Proving Existence of a Point in a Ball of Radius r

[Ryker]

Homework Statement

Show that for every a_* = (a₁, 1/a₁), there exists another point of the form (a, 1/a) in a ball (i.e. circle, since we're in R2) of radius r, centered at a_*, for any r > 0.

The Attempt at a Solution

This is actually only a part of the whole problem, but I just can't put it down properly. I've tried various things on squeezing a₁ < a < a₁ + r, but when I invert it I can't show that 1/a < (1/a₁) + r. My logic was that if I did show that, then I'd be able to show that the distance of such a point (a, 1/a) to the original one is less than r. I know just for r it wouldn't work, and I'd have to take r/2 or something, but for now I'm just trying to get to show it for r. Intuitively, I can see that this doesn't always happen for just any r, and that if a₁ < a < a₁ + r, it doesn't necessarily follow that 1/a < (1/a₁) + r.

In any case, help here would be much appreciated.

 

[LCKurtz]

Homework Statement

Show that for every (a₁, 1/a₁), there exists another point of the form (a, 1/a) in a ball (i.e. circle, since we're in R2) of radius r for any r > 0.

This is actually only a part of the whole problem, but I just can't put it down properly.

Exactly. And that's probably why there are no replies. The statement of the problem makes no sense.

[Edit] Do you mean that if there is a point (a,1/a) inside the circle then there is another (b,1/b) inside the circle? If so, consider the graph y = 1/x.

 

[Ryker]

Exactly. And that's probably why there are no replies. The statement of the problem makes no sense.

[Edit] Do you mean that if there is a point (a,1/a) inside the circle then there is another (b,1/b) inside the circle? If so, consider the graph y = 1/x.

Oh, sorry, I edited my original post now.

The ball is actually centered at (a₁, 1/a₁). The thing is I'm not sure what exactly we're allowed to use to prove this. For example, as my first step I've tried solving the equation (x - a₁)² + (1/x - 1/a₁)² = r² to show that there is always a solution to that, but since this yields a quartic equation with the x as a variable, I can't show that it therefore always has at least one solution. If I was able to show this, then it must also hold true that the equation is solvable for r/2, so that intersection point would then be the point I'm looking for.

Again, though, the problem is that I think I'm supposed to solve this only in terms of sets. That is, instead of y = 1/x, I have a set containing all points, such that x₁*x₂ = 1. Which I guess is the same, but still.

 

[LCKurtz]

I'm sorry, but you still haven't given a clear precise statement of the problem. Here's my guess at what the problem is:

Given a circle for radius r > 0 centered at (a,1/a), show there is another point (b,1/b) within the circle.

Is that the problem? Gotta run for now, lunch time.

[Edit] Back from lunch. If b is close enough to a, surely you can make the distance from (a,1/a) to (b,1/b) less than r. You can use the continuity of y = 1/x.

 

[Ryker]

I'm sorry, but you still haven't given a clear precise statement of the problem. Here's my guess at what the problem is:

Given a circle for radius r > 0 centered at (a,1/a), show there is another point (b,1/b) within the circle.

Is that the problem? Gotta run for now, lunch time.

[Edit] Back from lunch. If b is close enough to a, surely you can make the distance from (a,1/a) to (b,1/b) less than r. You can use the continuity of y = 1/x.

No, I think continuity is a no go, since I don't think we're allowed to use it. Any other suggestions? I'd be especially interested in something done by just manipulation of variables, not by more "advanced" methods. But like I said, I tried that, and couldn't come up with anything that would resolve this.

Oh, and yes, that is the problem. How is that different from what I said after the edit, though?

 

[SammyS]

No, I think continuity is a no go, since I don't think we're allowed to use it. Any other suggestions? I'd be especially interested in something done by just manipulation of variables, not by more "advanced" methods. But like I said, I tried that, and couldn't come up with anything that would resolve this.

Oh, and yes, that is the problem. How is that different from what I said after the edit, though?

I suppose since you did say "a* = (a1, 1/a1)" , we should have recognized a* as a point. Then a circle centered at a* makes sense. However, I didn't understand the question until LCKurtz clarified it.

I think the following will work.

If a ≥ 1, let [itex]\displaystyle b=a+\frac{r}{2}\,.[/itex]

If 0 < a < 1, then let [itex]\displaystyle \frac{1}{b}=\frac{1}{a}+\frac{r}{2}\,.[/itex]

Check details for a < 0 , but it should be similar.

 

[Ryker]

I suppose since you did say "a* = (a1, 1/a1)" , we should have recognized a* as a point. Then a circle centered at a* makes sense. However, I didn't understand the question until LCKurtz clarified it.

I think the following will work.

If a ≥ 1, let [itex]\displaystyle b=a+\frac{r}{2}\,.[/itex]

If 0 < a < 1, then let [itex]\displaystyle \frac{1}{b}=\frac{1}{a}+\frac{r}{2}\,.[/itex]

Check details for a < 0 , but it should be similar.

Yeah, but how do we know that the second b is the same as the first one? I mean, if I pick the first b as you suggested, then [itex]\displaystyle \frac{1}{b}= \frac{1}{a+\frac{r}{2}} = \frac{2}{2a + r}, [/itex] right? But why would that be less than, say, [itex]\frac{1}{a}+\frac{r}{2}\, [/itex] so that I could show that the distance to the center is less than r?

 

[LCKurtz]

No, I think continuity is a no go, since I don't think we're allowed to use it. Any other suggestions? I'd be especially interested in something done by just manipulation of variables, not by more "advanced" methods. But like I said, I tried that, and couldn't come up with anything that would resolve this.

Oh, and yes, that is the problem. How is that different from what I said after the edit, though?

The problem boils down to getting 1/b close to 1/a. For ease of discussion let's say a > 0 and b > 0.

|1/a - 1/b| = |b-a|/|ab|. Now if a > 0 and we can always choose b > a/2 and close to a. Then ab > a²/2 so 1/(ab) < 2/a².

|b-a|/|ab| < 2|b-a|/a².

So as long as b >a/2 you can choose it close enough to a to make that expression as small as you want. You can easily modify it if a < 0.

 

[SammyS]

Yeah, but how do we know that the second b is the same as the first one? I mean, if I pick the first b as you suggested, then [itex]\displaystyle \frac{1}{b}= \frac{1}{a+\frac{r}{2}} = \frac{2}{2a + r}, [/itex] right? But why would that be less than, say, [itex]\frac{1}{a}+\frac{r}{2}\, [/itex] so that I could show that the distance to the center is less than r?

Let's see,
[itex]\displaystyle \frac{1}{a}-\frac{1}{b}=\frac{1}{a}-\frac{2}{2a + r}[/itex]

[itex]\displaystyle =\frac{2a+r-2a}{a(2a+r)}[/itex]

[itex]\displaystyle =\frac{r}{a(2a+r)}<\frac{r}{2}[/itex]​

 

[Ryker]

Let's see,
[itex]\displaystyle \frac{1}{a}-\frac{1}{b}=\frac{1}{a}-\frac{2}{2a + r}[/itex]

[itex]\displaystyle =\frac{2a+r-2a}{a(2a+r)}[/itex]

[itex]\displaystyle =\frac{r}{a(2a+r)}<\frac{r}{2}[/itex]​

(there's no facepalm smiley) You are right, of course, I missed the a > 1 part, hence the confusion. Thanks for the help to both of you, I'll write it out tomorrow and hopefully I can deal with the other cases myself!​