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We all know that for every two real number $x,y$, the operation of addition $x+y$ satisfies the following conditions:

\begin{gather} x+y=y+x;\\ x+0=x;\\ (x+y)+z=x+(y+z);\\ x+(-x)=0. \end{gather}

Now, take a function $f\colon\Bbb R\times\Bbb R\to\Bbb R$ satisfying the following four conditions:

  1. $\forall x,y\in \Bbb R, f(x,y)=f(y,x);$

  2. $\forall x\in \Bbb R, f(x,0)=x;$

  3. $\forall x,y,z\in \Bbb R, f(f(x,y),z)=f(x,f(y,z));$

  4. $\forall x\in \Bbb R, f(x,-x)=0.$

Can we conclude that $f(x,y)=x+y$ for all $x,y\in \Bbb R$?

edit: Thanks to Jared, the answer is "NO".

think a function $g\colon\Bbb R\times\Bbb R\to\Bbb R$ satisfying the following four conditions:

a). $\forall x,y\in \Bbb R, g(x,y)=g(y,x);$

b). $\forall x\in \Bbb R, g(x,1)=x;$

c). $\forall x,y,z\in \Bbb R, g(g(x,y),z)=g(x,g(y,z));$

d). $\forall x\in \Bbb R,x\ne0, g(x,\frac1x)=1.$

Can we conclude that $g(x,y)=xy$ for all $x,y\in \Bbb R$?

again, two functions $f,g\colon\Bbb R\times\Bbb R\to\Bbb R$ satisfying 1,2,3,4, a),b),c),d),and

e). $\forall x,y,z\in \Bbb R, g(x,f(y,z))=f(g(x,y),g(x,z))$

Can we conclude that $f(x,y)=x+y,g(x,y)=xy$ for all $x,y\in \Bbb R$? It seems that the answer is NO, $f(x,y)=\sqrt[3]{x^3+y^3},g(x,y)=xy$ works

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  • $\begingroup$ @AhaanRungta Counterexample for multiplication $g(x,y)$ $\endgroup$ – ziang chen Dec 20 '13 at 19:46
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Counterexample:

$$f(x,y)=\sqrt[3]{x^3+y^3}$$

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    $\begingroup$ More generally, for any bijection g of $\mathbb R$ onto $\mathbb R$ which fixes the origin, the application $f(x,y)=g^{=1}(g(x)+g(y))$ will satisfies all the conditions. $\endgroup$ – Taladris Aug 11 '13 at 6:50
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    $\begingroup$ So try any bijection $g$ of $\mathbb R^+$ onto $\mathbb R^+$ extended with $g(0)=0$ and $g(-x)=-g(x)$ and then consider $f(x,y)=g^{=1}(g(x)+g(y))$ $\endgroup$ – Henry Aug 11 '13 at 9:16
  • $\begingroup$ @Andre Nicolas: yes, you are right. Thanks to Henry for the correction. $\endgroup$ – Taladris Aug 12 '13 at 2:32

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