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Can someone kind explain the mathematical quantity i to me (which is the square root of -1)?

Just to be clear, I'm not actually trying to understand i per se, I'm just trying to understand how it can possibly be defined as the square root of -1. I know that a square root is a number multiplied by itself to produce the original number e.g. 2 x 2 = 4. But I also recall that the only way a negative number can result from a multiplication is if exactly ONE of the two numbers is negative e.g. 2 x -2 = -4. But if the two numbers being multiplied are not the same magnitude and sign, then you are not actually squaring anything since a square root consists of a number of a given magnitude and sign multiplied by the exact same number with the exact same magnitude and sign. Therefore, it seems absurd to talk about the square root of a negative number.

I have never understood this and the teacher who taught us about i was never able to make it clear to me. It wouldn't surprise me if he didn't understand it himself.

I've wondered about this for many years now and never quite got around to tracking down a mathematician to explain it to me. I was wondering if anyone here could tackle this and help me understand?

I got pretty good marks in math right up until the end of high school - 40 years ago - but never did anything further in math at university or in my career. I'm just telling you that so you have some idea of who will be hearing your explanation.

I know that I learned about i in one of the three math classes I took in my final year of high school but I don't remember if it was algebra, calculus or geometry so I'm going to guess calculus. Please forgive me if I've tagged this incorrectly.

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marked as duplicate by Xander Henderson, Hans Lundmark, Lord Shark the Unknown, Cesareo, Namaste calculus Oct 17 '18 at 22:21

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10 Answers 10

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Your failure is in this sentence:

"But I also recall that the only way a negative number can result from a multiplication is if exactly ONE of the two numbers is negative e.g. 2 x -2 = -4"

That is true only for real numbers. That statement is not valid for complex numbers. If I give a similar reasoning, if I only know numbers in $\mathbb N$, I know that adding two numbers will always be greater or equal than any of my original numbers. So how can I define negative numbers, since they don't obey the above statement

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You have hit on exactly why there is no real number $x$ with the property that $x^2=-1$. You have confirmed that the number can't be negative, and that it can't be positive, and of course it can't be zero (since $0^2\neq -1$). So you are exactly right, there is no real number that is a square root of $-1$!

The explanation is, we propose the existence of a new number $i$ that is outside of the set of real numbers, which has the property that $i^2=-1$. This is precisely why we call such number "imaginary numbers".

In other words, we have extended the reals to include a whole batch of new numbers. We can't draw them on the familiar number line because that number line only contains real numbers. That's why we draw them on a different number line which is perpendicular to the "old" number line.

We also assume that the familiar rule "$0\cdot x = 0$ for any $x$" also holds for these new "imaginary" numbers. This means that $0i$ must really be the same as our familiar real number $0$, since $0i = 0\cdot i = 0$. This means that the two number lines meet at the number $0$, but nowhere else.

So far, we only have numbers on the real number line or on the imaginary number line (x-axis or y-axis). We can get all complex numbers $a+bi$ by adding a real number $a$ to an imaginary number $b$. This "addition" is an extension of the familiar addition of real numbers to the newly enlarged set which included the y-axis of pure imaginary numbers. In fact, this is how we define complex numbers, and this gives the correspondence of complex numbers with points in the plane: real numbers are $(a,0)$, pure imaginary numbers are $(0,b)$, and a general complex number is $(a,b)$.

We usually write "$a$" for "$a+0i$" or "$(a,0)$", "$bi$" for "$0+bi$" or "$(0,b)$", and "$a+bi$" for "$(a,b)$".

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I explained it to my daughter, 16, by quadratic equations which she is well aware of. Consider $$x^2+2x+4= (x^2+2x+1) + 3 = (x+1)^2+3 = 0.$$ This gives $x+1 = \pm\sqrt{-3}$ if the square root of -3 can be taken and then $x=\pm \sqrt{-3}-1$. If we expand $\sqrt{-3} = \sqrt{-1}\sqrt 3 = i\sqrt 3$, we obtain the usual notation. Now she is able to find the solutions of a quadratic equation even if the discriminant is negative, but she's not impressed.

