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For part b, I'm getting a little stuck. So I'm trying to show that if $(x,y,r)R(a,b,s)$ and $(a,b,s)R(x,y,r)$, then $(x,y,r)=(a,b,s)$

And so $(x,y,r)R(a,b,s)$ implies $\sqrt{(x-a)^2+(y-b)^2} \leq s-r$ and $(a,b,s)R(x,y,r)$ implies $\sqrt{(a-x)^2+(b-y)^2} \leq r-s$

After some algebra, here's what I get: For the first one, $x^2+y^2-s^2-2xa-2yb+2sr+b^2+a^2 \leq r^2$ and for the second one, $x^2+y^2-s^2-2ax-2by+2sr+b^2+a^2 \leq r^2$

and so they come out to be the same, but in an example from the notes, I think I should have came up with something like: $a \leq c$ and $c \leq a$, so $c=a$, and I didnt get anything of that form, so I'm wondering if my steps above are also correct. Have I proven anything, or if not, what have I done/assumed wrong?

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Prime: Please don't invent a tag as restrictive as "symmetric-relations". –  Arturo Magidin Dec 2 '10 at 5:56
    
Lol I cant do anything right in this place can I.. –  maq Dec 2 '10 at 5:59
    
Prime: Tags are meant to be (i) informative and to (ii) call attention to your question from people who are interested or knowledgeable in the topic. If you are inventing a tag, then nobody will be "watching" for the tag, so you will fail to achieve the second purpose of tags. Of course, if there are no tags that adequately describe your topic, you should create one that is informative; but here, you did have a tag ("relations"). –  Arturo Magidin Dec 2 '10 at 6:05

3 Answers 3

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Part (a) would give you the intuition. But the first thing to notice here (even before you notice that the square roots have the same value) is that since the square root is always nonnegative, the first condition gives you that $s-r\geq 0$ and the second gives you that $r-s\geq 0$. Those two together tell you what about $s$ and $r$?

Once you have that, you know something about the square root; and since the quantity inside is a sum of squares, the only way that something can happen is if both squares are -fill in the blank-. Which tells you something about $x$, $y$, $a$, and $b$.

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Well we know that s and r are both positive numbers, but I'm not getting your hint to your second paragraph.. –  maq Dec 2 '10 at 6:10
    
@f-Prime: Look at the first paragraph again. If $s-r\geq 0$, then can $r$ be strictly larger than $s$? What about $r-s\geq 0$? Once you get that, you will know exactly how much that big square root is. –  Arturo Magidin Dec 2 '10 at 6:12
    
Oh ok so in the first one $s \geq r$ and in the second one $r \geq s$, so they're equal. But now what about the rest of the variables? –  maq Dec 2 '10 at 6:16
    
@f-Prime: So, if they are equal, then $s-r=0$. So that means that the square root is less than or equal to zero. When can a square root be less than or equal to zero? Only if it is the square root of zero. So the expression inside the square root is equal to zero. But the expression inside the square root is a sum of two squares. When can a sum of two squares be equal to $0$? –  Arturo Magidin Dec 2 '10 at 6:20
    
When both are zero. So just about everything in this equation is equal to zero then? –  maq Dec 2 '10 at 6:23

Your two square roots are the same, so $r=s$ follows from the antisymmetry of $\leq$

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The relation is saying that the first circle is completely included inside the second circle. If two circles are completely included in one another, they must be the same circle.

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Wait so the answer for part a is what you said above? I put down that the relation means that the distance between (x,y) and (a,b) is less than the difference of their z coordinates, what would be wrong with that? –  maq Dec 2 '10 at 6:01
    
@f-Prime Your answer is syntactic, mine is semantic. –  Yuval Filmus Dec 2 '10 at 6:10
    
@f-Prime: Also, they aren't really "z"-coordinates, they are radii of circles. –  Arturo Magidin Dec 2 '10 at 6:16
    
Can you explain how you came up with that fact that that the circles must be included in one another? I'm not used to breaking down circle equations like this. I only know the basics.. –  maq Dec 2 '10 at 6:59
    
@f-Prime Now that I have claimed it, you can try to prove it. It's no well-known property, just something that can be deduced from the definition of circle (all points at distance at most the radius from the center). –  Yuval Filmus Dec 2 '10 at 16:04

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