{R}) \cup (\mathbb{R} \times \mathbb{Q})$.

  • $f(0, z) = f(0+0, z) = f(0, z) + f(0, z)$ using equation 1, so $f(0, z) = 0$. Likewise $f(z, 0) = 0$ using equation 2.
  • For any integer $n \ge 0$, $f(n x, z) = f(x, z) + f((n-1) x, z) = \ldots = n f(x, z) + f(0, z) = n f(x, z)$ by applying equation 1 $n$ times and elimitating $f(0, z) = 0$ thanks to the previous bullet point. Likewise using equation 2: $f(x, n z) = n f(x, z)$.
  • For any integer $n \gt 0$, $f(x/n, z) = \dfrac{n f(x/n, z)}{n} = \dfrac{f(n (x/n), z)}{n} = \dfrac{f(x, z)}{n}$ using the result in the previous bullet point. Combining those results together, for any rational number $p/q$ and any reals $x$ and $z$, $f((p/q) x, z) = (p/q) f(x, z) = f(x, (p/q) z)$.

In particular, if you plug in $x = 1$ in that last equation, you get $f(r, z) = r f(1, z) = r z$ for any rational $r$ and real $z$, and likewise $f(x, r) = r x$.

And more generally, that equation means that $f$ is a bilinear map from $\mathbb{R}$ to itself, with $\mathbb{R}$ considered a vector space over $\mathbb{Q}$. Conversely, the requirements on $f$ are satisfied by any bilinear map that satisfies the “piecewise identity” condition $\forall y, f(1, y) = f(y, 1) = y$. Can we construct such a bilinear map that isn't the identity?

Constructing other bilinear maps: finite-dimensional case

This is not at all obvious, so let's try a simpler case. Instead of working in $\mathbb{R}$ which is an infinite-dimensional vector space over $\mathbb{Q}$, let's try to construct such a map, but on a finite-dimensional space. The simplest case is 2-dimensional (a 1-dimensional space would be $\mathbb{Q}$ itself, where we already know that multiplication is the only possibility). Let's take a nice simple 2-dimensional vector space over $\mathbb{Q}$: $\mathbb{Q}(\sqrt{2})$, the algebraic extension generated by $\sqrt{2}$, i.e. the numbers of the form $r_0 + r_1 \sqrt{2}$ where $r_0$ and $r_1$ are rational. (You don't need to know anything about algebraic extensions to follow my answer, but they do provide some inspiration from looking at the problem in this particular way.) Let's write $B = \sqrt{2}$. If we write out how multiplication works on numbers decomposed as $r_0 + r_1 \sqrt{2}$, we get: $$ \begin{align} (x_0 + x_1 B) \cdot (y_0 + y_1 B) &= x_0 y_0 + x_0 y_1 B + x_1 y_0 B + x_1 y_1 B^2 & \text{} \\ &= (x_0 y_0 + 2 x_1 y_1) + (x_0 y_1 + x_1 y_0) B & \text{because \(B^2 = 2\)} \\ \end{align}$$ We already know that multiplication is a solution of the original functional equations, but let's prove it anyway, using this decomposition of number in $\mathbb{Q}(\sqrt{2})$ as $r_0 + r_1 \sqrt{2}$:

  • Equation 1: $((x_0 + x_1 B) + (y_0 + y_1 B)) \cdot (z_0 + z_1 B) = ((x_0 + y_0) + (x_1 + y_1 B)) \cdot (z_0 + z_1 B) = ((x_0 + y_0) z_0 + 2 (x_1 + y_1) z_1) + ((x_0 + y_0) z_1 + (x_1 + y_1) z_0) B = (x_0 + x_1 B) \cdot (z_0 + z_1 B) + (y_0 + y_1 B) \cdot (z_0 + z_1 B)$
  • Equation 2 is just like equation 1, only with the order flipped.
  • $(x_0 + x_1 B) \cdot 1 = (x_0 + x_1 B) \cdot (1 + 0 \cdot B) = (x_0 \cdot 1 + 2 x_1 \cdot 0) + (x_1 \cdot 1 + x_0 \cdot 0) B = x_0 + x_1 B$ and likewise for the symmetric equation.

That was a boring proof. But it has a remarkable property: we did not use the fact that $B^2 = 2$ at any point! So if we take any two-dimensional vector space over $\mathbb{Q}$ with a basis $(1, B)$ and define $f(\langle x_0, x_1 \rangle, \langle x_0, x_1 \rangle) := \langle x_0 y_0 + 2 x_1 y_1, x_0 y_1 + x_1 y_0 \rangle$, we get a solution of the functional equations. In $\mathbb{Q}(2)$, this happens to be multiplication, but in any other two-dimensional space over $\mathbb{Q}(2)$, for example $\{r_0 + r_1 \sqrt{3}\}$ or $\{r_0 + r_1 \sqrt[3]{2}\}$ or $\{r_0 + r_1 \pi\}$, this is a different function.

