2

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

Key Flex
  • 9,475
  • 7
  • 17
  • 34
Carlos Bacca
  • 638

1 Answers1

1

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.

Boshu
  • 1,476