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**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 $\displaystyle x\in A$ then $\displaystyle x\in B$

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

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

Since this works for all y, we get $\displaystyle x\in B$