Math Help - Proof

1. Proof

Using any property of numbers that may be needed, show that

$\{x\in\mathbb{R}|\text{for a real number y}, x=y^2\}=\{x\in\mathbb{R}|x\geq0\}$

OK, I'm very new to this, so bear with me. I think that I have one direction of the proof done...

For the sake of LaTeX, Let $A$ equal the left hand side.

Let $x\in{A}$

Since $x=y^2\Longrightarrow{x}\geq0$ $\forall{y}\in\mathbb{R}\Longrightarrow{x}\in{B}\Lo ngrightarrow{A}\subseteq{B}$.

So, I know that we've gotta do the other direction...

Did I even do the first part right?

2. that first part looks right, depending on how rigorous you need it to be, but thats the exact idea, the squares of every real number is greater than or equal to zero, so if $x\in A$ then $x\in B$

For the other direction, Let $y$ be a real number and let $x\in B$

Since $x \geq 0$, there is some $y\in\mathbb{R}$ s.t. $y^2=x$

Since this works for all x, we get $x\in A$

3. The above reply has logic error. (what are we supposed to prove?)

4. where's the logic error? the exercise is to prove to two sets are equal, meaning show 2 inclusions

5. Originally Posted by Shanks
The above reply has logic error. (what are we supposed to prove?)
That $A=B$. What else?

6. Originally Posted by artvandalay11
that first part looks right, depending on how rigorous you need it to be, but thats the exact idea, the squares of every real number is greater than or equal to zero, so if $x\in A$ then $x\in B$

For the other direction, Let $y$ be a real number and let $x\in B$

Since $y^2\geq 0$, there is some $x\in B$ s.t. $x=y^2$

Since this works for all y, we get $x\in B$
For the other direction, we need to show If x is in B, then x is in A.

7. good call, i fixed it... always remember which part you are assuming