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In the textbook I am reading, it says a dimension is the number of independent parameters needed to specify a point. In order to make a circle, you need two points to specify the $x$ and $y$ position of a circle, but apparently a circle can be described with only the $x$-coordinate? How is this possible without the $y$-coordinate also?

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    $\begingroup$ In intuitive terms: a creature living "on the circle" is like a train, it can only move forwards and backwards. That's true for any curve. $\endgroup$
    – orion
    Feb 2, 2015 at 18:48
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    $\begingroup$ Consider a clock with one hand (link to pictures). How many numbers $x,y,z,\ldots$ do you need to tell where the hand points? $\endgroup$ Feb 2, 2015 at 19:42
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    $\begingroup$ How about lines? You need at least two numbers to define a line first (the specific number of numbers may be greater, depending on the number of dimensions of your space), then at least one more number for a point on the line. So.... do you agree the line is 1-dimensional? $\endgroup$
    – CiaPan
    Feb 3, 2015 at 10:44
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    $\begingroup$ It's not about how many parameters you need to draw a curve... it's how many parameters you need to specify your position on the curve. And on a curve it's always one parameter (for instance, the arc length from some reference point). You are mixing the ambient space with the object you are measuring. You can have a line in 14-dimensional space and it's still 1d. Imagine a squiggle drawn with pen. You surely need many parameters to tell what shape it is on the paper, but it's still a curve, it still looks like a line in the close-up. Don't measure the dimension of the paper but the line. $\endgroup$
    – orion
    Feb 3, 2015 at 12:08
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    $\begingroup$ This is not probably the point of the question, but if can confuse some people: we are speaking here of the line (circumference), not of the region. "In everyday use, the term "circle" may be used interchangeably to refer to either the boundary of the figure, or to the whole figure including its interior; in strict technical usage, the circle is the former and the latter is called a disk" Wikipedia $\endgroup$
    – leonbloy
    Feb 3, 2015 at 12:36

17 Answers 17

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Suppose we're talking about a unit circle. We could specify any point on it as: $$(\sin(\theta),\cos(\theta))$$ which uses only one parameter. We could also notice that there are only $2$ points with a given $x$ coordinate: $$(x,\pm\sqrt{1-x^2})$$ and we would generally not consider having to specify a sign as being an additional parameter, since it is discrete, whereas we consider only continuous parameters for dimension.

That said, a Hilbert curve or Z-order curve parameterizes a square in just one parameter, but we would certainly not say a square is one dimensional. The definition of dimension that you were given is kind of sloppy - really, the fact that the circle is of dimension one can be taken more to mean "If you zoom in really close to a circle, it looks basically like a line" and this happens, more or less, to mean that it can be paramaterized in one variable.

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    $\begingroup$ I think it's quite unfair to say that a space-filling curve 'parameterizes' a square, since no space-filling curve is injective. $\endgroup$
    – user98602
    Feb 2, 2015 at 19:14
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    $\begingroup$ @cHao That's not true; you need a center and a radius to specify a circle, but once you know the circle, you can specify every point on it with one parameter. $\endgroup$ Feb 2, 2015 at 23:01
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    $\begingroup$ @cHao Unless you're proposing space-filling curves, I don't see how. The boundary of a square is one dimensional. The square is definitely not. To be precise in my question: given a square and a point within the square, how do you intend to specify the point with just two parameters? $\endgroup$ Feb 2, 2015 at 23:08
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    $\begingroup$ @cHao I think I see - are you saying that the boundary of a square is one-dimensional? Because that's certainly true. $\endgroup$ Feb 2, 2015 at 23:22
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    $\begingroup$ Last i checked, a square was what you're calling its "boundary". The space inside the square was not part of it. However we describe it, though, why do we get away with not having the square (or circle) itself as a parameter? $\endgroup$
    – cHao
    Feb 2, 2015 at 23:45
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Continuing ploosu2, the circle can be parameterized with one parameter (even for those who have not studied trig functions)... $$ x = \frac{2t}{1+t^2},\qquad y=\frac{1-t^2}{1+t^2} $$

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    $\begingroup$ For the reference of anyone unfamiliar with this, this is a stereographic projection. $\endgroup$ Feb 2, 2015 at 18:28
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    $\begingroup$ Minor comment: you have to let $t\to\pm\infty$ to reach $(0,-1)$ with this parametrization. $\endgroup$
    – 2'5 9'2
    Feb 2, 2015 at 18:30
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    $\begingroup$ @alex.jordan Problem solved: $$x=\frac{4t-4t^3}{(1+t^2)^2},\quad y=\frac{1-6t^2+t^4}{(1+t^2)^2}\qquad t\in(-1,1]$$ is a bijection onto the circle. (If you let $t\in\Bbb R$ it will cover the whole circle twice, except for one point.) $\endgroup$ Feb 3, 2015 at 23:17
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Officially, it's one-dimensional because at any point, the tangent space is a one-dimensional vector space.

