How Understanding Vectors Helps Resolve a Common Misconception in Mathematics

From elementary school, when we first learn about exponents, we’re introduced to the square root. Pass any real number into it, and you get its square root back; inversely, multiply the square root of a number by that square-rooted number again, and you get the original number. A question that always comes to the ever-so-curious child’s mind is “Well, what about negative numbers?” Of course, a teacher may tell a younger student that such cases are completely impossible, but later on, one is introduced to the concept of imaginary numbers: a set of numbers that expands one’s thinking to a plane beyond the limits of real numbers. The introduction of imaginary numbers is one of my favorite parts of early algebra, but it also comes with one of my biggest pet peeves in terms of how it’s taught. The imaginary number i is the main character of this topic, and with it, teachers can finally explain when a square root can be negative: because \( i = \sqrt{-1} \), we can say that for any positive number n , the square root of that negative number would be \( \sqrt{-n} = i\sqrt{n} \), solving our biggest mathematical plot hole since dividing by zero.

Except there’s a slightly-majorly important error here: i is not defined as the square root of -1. And here’s why.
Countless times I've seen people say that i is the square root of -1 and that that’s what allows for negative square roots. They’re not necessarily wrong, but such a definition hides a lot more of what there is to the story of complex numbers, and there’s more than a couple of ways to show that.

The first, and far easier, way I’ll describe here is actually quite simple- and if you don’t want to read the rest of this post, you can stop after this part- and relies on the definition of the square root: albeit a tad circular in its explanation, but if you multiply the square root of a number by itself, you will get the original number. At first glance, it seems trivial that this is definitely the case with i, but any closer look will show that this is not the case.
To say \( i = \sqrt{-1} \), it suffices that \( i^2 = \sqrt{-1} \cdot \sqrt{-1} = -1 \) must be the case as well. However, exponent rules state that \( a^n \cdot b^n = (a \cdot b)^n \). Since the square root is an exponent (n = 1/2), the rules apply here as well:
\( i^2 = \sqrt{-1} \cdot \sqrt{-1} \\ = (-1)^{1/2} \cdot (-1)^{1/2} \\ = (-1 \cdot -1)^{1/2} \\= \sqrt{-1 \cdot -1} \\= \sqrt{1} = 1 \) .
But, if what we stated before is true, then this would mean -1 = 1, which is a clear contradiction.
To circumvent this, mathematicians often say instead that \( i^2 = -1 \) when defining the imaginary constant i. It’s a very slight, but very important difference that allows i to hold under more involved mathematics; it is this description that I adhere to as well and will seek to demonstrate in a more rigorous manner.

One of the joys in mathematics is how several routes can lead to the same conclusion, illuminating different things along the way. To understand a more in-depth reason as to why the “square root” definition for i doesn’t hold, we can look at vectors, the operations of vector spaces, and how what they do (more precisely, what they don’t do) allows for the definition of the complex field, and a more rigorous proof of what i truly is.
But first, before we can get all complex, we’ve gotta keep things real- I promise this is needed later on too! Essentially, for the majority of calculations that the average person does day-to-day, we use real numbers: numbers that include both rational (any number that can be represented as \( \frac{p}{q} \), where p and q are both integers (0, 1, -1, 2, -2, 3, -3, etc; naturally, q is not equal to 0) and their greatest common divisor is 1) and irrational (numbers that specifically do not fit the prior description) numbers, and can describe anything from the number of meals someone had in the past week or the amount of weight one benched in their recent gym session.
Mathematically, one of the best things about real numbers is that the set of all of them forms what’s known as a field (rational numbers do as well, but not one that’s of much use analytically and won’t be of much use here): a set of numbers that follows certain axioms³, or accepted truths with no need to be proven, in addition and multiplication. While I won’t go into what exactly those axioms are (I’ve cited a link that explains those), I’ll be using this definition to describe a field vs a set from here on out; the fields of real and rational numbers are denoted \( \R \) and \( \mathbb Q \) respectively (integers and irrational numbers do not form a field). Another term I’ll define is closure: For any operation \( * \) (just a generalized way of denoting anything like addition, multiplication, exponentiation, etc) between elements a and b in any field \( \mathbb F \), if the result of that operation is also an element in \( \mathbb F \), then we say \( \mathbb F \) has closure under operation \( * \), or more specifically, \( \forall a, b \in \mathbb F \implies (a * b) \in \mathbb F \) (aka, “for all elements a and b in the field \( \mathbb F \), a \( * \) b is also in \( \mathbb F \)).
One reason why I wanted to define closure in a field is to highlight one of the key differences between squaring a number and taking the square root of a number. For any real or rational number, the square of that number (in fact, any non-negative integer exponent) is for sure to also be a real or rational number; that is to say, the field of real numbers (as well as rationals) has closure under the square operation (integers do as well; closure applies to any set, not just fields). On the contrary, the square root does not have this closure property: not every positive real/rational number has a real/rational square root, and no negative number at all has a real square root. This is just to offer more of a mathematical context behind why I’m writing this article.