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I've had some fun and some success several times with middle school students with this explanation.

The kids are comfortable with the (real) number line. There you interpret addition as translation: the map $x \to x+2$ just shifts everything to the right. Subtraction shifts to the left. Each shift is determined by how much it shifts $0$.

Multiplication is a little subtler. The map $x \to 2x$ rescales by a factor of $2$. Multiplication by $-1$ flips the line around $0$. (That helps convince kids that "minus times a minus is a plus".) Each multiplication is determined by how much it rescales $1$.

Now you can begin to think about square roots. What scaling can you perform so that the result of doing it twice is rescaling by a factor of $9$? Clearly it's $x \to 3x$.

Now what rescaling can you perform twice to end up with $x \to (-1)\times x$? With a little prompting you can see that you might just rotate the real line (counterclockwise) through a right angle, and then do it again. That's motivation for thinking about "numbers" as filling the plane, rather than just the line. The image of $1$ in that rotation is the "amount of rescaling". It needs a name. The symbol $i$ happens to be traditional.

Then you're off to the races. The map $x \to x + i$ is a vertical shift. Then vector addition of "complex numbers" $a+bi$ follows immediately: it's just translation.

It's more work than middle school kids can manage but possible to get to polar form and the "add angles, multiply lengths" rule for multiplication.

This is a version of the invention of complex numbers due to the Danish surveyor Caspar Wessel at about the same time mathematicians were learning how to interpret their algebraic success positing square root of $-1$ as the geometry of the complex plane.

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  • $\begingroup$ Tristan Needham's book demonstrates the multiplication rule for complex numbers without ever referring to the angles and using the addition formulas. $\endgroup$ – Ovi Oct 23 '18 at 12:52
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Well, a better, unambiguous definition is the quantity $i$ such that $i^2 = -1$, not that $\sqrt{-1} = i$. They are two extremely different statements. (N.B. Technically speaking $\sqrt{-1} = i$ is the principal branch of the square root).

usually square roots can only be really defined (pardon the pun) for real numbers. This has a number of reasons, one is the ease of simplification of

$$\sqrt{mn} = \sqrt{m}\sqrt{n}$$

If we assume this also holds for negative numbers, we get:

$$1 = \sqrt{1} = \sqrt{(-1)\cdot(-1)} = \sqrt{-1}\cdot \sqrt{-1} = \left(\sqrt{-1}\right)^2 = -1$$

Which clearly is false. Thus defining square roots of neagtive numbers is ill-advised, but we can assign $i$ to something....

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  • $\begingroup$ When you write "the quantity $i$", that is obviously not an "unambiguous definition": the quantity $-i$ also satisfies your definition. So you are back to where you started! $\endgroup$ – TonyK Oct 17 '18 at 15:55
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Formally, $i$ isn't defined as $\sqrt{-1}$ at all. Instead, the following is the actual definition of $\Bbb C$ as used in field theory and algebra in general:

Start with the set $\Bbb R[i]$ of polynomials with the variable $i$ and real numbers as coefficients, and add onto that the standard polynomial addition and multiplication. At this stage there is nothing special about $i$, and it acts just as any other letter would in algebra. You have polynomials like $i^5-3i^3+\sqrt2i-\pi i-\ln(2)$, and you add and multiply these polynomials the way you learned in school.

Now comes the complex numbers: Every time you see $i^2$, you're allowed to exchange that for $-1$. That includes $i^3=i^2\cdot i$ and any higher degree terms as well. We say $i^2=-1$.

That is what the complex numbers are, algebraically. That is how any mathematician, from graduate student and up, should have learned to think about them, and to be honest I believe it's how they should be introduced in the first place. No square roots, no new multiplication rules to memorise. Just polynomials the way you're used to, and a certain substitution.