Unofficially, it's one-dimensional because if you zoom in enough on a tiny little piece of it, it is indistinguishable from a segment of the real line.


Sorry, I have a bad habit of answering a question title and not exactly answering the question itself. You say you need an $x$ and $y$ value to identify a circle. So you are identifying its center. Even though you've just named a point with two numbers in its address, you have only identified a $0$-dimensional thing at this point. It's just a point, there is no direction to travel in.

To identify the circle, you'd need a third number: the radius. And the circle will consist not of the point you had earlier identified, but rather of all points that are that radius from your identified center.

This collection of points can be parametrized using only one independent variable, as a few other answers have shown.

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    $\begingroup$ Another way of saying this is: a line is a circle with an infinite radius. $\endgroup$ Feb 3, 2015 at 23:19
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    $\begingroup$ @BrentWashburne: a line is the limiting case of a circle as the radius tends to infinity $\endgroup$
    – smci
    Feb 5, 2015 at 8:43
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When the book says, "a dimension is the number of independent parameters needed to specify a point" it's talking about the parameters needed to specify a point on the circle. Only one is needed to describe which point on a certain circle.

That's different from what's needed to describe a circle itself. A circle spans two dimensions, but a single point on it can be described with a single value.

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    $\begingroup$ it's talking about the parameters needed to specify a point on the circle. [...] That's different from what's needed to describe a circle itself. A circle spans two dimensions, but a single point on it can be described with a single value. Cleared it up for me! $\endgroup$
    – legends2k
    Feb 5, 2015 at 6:44
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It can be parametrized with the angle parameter $t$ as $(\sin(t), \cos(t))$.

If you want to use the $x$-coordinate as a parameter, then you can solve $y$ from $x^2+y^2 = r^2$. But then you only have a halve of the circle since you must choose the sign of $y$.

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  • $\begingroup$ So, are you saying that the unit circle that's centered on the origin is one-dimensional whereas all other circles are something else (presumably, two-dimensional)? $\endgroup$
    – Mico
    Feb 6, 2015 at 10:44
  • $\begingroup$ No, no. All circles with all radii are 1-dimensional. For the circle with center $(x_0, y_0)$ and radius $r$, we have the parametrization $r(x_0+\sin(t), y_0+\cos(t))$. $\endgroup$
    – ploosu2
    Feb 6, 2015 at 11:25
  • $\begingroup$ With positive radii, I should say. $\endgroup$
    – ploosu2
    Feb 6, 2015 at 11:26
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That is a bad definition of dimension. By this definition, the (filled-in) unit square in the real coordinate plane is one-dimensional, since you only need one number to describe a point $(x,y)$. An easy way to do this is to write out the decimal expansions of $x$ and $y$:

$$x=0.a_1a_2a_3a_4a_5\dots$$

$$y=0.b_1b_2b_3b_4b_5\dots$$

Now the point $(x,y)$ can be described with just one number:

$$0.a_1b_1a_2b_2a_3b_3\dots$$


But suppose we overlook this shortcoming of your definition, and we apply it to a circle. Then we can use the polar coordinate map

$$(x,y) \rightarrow (r,\theta)$$

This lets us write every point $(x,y)$ in terms of just one variable: $\theta$ (since $r$ is constant).

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Alex Jordan gives the official answer if you officially define the circle as a differential manifold. If you officially define it as a topological space then it is one dimensional because you can cut it into separate parts by removing single points. Points are 0 dimensional, and connected spaces that can be disconnected by removing points are 1 dimensional.

Actually this is local. Two cones joined at the vertex can be disconnected by removing one point -- the vertex. You could say a circle is one dimensional at every point, while those two cones have one dimension at the vertex but two dimensions elsewhere.

