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Define $\oplus$: if for any real numbers $a,b,c$ there have $$\left(a\oplus b\right)\oplus c=a+b+c$$

show that $$a\oplus b=a+b$$

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closed as off-topic by Masacroso, Juniven, José Carlos Santos, user26857, Namaste Oct 29 '17 at 15:52

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  • $\begingroup$ what had you tried? $\endgroup$ – Masacroso Oct 29 '17 at 14:23
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Let $k =0\oplus 0$.

Since $(0\oplus 0)\oplus c=c$ we see that $k\oplus c=c$ for each $c$, so $k$ is left neutral.

Next we see that $\oplus $ is commutative: $$b\oplus c=(k\oplus b)\oplus c=k+b+c=k+c+b=(k\oplus c)\oplus b=c\oplus b$$

Now we see that $k=0$: $$(a\oplus k)\oplus k=a+2k \Rightarrow a=a+2k \Rightarrow k=0 $$

So $$a+b=a+b+0=(a\oplus b)\oplus 0=a\oplus b$$

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    $\begingroup$ $(0\oplus 0)\oplus c=0+0+c=c$ $\endgroup$ – Aqua Oct 29 '17 at 14:47
  • $\begingroup$ I just picked $1$ for the neutral element and I chose for $\oplus$ to be the multiplication operator. Your answer works, does it not. $\endgroup$ – zoli Oct 29 '17 at 15:10
  • $\begingroup$ You mean you observed $(a\odot b)\odot c= a\cdot b\cdot c$? $\endgroup$ – Aqua Oct 29 '17 at 15:13
  • $\begingroup$ Yes. But why would I use another notation? The only thing that I wanted t say was that there were two answers: $+$ and $\cdot$ with the choice of $0$ and $1$, respectively. $\endgroup$ – zoli Oct 29 '17 at 15:16
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    $\begingroup$ @zoli By assumption we have $(a \oplus b)\oplus c=a+b+c$. How do you argument that the expression $(1 \oplus 1) \oplus c=1+1+c=2+c$ is equal to $c$? I don't see that. $\endgroup$ – Fritz Oct 29 '17 at 15:41

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