# How prove this definition $a\oplus b=a+b$ [closed]

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

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

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Masacroso, Juniven, José Carlos Santos, user26857, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.

• what had you tried? – Masacroso Oct 29 '17 at 14:23

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

• $(0\oplus 0)\oplus c=0+0+c=c$ – Aqua Oct 29 '17 at 14:47
• I just picked $1$ for the neutral element and I chose for $\oplus$ to be the multiplication operator. Your answer works, does it not. – zoli Oct 29 '17 at 15:10
• You mean you observed $(a\odot b)\odot c= a\cdot b\cdot c$? – Aqua Oct 29 '17 at 15:13
• 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. – zoli Oct 29 '17 at 15:16
• @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. – Fritz Oct 29 '17 at 15:41