Or you could limit the definition to spaces that are locally isomorphic to $\mathbb{R}^n$. That use of $n$ does not make the idea trivial. Rather it shows that the topological criterion of connectedness agrees with the linear algebra criterion of a basis in defining dimension of the $\mathbb{R}^n$.

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  • $\begingroup$ Although I appreciate complicated answers to simple questions, the tags are "algebra-precalculus" and "geometry". This answer clearly goes beyond both the approximate level of education and the explicit mathematical domain(s) the question is looking for. Or to put it a different way: do you think it's likely that people who know what a differential manifold is are unable to answer this question? $\endgroup$ Feb 5, 2015 at 15:15
  • $\begingroup$ @EnvisionAndDevelop You consider the idea that a circle can be disconnected by removing two points "complicated"? Or is it the cones that are too complicated? Or is it just the word "topology"? In fact most answers correctly point out that the definition in OP's book is just wrong and can only be made right by using the notion of continuity. Would you claim on this ground every correct answer to the question is off topic? $\endgroup$ Feb 5, 2015 at 15:35
  • $\begingroup$ What I'm telling you is that, although the "the word 'topology'" is just a word, the field of topology and the concept of a topological space are far more complicated than what you'd expect to see in a precalculus class. In any case, the definition you gave does not fit the areas of mathematics that the question is addressing. The bottom line is that I don't think this is a helpful answer to the question, no matter how big words you use. I appreciate big words, especially when they're accurate, they're not enough, especially when you're trying to make someone understand something else. $\endgroup$ Feb 5, 2015 at 16:06
  • $\begingroup$ Oh my. If my answer seemed to you to require mastery of the field of topology I don't know whether to be flattered or chagrined. It really only requires that you picture a circle with a couple of points erased, and a pair of cones meeting at the vertex -- then erasing the vertex. $\endgroup$ Feb 5, 2015 at 16:37
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You need not getting too sophisticated.

One dimension means "forward or backward" - these are your options if you sit on a line. A circle is just a bent line in a way that you reach the same spot after some distance.

So a circle is finite but unlimited.

If you think about the "bending" it is quite interesting that this bending takes place in the second dimension.

Take this one level higher: in two dimensions your options are "forward, backward, left and right" so on a plane you have these choices. Now bend your plane in a special way... the corresponding thing to the circle is a sphere.

Next level... out of our imagination ;-)

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  • $\begingroup$ The next level was imagined in a story by Robert Heinlein, "And He Built a Crooked House." $\endgroup$
    – David K
    Feb 3, 2015 at 12:29
  • $\begingroup$ thanks for the reference - I knew there is a book around these ideas. $\endgroup$
    – balzambass
    Feb 3, 2015 at 14:47
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There are several ways in which the dimension of an object may be defined. A very general one, and the de-facto way to define "dimension" (I.e: with no extra adjectives) is the topological dimension. From the linked page of Wikipedia:

Any given open cover of the unit circle will have a refinement consisting of a collection of open arcs. The circle has dimension one, by this definition, because any such cover can be further refined to the stage where a given point x of the circle is contained in at most two open arcs. That is, whatever collection of arcs we begin with, some can be discarded or shrunk, such that the remainder still covers the circle but with simple overlaps.

This means roughly and in more intuitive terms that if you have an unlimited collection of colors of paint, but only a very small amount of each color, to cover a circle (bigger than what you can cover with a single color), then you can color the circle such that at most, 2 paint blobs of different colors meet on each point (the dimension is 1 less than this number, therefore 1 for a circle).

See the image included in the same Wikipedia article for an illustration of this concept.

Note that the topological dimension described above is just one possible way to define what a dimension is in mathematical terms. Mathematics only give results about specific definitions, not natural language concepts (other than how to construct a mathematical object that models some of those properties). Only some mathematical constructions have names corresponding to their intuitive meaning (Such as this, the topological dimension). Other mathematical constructions have names which have only a very abstract correspondence with the daily-life object bearing the sane name (E.g: a [manifold][3] needs not resemble a intake or exhaust manifold of an internal combustion engine, or an object which could exist physically at all) and others have no real-life counter part (Such as groups, only some physically existing systems have a behavior that can be modeled by groups, such as the Rubik's cube, and clocks).

