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(Sorry for any spelling mistake, English isn't my mothertongue)

Today at school we took a logic exam. One of the questions asked us to find the domain and range of the relations xRy and xSy, where R = y^2 + x^2 < 4 and S = x - 2 < y^2.

Now, I know the first inequality is a circle and the second one is a parabole. I found the domain and range of both binary relations easily by graphing them. However, when I was asked to find the domain and range of the intersection between R and S, I was completely lost. The professor taught us that we can find the intersection by plugging one of the variables of one relation into the other relation if we were working with equations, but is that correct if both are inequalities?

Thanks in advance.

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  • $\begingroup$ A relation cannot equal an inequality. R = x + y < 3 is nonsense unless you are defining R to be x + y. However x + y is not a relation. $\endgroup$ – William Elliot Jun 28 at 1:19
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You give us $R=\{(x,y):x^2+y^2<4\}$ and $S=\{(x,y):x-2<y^2\}$

I found the domain and range of both binary relations easily by graphing them.

Look at those graphs. Better yet, plot them on the same plane.

$R$ is a disc; the region inside the circle centred on the origin with a radius of 2.

$S$ is the semi-infinite region to the left of a parabola, which has a vertex at $(2,0)$ and pivots around the positive x-axis.

$R$ is thus a proper subset of $S$, and hence the intersection of $R$ with $S$ is exactly $R$.

$$R\cap S=R$$

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  • $\begingroup$ Perfectly correct, +1, but maybe the OP would benefit from the same proof in algebraic terms: The inequality defining $R$ implies $x<2$ (because if $x\geq 2$ then $x^2+y^2\geq x^2\geq2^2=4$), which in turn implies $x-2<0\leq y^2$. So (as in the geometric proof) $R\subseteq S$ and therefore $R\cap S=R$. (I must admit that the geometric argument is probably easier to find than the algebraic one.) $\endgroup$ – Andreas Blass Jun 28 at 2:20
  • $\begingroup$ This is an amazing answer. I plotted both relations on the same graph and I could see what you mean easily. However, I wonder how this works if the answer is not evident on the graphic. For example, if the parable had the vertex at (-2;0), would the intersection be empty? In that case, both relations would share the point on x = -2 but this is < and not ≤, so -2 is empty. $\endgroup$ – Augusto Jun 28 at 3:03
  • $\begingroup$ @Augusto No, not quite that. $(-2,0)$ would not be in the intersection, but other points would. Look at the plot: here or $x+2<y^2<4-x^2$ $\endgroup$ – Graham Kemp Jun 28 at 3:09

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