- #1
slipperypete
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I started my way through baby Rudin recently, been able to answer every question so far, but I just can't get #16 from Chapter 1.
I (think) understand the geometry of it perfectly. I just can't prove it rigorously/analytically. I haven't been able to do anything for 3 days now, except think about this problem. So any help would be appreciated.
Suppose k\geq 3, |\mathbf{x}-\mathbf{y}|=d>0,\text{ and }r>0. Prove:
(a) If 2r>d, there are infinitely many \mathbf{z}\in\mathbb{R}^k such that \left|\mathbf{z}-\mathbf{x}\right|=\left|\mathbf{z}-\mathbf{y}\right|=r.
I have:
Since \left|\mathbf{z}-\mathbf{x}\right|=\left|\mathbf{z}-\mathbf{y}\right|=r, then 2r=\left|\mathbf{z}-\mathbf{x}\right|+\left|\mathbf{z}-\mathbf{y}\right|. Also, so \left|\mathbf{z}-\mathbf{x}\right|=\left|\mathbf{x}-\mathbf{z}\right|, so 2r=\left|\mathbf{z}-\mathbf{x}\right|+\left|\mathbf{z}-\mathbf{y}\right|=\left|\mathbf{x}-\mathbf{z}\right|+\left|\mathbf{z}-\mathbf{y}\right|. By Rudin's Proposition 1.37(f), \left|\mathbf{x}-\mathbf{y}\right|\leq\left|\mathbf{x}-\mathbf{z}\right|+\left|\mathbf{z}-\mathbf{y}\right| for any \mathbf{x},\mathbf{y},\mathbf{z}\in\mathbb{R}^{k}. Since \left|\mathbf{x}-\mathbf{y}\right|=d, this proposition makes it is clear that, given \mathbf{x},\mathbf{y}\in\mathbb{R}^{k}, there exists infinitely many \mathbf{z}\in\mathbb{R}^{k} satisfying the equations 2r>d,\left|\mathbf{z}-\mathbf{x}\right|=\left|\mathbf{z}-\mathbf{y}\right|=r.
I'm not sure this is a valid proof, because it would hold in \mathbb{R}^2. But if I'm imagining the geometry correctly, there should be exactly 2 solutions in 2-space (i.e., the 2 points at which the circles of radius r, drawn about \mathbf{x} and \mathbf{y}, intersect), not infinitely many. So I'm sure I'm missing something, but I don't see any holes in the proof.
I've also observed that under 2-space, if you square these equations: \left|\mathbf{z}-\mathbf{x}\right|=r,\left|\mathbf{z}-\mathbf{y}\right|=r and apply the definition of norm, you get a system of two equations and two variables (z_1,z_2). I know there's a solution there, which meshes perfectly with my understanding of the geometry behind this problem. Moreover, if you move on to 3-space and proceed in the same way, you get a system of two equations and three variables (z_1,z_2,z_3). I know that this system will have infinitely many solutions. But I don't know how to prove rigorously that the 2-variable system is solvable or that the 3-variable system is unsolvable. Also, the way I understand the problem, r is fixed. But it is subject to some constraint (e.g., 2r>d). I don't know how to incorporate this constraint into the proof.
(b) If 2r=d, there is exactly one such \mathbf{z}.
I can't even figure out where to begin. I've noticed that if you set \mathbf{z}=\tfrac{1}{2}\left(\mathbf{x}+\mathbf{y}\right), then \left|\mathbf{z}-\mathbf{x}\right|=\left|\mathbf{z}-\mathbf{y}\right|. Also, this makes 2r=2\left|\mathbf{z}-\mathbf{x}\right|=2\left|\tfrac{\mathbf{x}+\mathbf{y}}{2}-\mathbf{x}\right|=\left|\mathbf{y}-\mathbf{x}\right|=d.
I can derive this "solution" for \mathbf{z} by squaring \left|\mathbf{z}-\mathbf{x}\right|=\left|\mathbf{z}-\mathbf{y}\right| and going from there. But no relationship between 2r and d is needed to figure this out, so it would seem that \mathbf{z}=\tfrac{1}{2}\left(\mathbf{x}+\mathbf{y}\right) should be true in the cases of 2r>d and 2r<d. So I'm stuck on this one.
(c) If 2r<d, there is no such \mathbf{z}.
This is the one I think I've made the most progress on. I think Rudin's Theorem 1.37(f) should apply. Let me know if this proof works/doesn't work:
Assume a solution does exist. Since \left|\mathbf{z}-\mathbf{x}\right|=\left|\mathbf{z}-\mathbf{y}\right|=r and \left|\mathbf{z}-\mathbf{x}\right|=\left|\mathbf{x}-\mathbf{z}\right|, then 2r=\left|\mathbf{x}-\mathbf{z}\right|+\left|\mathbf{z}-\mathbf{y}\right|. It's given that \left|\mathbf{x}-\mathbf{y}\right|=d, so if 2r<d, then \left|\mathbf{x}-\mathbf{z}\right|+\left|\mathbf{z}-\mathbf{y}\right|<\left|\mathbf{x}-\mathbf{y}\right|. But this contradicts Rudin's Theorem 1.37(f), which says that \left|\mathbf{x}-\mathbf{z}\right|+\left|\mathbf{z}-\mathbf{y}\right|\geq\left|\mathbf{x}-\mathbf{y}\right|. Therefore, no \mathbf{z} exists.