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As others have pointed out, the definition in your textbook is not really as clear as it could be. Since you are reading this in a precalculus textbook, perhaps a better way to think about that sentence is not as a definition of 1-dimensional, but rather to mean: "We can usefully work with a circle in the same way we work with a line, by using only 1 number to specify a point on it."

For example, take the classic calculus problem of finding a line that just touches the circle (the tangent line), say at the topmost point of a radius 1 circle. We could approach this problem by using 2 variables, x and y, to specify points on the circle:

$$x^2 + y^2 = 1$$

$$y = \sqrt { 1 - x^2 }$$

$$slope = \frac{dy}{dx} = \frac{1}{2} (2x) (1 - x^2)^{-\frac{1}{2}}$$

$$ = \frac{dy}{dx} = \frac{1}{2} (2 \cdot 0) (1 - 0^2)^{-\frac{1}{2}} = 0$$

But, we could also approach the same problem using just one number $t$ to specify a point:

$$x = cos(t),~ y = sin(t)$$

$$slope = {\frac{dy}{dt}}/{\frac{dx}{dt}} = -\frac{sin(t)}{cos(t)} = -\frac{sin(\frac{\pi}{2})}{cos(\frac{\pi}{2})} = 0$$

Since we can do these calculations easily using just one parameter $t$, we say a circle "is" one dimensional, not as a definition, just as an observation about how we can work with it.

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To put the question into a more dramatic scenario let's imagine you are not talking about a circle but a curve in 3D, like the trajectory of a bee. The bee itself is at a given point on that route on a given time. So, you could assume that the parameter t (=time) is enough to specify the position of the bee, right?
The trick is that you have to separate the definition of the curve (which is n-dimensional) from the definition of a point on that particular curve (that might just be one parameter).
From my point of view, the definition from your book is not that good at all.

(sorry for my bad English. It is my third language)

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Circle is one-dimensional also according to the fractal dimension, for example using box counting dimension (Minkowski–Bouligand dimension) definition:

Suppose that N(ε) is the number of boxes of side length ε required to cover the set. Then the box-counting dimension is defined as:

$$\dim_{\rm box}(S) := \lim_{\varepsilon \to 0} \frac {\log N(\varepsilon)}{\log (1/\varepsilon)}$$

Instead of number of boxes it can be minimal number of open balls covering the set in question (circle here).

In other words number of pixels needed to draw a circle on the screen grows linearly ($d=1$) with increasing resolution ;-)

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It's possible that your confusion comes from misunderstanding the term "circle". In mathematics, a circle is a curved line and the area within that line is a disc.

The disc is indeed two-dimensional: you need to specify, e.g., $x$- and $y$-coordinates or an angle and distance to identify a point. But, as the other answers explain, a circle is one-dimensional: given that the radius is fixed, you can specify any point by giving its angle or, equivalently, the distance around the circle from some fixed reference point.

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Your mixing two things. The circle that you describe is a set in a plane (two dimensional) with inside and outside. Now generically the boundary has a dimension 1 less than an enclosed volume (in this case area).

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I like to think about it as follows, a point is 0-dimensional, but anything that has a "length" associated with it, is 1-dimensional, and anything that has an "area" associated with it as 2-dimensional, and anything with a "volume" as 3-dimensional, etc.

Be careful not to confuse a circle with a disc, as others have mentioned in mathematics a circle is just a curved line, hence it's 1-dimensional. A disc on the other hand includes all the points inside as well. So the number of parameters is not what matters, you want to look is the set of points itself.

Formally speaking the points on a circle of radius r centered at the origin are the set of all points (x,y) in this set: $\{(x,y)\ |\ x^2 + y^2 = r^2\}$ whereas a disc of radius r centered at the origin is the set of all points (x,y) in this set: $\{(x,y)\ |\ x^2 + y^2 \le r^2\}$.

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You can follow the circle riding a bike with jammed handlebars. Counting the number of wheel turns, a single parameter, you can always tell where you are.

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You are thinking on the area of the circle. It is 2 dimensional, because you can move on it on two direction: in $x$ and $y$, or to $\phi$ and $r$, or any other. The essence is, that you need 2 real numbers to describe the position of a point.

But the circumference of the circle is only 1 dimensional. You can describe the place of a point by $r$ or by $\phi$.

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    $\begingroup$ One can use the term disk. $\endgroup$
    – user65203
    Apr 28, 2015 at 7:09

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