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I'm not sure how to say, using proper mathematical notation, a sphere $S^2$ is equal to the basic unit sphere $1=x^2 + y^2 + z^2$. Just looking for a quick answer, thank you.

John Miller
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    How can you define it? You just have defined it! – Dietrich Burde Aug 09 '18 at 19:10
  • @JohnMiller Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Sep 06 '18 at 23:51

3 Answers3

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Using set builder notation:

$$S=\{(x,y,z)\in \Bbb R^3\mid x^2+y^2+z^2=1\}$$

If it is clear that the ambient space is $\Bbb R^3$ and you don't feel that it's necessary to specify it, a possible shortcut is simply

$$S=\{x^2+y^2+z^2=1\}$$

Arnaud Mortier
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  • Would it not be $x,y,z \in \Bbb R$? – Tyler6 Aug 09 '18 at 19:10
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    @Tyler6 Not if you write brackets around, then the object is the whole triple, and as a triple it lives in $\Bbb R^3$. – Arnaud Mortier Aug 09 '18 at 19:11
  • @Tyler6 there is no difference between saying $(x,y,z)\in\Bbb R^3$ and saying $x\in \Bbb R$ and $y\in\Bbb R$ and $z\in \Bbb R$. Saying $x,y,z\in\Bbb R$ is just shorthand for the second. ${(x,y,z)~|~x,y,z\in\Bbb R,~x^2+y^2+z^2=1}$ is the same set as written in the post above. – JMoravitz Aug 09 '18 at 19:11
  • Ah ok, I was just mistaken as I thought that since they were being treated as variables we couldn’t define them as a vector – Tyler6 Aug 09 '18 at 19:12
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    @Tyler6 (continuation of JMoravitz and Dietrich's comments): but here you have to use this notation because each element of your sphere is represented by a triple $(x,y,z)$. – Arnaud Mortier Aug 09 '18 at 19:13
  • Thank you, currently doing an assignment/essay for my teacher and he's asked to write it using no English only mathematics. It's becoming extremely tedious. But it is useful in understanding set builder notation/first-order logic etc. – John Miller Aug 09 '18 at 19:25
  • @JohnMiller You're welcome. Then you should not use the shortcut version but the full one to be as clear as possible. – Arnaud Mortier Aug 09 '18 at 19:26
  • @ArnaudMortier Yes that's what I'm using. However, I'm not sure how to reference you in my bibliography? As in a user from a specific StackExchange question. – John Miller Aug 09 '18 at 19:35
  • @JohnMiller I don't think you need to reference anyone for that :) no one is going to ask. They will assume you knew or looked it up somewhere and that's fine. – Arnaud Mortier Aug 09 '18 at 19:36
  • @ArnaudMortier Well thank you again :) – John Miller Aug 10 '18 at 08:32
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Yes for $(x,y,z) \in \mathbb{R^3}$ the cartesian equation for a sphere centered at the origin and with radius $R$ is

$$x^2+y^2+z^2=R^2$$

and more in general with center in $C=(a,b,c)$

$$(x-a)^2+(y-b)^2+(z-c)^2=R^2$$

Note that the equation is derived from Pytagorean Theorem, that is

$$R= \sqrt{(x-a)^2+(y-b)^2+(z-c)^2}$$

user
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There seems to be a difference in notation between geometers (the number denoting the dimension of the coordinate space) and topologists (the number denoting the dimension of the surface).

See introduction here.

mvw
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