If A is the set of ordered pairs defined by $x^2 +4y^2 < 1$ and B is the set of all pairs defined by $y< 1/2$ then prove that A is a proper subset of B.
Im new to proof and not sure how this should be written, but i aimed to show all the solutions to the first inequality are solutions to the second inequality and "within" them solutions.
so i started with
$x^2 + 4y^2 <1$
$y^2 < (1-x^2)/4$
$-\sqrt{(1-x^2)/4}<y<\sqrt{(1-x^2)/4}$
not really sure where to go from here just said max value of $\sqrt{(1-x^2)/4}$ is $1/2$ i think so
$-1/2<y<1/2$
Have i done this correct, and is there anything which needs to be done to make the prrof more clear, or a better method. thanks
Notice that $A$ is an ellipse, which is entirely below the line $y=0.5$. So you should be able to see that $A$ is indeed a proper subset of $B$. To formally write the proof, show that any point in $A$ must have $y$ coordinate less than $0.5$; this follows simply from the fact that all terms on the LHS are positive. To show this is proper, find a point in $B$ not in $A$.
Your solution seems fine as well, don’t worry. I just pointed out some things along with it which you might have missed.