Any input would be appreciated.
I (think) understand the geometry of it perfectly. I just can't prove it rigorously/analytically. I haven't been able to do anything for 3 days now, except think about this problem. So any help would be appreciated.
Suppose k\geq 3, |\mathbf{x}-\mathbf{y}|=d>0,\text{ and }r>0. Prove:
(a) If 2r>d, there are infinitely many \mathbf{z}\in\mathbb{R}^k such that \left|\mathbf{z}-\mathbf{x}\right|=\left|\mathbf{z}-\mathbf{y}\right|=r.
I have:
Since \left|\mathbf{z}-\mathbf{x}\right|=\left|\mathbf{z}-\mathbf{y}\right|=r, then 2r=\left|\mathbf{z}-\mathbf{x}\right|+\left|\mathbf{z}-\mathbf{y}\right|. Also, so \left|\mathbf{z}-\mathbf{x}\right|=\left|\mathbf{x}-\mathbf{z}\right|, so 2r=\left|\mathbf{z}-\mathbf{x}\right|+\left|\mathbf{z}-\mathbf{y}\right|=\left|\mathbf{x}-\mathbf{z}\right|+\left|\mathbf{z}-\mathbf{y}\right|. By Rudin's Proposition 1.37(f), \left|\mathbf{x}-\mathbf{y}\right|\leq\left|\mathbf{x}-\mathbf{z}\right|+\left|\mathbf{z}-\mathbf{y}\right| for any \mathbf{x},\mathbf{y},\mathbf{z}\in\mathbb{R}^{k}. Since \left|\mathbf{x}-\mathbf{y}\right|=d, this proposition makes it is clear that, given \mathbf{x},\mathbf{y}\in\mathbb{R}^{k}, there exists infinitely many \mathbf{z}\in\mathbb{R}^{k} satisfying the equations 2r>d,\left|\mathbf{z}-\mathbf{x}\right|=\left|\mathbf{z}-\mathbf{y}\right|=r.
I'm not sure this is a valid proof, because it would hold in \mathbb{R}^2. But if I'm imagining the geometry correctly, there should be exactly 2 solutions in 2-space (i.e., the 2 points at which the circles of radius r, drawn about \mathbf{x} and \mathbf{y}, intersect), not infinitely many. So I'm sure I'm missing something, but I don't see any holes in the proof.
I've also observed that under 2-space, if you square these equations: \left|\mathbf{z}-\mathbf{x}\right|=r,\left|\mathbf{z}-\mathbf{y}\right|=r and apply the definition of norm, you get a system of two equations and two variables (z_1,z_2). I know there's a solution there, which meshes perfectly with my understanding of the geometry behind this problem. Moreover, if you move on to 3-space and proceed in the same way, you get a system of two equations and three variables (z_1,z_2,z_3). I know that this system will have infinitely many solutions. But I don't know how to prove rigorously that the 2-variable system is solvable or that the 3-variable system is unsolvable. Also, the way I understand the problem, r is fixed. But it is subject to some constraint (e.g., 2r>d). I don't know how to incorporate this constraint into the proof.
(b) If 2r=d, there is exactly one such \mathbf{z}.
I can't even figure out where to begin. I've noticed that if you set \mathbf{z}=\tfrac{1}{2}\left(\mathbf{x}+\mathbf{y}\right), then \left|\mathbf{z}-\mathbf{x}\right|=\left|\mathbf{z}-\mathbf{y}\right|. Also, this makes 2r=2\left|\mathbf{z}-\mathbf{x}\right|=2\left|\tfrac{\mathbf{x}+\mathbf{y}}{2}-\mathbf{x}\right|=\left|\mathbf{y}-\mathbf{x}\right|=d.
I can derive this "solution" for \mathbf{z} by squaring \left|\mathbf{z}-\mathbf{x}\right|=\left|\mathbf{z}-\mathbf{y}\right| and going from there. But no relationship between 2r and d is needed to figure this out, so it would seem that \mathbf{z}=\tfrac{1}{2}\left(\mathbf{x}+\mathbf{y}\right) should be true in the cases of 2r>d and 2r<d. So I'm stuck on this one.
(c) If 2r<d, there is no such \mathbf{z}.
This is the one I think I've made the most progress on. I think Rudin's Theorem 1.37(f) should apply. Let me know if this proof works/doesn't work:
Assume a solution does exist. Since \left|\mathbf{z}-\mathbf{x}\right|=\left|\mathbf{z}-\mathbf{y}\right|=r and \left|\mathbf{z}-\mathbf{x}\right|=\left|\mathbf{x}-\mathbf{z}\right|, then 2r=\left|\mathbf{x}-\mathbf{z}\right|+\left|\mathbf{z}-\mathbf{y}\right|. It's given that \left|\mathbf{x}-\mathbf{y}\right|=d, so if 2r<d, then \left|\mathbf{x}-\mathbf{z}\right|+\left|\mathbf{z}-\mathbf{y}\right|<\left|\mathbf{x}-\mathbf{y}\right|. But this contradicts Rudin's Theorem 1.37(f), which says that \left|\mathbf{x}-\mathbf{z}\right|+\left|\mathbf{z}-\mathbf{y}\right|\geq\left|\mathbf{x}-\mathbf{y}\right|. Therefore, no \mathbf{z} exists.
Any input would be appreciated.