Why is this equation of what appears to be a circle not a function? How do show this algebraically.

I know this is a circle (right?) and so for every x, there are 2 values of y (so y is not a function of x), but how do I algebraically show this. This is the question: If I square root both sides, I get:

$$y = \sqrt{4 - x^2}$$

But that seems to get rid of a y answer.

If I reduce it, I get:

$$y^2 = (2+x)(2-x)$$

But I still feel like I haven't algebraically shown this NOT to be a function.

• If I square root both sides, I get ... No, that's only the half of the circle above the $x$ axis. What you should be actually getting is $\,|y| = \sqrt{4-x^2} \iff y = \pm \sqrt{4-x^2}\,$ which does not define a function.
– dxiv
Jan 23 '18 at 4:59
• Showing that something is not a function of $x$ is relatively easy--all you need is one counterexample, which can be just two points with the same $x$ and two different $y$ values. If the domain of the function is stated then an $x$ that has no $y$ could also be a counterexample. Jan 23 '18 at 12:15
• An equation (alone) never defines a function. At least you need to define what the input and output set of your function shall be.
– chtz
Jan 24 '18 at 9:56

Hint:

For $x=0$, how many $y$'s can you find to satisfy this equation?

I know this is a circle (right?)

Yes, that's right. If you move $x^2$ to the other side (in other words, add $x^2$ to both sides), then you'll get $x^2+y^2=4$, which is the standard equation of a circle centered at the origin with radius …. Since a circle fails the Vertical Line Test, it is not the graph of a function.

and so for every $x$, there are $2$ values of $y$ (so $y$ is not a function of $x$)

No, that doesn't have to be true. What is true is that for SOME values of $x$ (at least for one such value, although for the circle we have many) there are multiple (two or more) values of $y$.

Say, in this case, some values of $x$ have multiple values of $y$ that correspond to them. For example, if $x=0$, then $y^2=4-x^2=4$ and therefore $y=\pm2$, two values. The same is true for many other values of $x$, but not for all. To $x=2$ corresponds only one value of $y$, and for $x=3$ there are no corresponding values of $y$ at all.

If I square root both sides, I get: $y=\sqrt{4−x^2}$

No, not true. Learn to be extra careful whenever you try to take square roots! Since two different numbers can have the same square, the correct consequence of $y^2=4−x^2$ is $y=\pm\sqrt{4−x^2}$. This already indicates that to a single value of $x$ often correspond multiple (two) values of $y$.

But I still feel like I haven't algebraically shown this NOT to be a function.

Even a single violation of the function property "to each $x$ (in the domain) corresponds only one $y$" is sufficient to conclude that an equation does not define a function. So the example that I showed above with $x=0$ is sufficient as an algebraic proof.

To show that this is not a function, it is sufficient to show that there are two values of $y$ that correspond to the same value of $x$. In other words, we want to find two pair $(x,y_1)$ and $(x,y_2)$ that both solve the equation. This is, essentially, the algebraic version of the "vertical line test." In this case, we might consider the example of $(0,2)$ and $(0,-2)$.

Something that for some reason seems to not be taught by any high school teacher:

It is not true in general that $\sqrt{y^2}=y$. The actual relation is $\sqrt{y^2}=|y|$.

The symbol $\sqrt{x}$ is used to denote the non-negative square root of a non-negative real number. You can get the other root with $-\sqrt x$.

That said, you don't need square roots to discuss the problem in your question.

In the case of the relation $y^2=4-x^2$, it does not represent a function, because in a function you should have, for each value of $x$, a single value of $y$; and that's not the case here. For instance, if you take $x=0$, then both $y=2$ and $y=-2$ satisfy the relation.

You should have $$y = \pm \sqrt{4 - x^2}$$ instead. Then it is not one function but two glued together to make a circle.

Taking square root on both sides, you would get:

$$y =± \sqrt{4-x^2}$$

It means that there are two values of$y$ for a single value of $x$. If you recall the definition of a function, then you should know that an input ($x$) can have one and only one output($y$), but the vice versa is not true.

I hope, now you know why it is not a function. And to prove that, maybe you can take any value of $x$, say $0$ and then show that there are two values of $y$ for that value of $x$.

The Cartesian coordinate system links algebra and geometry...

A circle fails the vertical line test... so is not the graph of a function...

A function should have one output for every input. The equation for a circle does not satisfy that condition.

I feel like other answers lack the strictness necessary to reason about mathematical notions and objects.

When we try to determine, whether object A belongs to a class C, we have to understand 1) what defines class C? 2) does A satisfy the definition?

An algebraic function is defined as a bijection, i.e., a one-to-one mapping between two sets. One set is called a set of inputs (we usually denote an element of this set as x). The other one is called a set of outputs, which we denote as y.

Our object does define a mapping from a set of real numbers into a set of real numbers. Now we need to check whether it is a bijection, i.e., every x should have exactly one corresponding y. A single example, where this doesn’t hold is enough to declare the object as not belonging to the class of algebraic functions.

x = 0 is an example of two corresponding y’s; x = 5 is an example of no corresponding y’s

Therefore, the object is not an algebraic function.

PS: Proving that something is of a certain class is more challenging and more interesting :)