# Relations Problem 3

Reminder of how we defined the relation:

(x,y,r) R (a,b,s) if and only if $$\sqrt{(x-a)^2 + (y-b)^2} \le s- r$$

### Solution to (b)

Let (x,y,r) and (a,b,s) be elements of $$C = \mathbb{R} \times \mathbb{R} \times \mathbb{R}^+$$. Suppose that (x,y,r) R (a,b,s) and (a,b,s) R (x,y,r).

Applying the definition of R, we get that $$\sqrt{(x-a)^2 + (y-b)^2} \le s- r$$ and $$\sqrt{(a-x)^2 + (b-y)^2} \le r- s$$.

Notice that the lefthand sides of these equations can't be negative. So both s-r and r-s must be non-negative. But this can be true only when s-r = 0, i.e. r=s.

So now we know that $$\sqrt{(x-a)^2 + (y-b)^2} = 0$$. So $$(x-a)^2 + (y-b)^2 = 0$$. So $$(x-a)^2 = (y-b)^2 = 0$$. So $$x-a = y-b = 0$$. So x=a and y=b.

Since r=s, x=a, and y=b, (x,y,r) = (a,b,s), which is what we needed to show.