So in spaces that are larger than $\mathbb{Q}$ but smaller than $\mathbb{R}$, there are alternative solutions. What about $\mathbb{R)$?

Constructing other bilinear maps on $\mathbb{R)$

The natural question at this point is if we can extend the example we already have. Can we define a bilinear function that maps $(x_0 + x_1 \sqrt{3}, y_0 + y_1 \sqrt{3})$ to $x_0 y_0 + 2 x_1 y_1 + (x_0 y_1 + x_1 y_0) \sqrt{3}$, and is boring old multiplication on numbers that aren't “related” to $\sqrt{3}$? We'd like to have a unique decomposition of real numbers as $x = \langle x_0, x_1, x_2 \rangle = x_0 + x_1 \sqrt{3} + x_2$ where $x_0$ and $x_1$ are rational. We could do that if we could extend $(1, \sqrt{3})$ into a basis of $\mathbb{R}$ as a vector space over $\mathbb{Q}$: then $x_0$ would be the projection on the $1$-axis, $x_1 \sqrt{3}$ would be the projection on the $\sqrt{3}$-axis, and $x_2$ would be the projection on the complement subspace of $\mathbb{Q} + \sqrt{3}$.

It turns out that the existence of such a basis is not provable using the “everyday” axioms of mathematics. (And I think the existence of an alternate $f$ is not provable either, but I'm not sure.) The existence of such a basis is equivalent a form of the axiom of choice (the axiom of choice is equivalent to stating that every vector space has a basis).

Multiplication is the only continuous solution

We've seen that all solutions must be identical to multiplication over $(\mathbb{Q} \times \mathbb{R}) \cup (\mathbb{R} \times \mathbb{Q})$. This is a dense subset of $\mathbb{R} \times \mathbb{R}$. A continuous function is uniquely defined by its values on a dense subset of its domain. Therefore multiplication is the only continuous solution.

Multiplication is the only solution obeying the sign rule

Multiplication is distributive over addition. It also has an important algebraic property in relation to the order over numbers: the sign rule, that states that multiplying two numbers of the same sign gives a positive result, and multiplying two numbers of opposite signs gives a negative result.

In addition to the original equations, let's require that $f(x, z) \ge 0$ if $x \ge 0$ and $y \ge 0$ (positive sign rule).

Take $x \le y$ and $z \ge 0$. By linearity, $f(y - x, z) = f(y, z) - f(x, z)$. Since $y - x \ge 0$, by the positive sign rule, $f(y - x, z) \ge 0$. Therefore $f(y, z) \ge f(x, z)$.

Let's put everything together now. Take $r \le x \le s$ with $r$ and $s$ rational, and take $z \ge 0$. We know that $f(r, z) = r z$ and $f(s, z) = s z$, and now we know that $f(r, z) \le f(x, z) \le f(s, z)$, so $r z \le f(x, z) \le s z$. Now, given a real number $x$, let's consider its decimal approximations: let $r_n$ be the decimal approximation of $n$ to $n$ decimals rounded down, and $s_n$ rounded up. We have $r_n z \le f(x, z) \le s_n z$ for all $n$. The only number that has this property is $x z$, because $\lim_{n\to\infty} r_n z = \lim_{n\to\infty} s_n z = x z$. So $f$ is multiplication over all of $\mathbb{R} \times \mathbb{R}_+$. And since $f(x, -z) = - f(x, z)$, $f$ is multiplication over $\mathbb{R} \times \mathbb{R}_-$ as well.

A shorter proof if you have more advanced tools is that a real number is uniquely determined by the set of rationals that are less or equal to it (that set and its complement form a Dedekind cut of the rationals). Since $f(\bullet, z)$ is monotonic, it's fully determined by its values over the rational.

The fact that the sign rule and continuity each on its own determine $f$ uniquely is not a coincidence. You may even have observed some similarities in the proofs: with continuity, the function is uniquely determined by its value on a dense subspace (for the chosen topology), which you can show by constructing convergent sequences. With order, the function is uniquely determined by its value on a dense subset (for the chosen order), which you can show by constructing a series of approximations. The two conditions are effectively a restatement of each other plus a little algebra to transform the sign rule into a monotonicity condition, because the usual topology on the reals is the order topology.

  • $\begingroup$ Thank you for these eye-opening arguments, esp. the proof that the sign-rule via monotonicity implies the continuity of $f$. $\endgroup$
    – exp8j
    Commented Aug 11, 2019 at 10:42

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