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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.

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    $\begingroup$ How can you define it? You just have defined it! $\endgroup$ – Dietrich Burde Aug 9 '18 at 19:10
  • $\begingroup$ @JohnMiller Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… $\endgroup$ – gimusi Sep 6 '18 at 23:51
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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

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

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  • $\begingroup$ Would it not be $x,y,z \in \Bbb R$? $\endgroup$ – Tyler6 Aug 9 '18 at 19:10
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    $\begingroup$ @Tyler6 Not if you write brackets around, then the object is the whole triple, and as a triple it lives in $\Bbb R^3$. $\endgroup$ – Arnaud Mortier Aug 9 '18 at 19:11
  • $\begingroup$ @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. $\endgroup$ – JMoravitz Aug 9 '18 at 19:11
  • $\begingroup$ 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 $\endgroup$ – Tyler6 Aug 9 '18 at 19:12
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    $\begingroup$ @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 $(x,y,z)$. $\endgroup$ – Arnaud Mortier Aug 9 '18 at 19:13
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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

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

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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.

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