Now, sometimes it’s simply not enough to just have a number when you want to attribute some quantity. Say, for instance, a projectile of any kind is being shot in some direction- I know the speed of the object, but no idea where it’s going. As any physics class would tell you, this is where vectors come in! We’ll come back to a more general definition for them in a bit, but for a physical interpretation, vectors have a direction and a magnitude (a number that describes how far the vector extends in that direction). Velocity, for example, is a vector that has a magnitude (speed) and a direction (47 degrees north of east), and can perfectly describe the state of motion of our flying projectile in this example. A vector can be broken into discrete components by directions- x, y, and z coordinate axes are a perfect case of these- and are generally denoted just like a variable, but with an arrow on top: \( \vec v =\lang v_x, v_y, v_z \rang = v_x \hat i + v_y \hat j + v_z \hat k \), where the x, y, and z subscripts denote individual magnitudes along unit vectors (vectors with magnitude 1) in their respective directions from some arbitrary origin. Unit vectors are perpendicular to each other, so by adding them we can directly find the vector \( \vec v \) with minimal effort.
What I described above was a more geometric definition for a vector. More abstractly/algebraically, a vector is defined as any element of a set that follows certain vector axioms⁴ under vector addition and scalar multiplication, and such a set is called a vector space (sound familiar? Note the use of the word “set” instead of “field,” as it is an intended wordage). The entirety of these axioms and what they mean are again explained in another source I’ve linked, but I want to focus more specifically on what “scalar multiplication” means. A vector space \( V \) is not itself a field, but simply a set of vectors. Thus, in order to scale a vector by some number (i.e. doubling in length), vector spaces are defined over a field \( \mathbb F \) to “take” a scalar (basically an element of that field) from- usually this is the field of real numbers \( \R \).

However, just because a vector space uses scalars from \( \R \) does not mean the vector space itself is \( \R \). In fact, vector spaces in particular are specific in exactly which kinds of vectors they have; a vector \( \vec v_1 = \lang3, 6 \rang \) is not necessarily in the same category as another vector \( \vec v_2 = \lang-4, 9, 0, 0 \rang \), even if they both have two non-zero coordinates. In fact, the number of coordinates a vector has dictates the minimum number of basis vectors- these are a set of vectors that are linearly independent (no other vector in the basis can be made from these vectors) and span (by combining scalar multiples of the basis vectors) the entire vector space- that can describe it; for the first vector, this is 2, and 4 for the second. This minimum number of basis vectors, for any such collection in a vector space, is called the dimension of that space (we’ll just say n for any n-dimensional space). For \( \vec v_1 \) and \( \vec v_2 \), their vector spaces would be 2- and 4-dimensional. Because both these vectors’ coordinates are strictly real numbers, we can basically say their respective vector spaces are \( V_1 \sube \R^2, V_2 \sube \R^4 \) (the \( \sube \) symbol means an improper subset, which itself just means the sets might not necessarily be the entire set of (n-dimensional) real numbers, but are for sure within that field; think of it like saying “you live in Massachusetts” when specifically you’re in Boston).
Now, I’m gonna stick with \( \R^2 \) here because it highlights something in particular that transitions to complex numbers quite well, but the following is true for any vector space \( \R^n \) (with n > 1 of course): it actually is not a field. For all of these one-dimensional fields thus far, we’ve used the default addition and multiplication rules we all know and love, which perfectly fit the field axioms necessary for \( \R \), but this doesn’t work when we have a vector. The only forms of vector multiplication² we have are the dot product, which (for our intents and purposes) returns only a real number, or the cross product, which does return a vector, but not one in the set of our original vectors (in fact, it will be orthogonal to the plane containing both original vectors). So how can we find a form of multiplication that returns a vector still in our original set? There is one way…
With any set that may not necessarily be a field, they can be equipped with certain addition and multiplication rules (the vast majority of the time) to be made into what’s called a ring¹. Rings here are the triple \( (S, +, \cdot) \) where \( S \) is any nonempty set and for any elements a, b, c in \( S \), addition and multiplication follows certain properties: For addition, standard rules of associativity and commutativity apply, as well as the identity element (essentially the zero of the set, where adding any element a to it brings back a), and the inverse element (there exist some element -a where adding a to it returns the zero element), both in \( S \) (and naturally, closure in \( S \) as well)- a set that follows these properties is interestingly called an abelian group. For multiplication, all that’s required is associativity and distributivity over addition (i.e. \( a \cdot (b + c) = ab + ac = ba + ca = (b + c) \cdot a \), note from both left and right sides), and also has closure. Note that rings are not fields because a field requires an inverse element in multiplication as well; inversely though, this in fact means all fields are rings.
The issue here is that the vector space \( \R^2 \) doesn’t have a standard multiplication closed in itself, so we can introduce a special kind of element-wise multiplication \( \cdot_{\R^2} \) where \( \forall \vec a, \vec b \in \R^2, \vec a \cdot_{\R^2} \vec b = \lang a_1b_1, a_2b_2\rang \) gets a result also in \( \R^2 \). Accompanying it with element-wise addition \( +_{\R^2} \) as usual (subscript is just for convenience), we can turn the vector space into a ring \( (\R^2, +_{\R^2}, \cdot_{\R^2}) \) which solves that problem. However, it’s still not a field- not every nonzero element \( \vec v = \lang v_1, v_2\rang \) has an inverse that, when multiplied by it, returns the identity element (for this case, \( \lang1, 1\rang \) is the multiplicative identity). For that, we need a slightly more robust operator for multiplication: for two vectors in \( \R^2 \) just as before, let \( \cdot_{\mathbb C} \) be the multiplication (the subscript \( \mathbb C \) used here will be important very soon) where \( \vec a \cdot_{\mathbb C} \vec b = \lang a_1b_1 - a_2b_2, a_1b_2 + a_2b_1 \rang \) is also a vector in \( \R^2 \). With this, one can see that with the multiplicative identity \( \lang1, 0\rang \) it’s possible to find an inverse for any nonzero \( \vec v \): \( \lang \frac{v_1}{v_1^2 + v_2^2}, \frac{-v_2}{v_1^2 + v_2^2} \rang \), where showing that multiplying this with \( \vec v \) gives \( \lang1, 0\rang \) is a succinct enough exercise to leave to the reader.

