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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?

Arturo Magidin
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Snowman
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  • Prime: Please don't invent a tag as restrictive as "symmetric-relations". – Arturo Magidin Dec 02 '10 at 05:56
  • Lol I cant do anything right in this place can I.. – Snowman Dec 02 '10 at 05:59
  • 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 02 '10 at 06:05

3 Answers3

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

Ross Millikan
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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 $b$.

Arturo Magidin
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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.. – Snowman Dec 02 '10 at 06:10
  • @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 02 '10 at 06:12
  • 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? – Snowman Dec 02 '10 at 06:16
  • @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 02 '10 at 06:20
  • When both are zero. So just about everything in this equation is equal to zero then? – Snowman Dec 02 '10 at 06:23
  • @f-Prime. Which equation? There are lots of equations here. "$s=r$" is an equation. But the point is not whether they are equal to zero or not, it's what you can conclude from that. Since both squares have to be zero, that means that $(x-a)^2 = 0$; that tells you something about $x$ in relation to $a$. And $(y-b)^2 = 0$; that tells you something about $y$ in relation to $b$. Remember you were asking about "the other variables"? Well, there you are. – Arturo Magidin Dec 02 '10 at 06:24
  • Well I mean all the variables would be equal to 0. If $(x-a) +...$ is equal to zero, and $(a-x)+...$ is equal to zero, then a and x must be zero, right? And same goes for y and b. So all the variables turn out to be zero. – Snowman Dec 02 '10 at 06:26
  • @f-Prime: You're not really thinking much about this. Are you really claiming that the only way for $x-a$ to be zero is if both $x$ and $a$ are zero? Really? So, if you subtract 3 from 3, you don't get 0? – Arturo Magidin Dec 02 '10 at 06:30
  • Oh wow I missed that :/ So I'm not seeing the 'big picture'. Where are we going with this? – Snowman Dec 02 '10 at 06:34
  • @f-Prime: I thought you were trying to prove that $x=a$, that $y=b$, and that $r=s$ is the necessary conclusion you derive from $(x,y,r)R(a,b,s)$. If you forgot that this is where you were going, then you've certainly lost sight of the "big picture". – Arturo Magidin Dec 02 '10 at 06:40
  • Oh ok so with all we've done above, we've shown that x=a, y=b, and r=s huh? – Snowman Dec 02 '10 at 06:52
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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.

Yuval Filmus
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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? – Snowman Dec 02 '10 at 06:01
  • @f-Prime Your answer is syntactic, mine is semantic. – Yuval Filmus Dec 02 '10 at 06:10
  • @f-Prime: Also, they aren't really "z"-coordinates, they are radii of circles. – Arturo Magidin Dec 02 '10 at 06:16
  • 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.. – Snowman Dec 02 '10 at 06:59
  • @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 02 '10 at 16:04