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It can be seen easy but i'm really stack.I don't know how to start.

Let $A \neq \emptyset $ is a subset of $ \mathbb{R} $.If for every real $x$,$y$ sum $x+y$ belongs to $A$,then $xy$ also belongs to the set $A$.Then prove that $A= \mathbb{R} $

$\mathbb{R}$- here is the set of real numbers. Thanks for help in advance!

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  • $\begingroup$ hi,i'm new here,i don't know how to use here,i will be thankful to you if you help me $\endgroup$ – perelman Sep 5 '15 at 18:31
  • $\begingroup$ Where you have considered $x,y$ in $A$ or in $\mathbb{R}$? If in A this is not true, if in $\mathbb{R}$, these assumptions are not enough to say that $A=\mathbb{R}$ $\endgroup$ – Dr. Mohammad Alomari Sep 5 '15 at 18:56
  • $\begingroup$ As I said for every real($R$) x,y $\endgroup$ – perelman Sep 5 '15 at 18:59
  • $\begingroup$ Your treatment of quantifiers is kind of sloppy. As it stands, $A$ can be any subset of reals (because either $A={\bf R}$ or there is some $x+y\notin A$. $\endgroup$ – tomasz Sep 5 '15 at 19:42
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First, since $A\ne \emptyset$, there exists a $y=0+y\in A$, so $0\times y=0\in A$.

Then, for all $y\in\mathbb R$, $0=y+(-y)\in A$, so $y\times(-y)=-y^2\in A$, i.e. all negative real numbers are in $A$.

Finally, for all $y\in\mathbb R_{<0}$, $y=(y/2)+(y/2)\in A$, so $(y/2)(y/2)=y^2/4\in A$, i.e. all positive real numbers are in $A$.

As a result, $A=\mathbb R$.

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Let $a\in A$ be an element. Then to show that $x\in\mathbb{R}$ is in $A$ (for $x$ arbitrary) amounts to solving the system $$\cases{y+z = a, &\\yz = x.}$$ Can you take it from here?

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