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I'm a computer science-type, and trying to get a deeper grip of "math-speak."

The use of the equal sign is not clear to me in terms of when it's being used as "assignment" vs. "equality."

Where $y=x$, it's intuitive that as $x$ varies over $\mathbb{R} $ (although not sure how "varies" is to be understood — traverse or crawls time, or space or what) it maps its value to $y$, and then you have pairs of $(x,y)$ points that form a continuous line through the origin of the Cartesian plane. This seems like a function to me, not an equation — because $y$ just seems synonymous with $f(x)$ — it's simply a "conversion" of x to a "new x"."

But with:

$$x^2 + y^2 = r^2$$

Equation of a circle at the origin—clearly a different beast. Intuition on how to read this is gone here. Now the "dependent variable is on the same side as (what was) the independent" and the sense of assignment or "function" is entirely lost, and $r$ is called a constant, but in reality acts as some kind of user-defined variable (not in the sense of "varies over," but as something you can manually change or "slide" yourself technically over $\mathbb{R} $).

So how do you intuitively describe this with the same intuitive clarity as "the line" above? Are there two independent variables, and the "$r$" constant is acting as a "dependent" on the right side? My intuition wants to write it as:

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

To keep the sense of "assignment" from "one side" to the other intact. But this doesn't work — it graphs as a half-circle and only when $r$ is parameterized.

Any help is appreciated. Thank you!

Jordan Fine
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    $x^2+y^2=r^2$ just means that $x^2+y^2$ is the same number as $r^2$. So a point $(x,y)$ is on the curve iff $x^2+y^2$ is the same as $r^2$. – Angina Seng Aug 11 '20 at 03:05
  • Don't forget that for a positive real number $t$, $-\sqrt t$ is also a square root of $t$. – Angina Seng Aug 11 '20 at 03:06
  • Another thing to add: the equation $x^2+y^2=r^2$ doesn't represent a function. I could, however, be misinterpreting what you mean by function. – Kman3 Aug 11 '20 at 03:27
  • Angina — so in this case, you need to know x AND y to know r, which means x and y are both independent here, yes? It’s just a pythagorean theorem right triangle revolving around the origin...why is r not considered a dependent variable? – Jordan Fine Aug 11 '20 at 05:00
  • kman3 — why is not a function, if the output of x + y = r^2? Is not y=x a function? (f(x)=x?) – Jordan Fine Aug 11 '20 at 05:05
  • @JordanFine A function is a relation where every $x$ value has only one corresponding $y$ value. You can test this by drawing a vertical line through the graph (called the vertical line test). On the line $y=x$, every $x$ value has only one $y$ value, which is why $y=x$ is a function. If you isolate $y$ in $x^2+y^2=r^2$, you get $y=\pm \sqrt{r^2-x^2}$. This shows that every $x$ which doesn't satisfy $|x|=r$ will have two corresponding $y$ values, which means it isn't a function. – Kman3 Aug 11 '20 at 20:32

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You are correct to be confused about this; it's legitimately confusing!

In math-speak using the equation "$x^2 + y^2 = r^2$" is shorthand. Formally one should be using set-builder notation to describe the set of points on the circle as

$$C_r = \{ (x, y) \in \mathbb{R}^2 : x^2 + y^2 = r^2 \}$$

where $r$ is fixed. Neither $x$ nor $y$ is "dependent" or "independent" here and we have not introduced any kind of "time" variable (although we could if we wanted to).

If you like, you can think of the equation $x^2 + y^2 = r^2$ as defining a Boolean function called isAPointOnTheCircle(x, y) which returns true if $x^2 + y^2 = r^2$ (that is, if $(x, y)$ are the coordinates of a point on the circle of radius $r$ centered at the origin) and false otherwise. The set-builder notation picks out the subset of $\mathbb{R}^2$ consisting of the points on the circle, which is to say, the points satisfying this condition.

Here I'm thinking of $r$ as being something like a global variable we defined earlier but of course we can also consider a Boolean function isAPointOnTheCircle(x, y, r) taking three inputs. It depends on whether we're planning on saying something about different values of $r$ or not.

Qiaochu Yuan
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  • Thanks! as I asked Angina above, in this case, you need to know x AND y to know r, which means x and y are both independent here, yes? It’s just a pythagorean theorem right triangle revolving around the origin...why is r not considered a dependent variable? It too varies, just in relation to x and y... how is it fixed if it is simply the hypotenuse that’s shifting along with x and y? – Jordan Fine Aug 11 '20 at 05:02
  • @Jordan: I maintain that neither $x$ nor $y$ are "independent" or "dependent" here, there simply is no independence / dependence being imposed on the situation. Whether $r$ is itself varying depends on further details of the context in which you're talking about the equations for circles, but you said "equation of a circle" and not "Pythagorean theorem" and the easiest interpretation of that as a context is that $r$ is fixed. – Qiaochu Yuan Aug 11 '20 at 05:25
  • If you really want to consider $r$ a third variable you can consider the graph ${ (x, y, r) \in \mathbb{R}^3 : x^2 + y^2 = r^2 }$ which is a pair of cones. Picking a particular value of $r$ amounts to slicing these cone in a particular way to form circles, like this: https://en.wikipedia.org/wiki/Conic_section – Qiaochu Yuan Aug 11 '20 at 05:27
  • I guess I just don’t get why, in $y=x,$ x is independent and y is dependent, no? Is not $y=x$ considered a function, where $y’s$ values are dependent on $x?$ The equal sign fundamentally carries a different semantic meaning then(!) no? – Jordan Fine Aug 11 '20 at 06:35
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    @Jordan: that's an extra layer of interpretation it's common but not necessary to put on top of the bare equation. Note that, as an equation, it's identical to saying $x = y$, but there's a layer of convention on top concerning what it means to write an equation of the specific form $y = f(x)$ where $f$ is some function. – Qiaochu Yuan Aug 11 '20 at 10:01
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    The equation $x = y$ can be analyzed in the style I describe above, as simply describing a particular subset ${ (x, y) \in \mathbb{R}^2 : x = y }$ of the plane $\mathbb{R}^2$, without any need to impose a notion of dependent or independent variable. – Qiaochu Yuan Aug 11 '20 at 10:02
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    You are basically correct that the equal sign carries a different meaning when you impose the usual layer of interpretation on top of $y = f(x)$, it behaves more like an "assignment" as you say in the OP. It's tricky to talk about this in math-land because math takes place in what you might call a language with no side effects; every mathematical object just is the way it is forever, they never get "modified" as such. – Qiaochu Yuan Aug 11 '20 at 10:03
  • As I say, this is legitimately confusing, not least because it's hard to find anyone willing to explain this sort of thing in detail. Mostly people either don't get it or pick it up implicitly and don't know how to talk about it. – Qiaochu Yuan Aug 11 '20 at 10:05
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    This is ironic and funny: “Mostly people either don't get it or pick it up implicitly and don't know how to talk about it.” Thanks for taking the time to explain. Von Neumann quote: “In math, you don’t understand things, you just get used to them” I suppose. That subset notation seems like “the only way it should be taught” and the rest is seriously just inconsistent, confusing, and worthless! No wonder people walk away from math early on... it’s truly taught in a non-optimal manner. – Jordan Fine Aug 11 '20 at 10:35
  • Let us continue this discussion in chat. – Jordan Fine Aug 11 '20 at 10:45