But what does any of this have to do with complex numbers? Well what if I said that with the structure \( (\mathbb{R}^2, +_{\mathbb{R}^2}, \cdot_{\mathbb{C}}) \) as we’ve defined, we’ve actually just essentially made the complex number field \( \mathbb{C} \)? Just as two unit vectors in x and y are on orthogonal axes, let the set of real numbers \( \mathbb{R} \) be an axis, and the set of (purely) imaginary numbers \( \mathbb{I} \) be its own axis, completely perpendicular to \( \mathbb{R} \), but both intersect at the origin 0. This setup actually becomes equivalent to \( \mathbb{R}^2 \) once we let the ordered pair \( (a, b) \) describe any point on this grid. If we then take 1 and i to be the “unit vectors” along the real and imaginary axis specifically, the pair is equal to \( a + bi \), and by incorporating the rest of the ring \( (\mathbb{R}^2, +_{\mathbb{R}^2}, \cdot_{\mathbb C}) \), we’ve officially defined complex numbers in this new field \( \mathbb{C} \)! When two such algebraic structures are functionally and behaviorally equivalent like this, we say they are isomorphic to each other.
So we have complex numbers, now what? Where do we show that \( i^2 = -1 \) works in all cases here? First, note that with our definition of \( \mathbb C \), i is equivalent to the ordered pair \( (0, 1). \) Thus, it becomes a very doable bit of multiplication to show that
\( i^2 = (0, 1)^2 \\ = (0, 1) \cdot_{\mathbb C} (0, 1) \\ = (0 - 1, 0 + 0) \\ = (-1, 0) = -1 \).
This is also the case with \( -i = (0, -1) \), which I’ll leave for the esteemed reader to show. On the contrary, saying \( i = \sqrt{-1} \) only gets half of the story; you can’t directly define the square root without first already knowing the former case (of \( i^2 = -1 \)), not to mention the fact that even carrying out \( i^2 \), starting from \( \sqrt{-1} \), ends up in a fundamental error, as explained in the very beginning of this piece.

So there we have it: from elementary algebra to ring theory (and a dash of topology), it not only makes more sense to say that \( i^2 = -1 \), but such a definition brings with it all of what complex numbers have to offer, and the ambiguity in the square root leaves just too many loose ends to use in its stead.

Reference List

Thangarajah, Pamini. "5.1: Introduction to Rings." LibreTexts Mathematics. Last modified July 7, 2023. libretexts.org.
Weideman, Tom. "1.2: Vector Multiplication." Physics LibreTexts. University of California, Davis. Accessed July 22, 2026. https://phys.libretexts.org/.
Weisstein, Eric W. "Field Axioms." From MathWorld--A Wolfram Resource. https://mathworld.wolfram.com/FieldAxioms.html
Weisstein, Eric W. "Vector Space." From MathWorld--A Wolfram Resource. https://mathworld.wolfram.com/VectorSpace.html