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Only a bad teacher defines $i$ as the square root of $-1$. It turns out that at some point, in the XVIth century, it became clear that there were good reasons to extend the real numbers to a larger set in which the usual algebraic operations were still defined but in which there are also square roots of negative real numbers.

One possible way of doing this consts in taking the set $C$ of all maps $f_{a,b}\colon\mathbb{R}^2\longrightarrow\mathbb{R}^2$ of the type $f_{a,b}(x,y)=(ax+by,-bx+ay)$. You can add and subtract any two of these maps, and then you will get another element of $C$. And you can also compose them; you'll get again an element of $C$. Now, call multiplication to the composition. Furthermore, if $a\in\mathbb R$, identify the real number $a$ with the map $f_{a,0}$ (which is just the multiplication by $a$). If you do all this and if you call $i$ to $f_{0,1}$, it is easy to check that $f_{0,1}\circ f_{0,1}=f_{1,0}$. In other words, $i\times i=-1$ or, to put it shorter, $i^2=-1$.

This way, you will achieve the goal of extending the real numbers to a larger set in which the usual algebraic operations were still defined but in which there are also square roots of negative real numbers.

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  • $\begingroup$ Just curious, how would define $i$ and the complex numbers to highschool students just being introduced to basic algebra? (here in the U.S. it's part of the curriculum in basic algebra class) $\endgroup$ – Ovi Oct 18 '18 at 14:09
  • $\begingroup$ Perhaps this way: a complex number is a first degre polynomial $a+bx$ with real coefficients. Adding complex numbers is just adding polynomials. And the product of $a+bx$ by $c+dx$ is the remainder of the sivision of $(a+bx)(c+dx)$ by $1+x^2$. With this definition, $x\times x=-1$. So, define $i$ as being $x$. $\endgroup$ – José Carlos Santos Oct 18 '18 at 14:12
  • $\begingroup$ Ah okay thanks ${}{}$ $\endgroup$ – Ovi Oct 18 '18 at 16:57
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This is conventional. $i$ is not a real number, so that the rule "a square is non-negative" doesn't need to hold.

We define a "constant" $i$ and the operations

  • multiply $i$ by a real: $ib$,

  • add a real to $i$ multiplied by a real: $a+ib$ (this is called a complex number),

  • multiply $i$ by itself: $i^2$,

and add the rule $i^2=-1$.

Doing this we define a complete algebra on the complex numbers, which turns out to be extremely powerful and useful.

The famous expression "square root of $-1$" comes from the last rule I stated, that has nothing contradictory.


Complex numbers have a simple geometric interpretation that says "multiplying by $i$ amounts to rotating by $90°$". So multiplying twice by $i$ amounts to rotating by $180°$, or multiplying by $-1$.

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  • 1
    $\begingroup$ I thank the anonymous downvoter for illustrating $i^2$. $\endgroup$ – Yves Daoust Oct 17 '18 at 15:06
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Here is one way to understand what is going on:

Suppose that there were some new kind of number $i$ with the property that $i^2=-1$. What could we do with such a strange number? Would it be useful? The answer turns out to be: YES! Enormously useful. Astonishingly useful!

The question of whether such a number "really exists" is irrelevant, just as the question of whether negative numbers really exist is irrelevant. We have defined them, and we find that these definitions lead to useful mathematics.

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Try thinking of it this way: When we start with counting numbers and extend them to signed numbers, we model this by laying them out on the number line. Now a number is something with both magnitude and sign, and we can interpret 'sign' as direction and so interpret the combination as an 'arrow'. I.e when we say +1, we mean 'an arrow of size 1, pointed to the right'; and when we say -1 we mean 'an arrow of size 1, pointed to the left'. When we extend to the complex numbers, we add a dimension [i.e we add the 'imaginary part' to the 'real part']; complex numbers are now 'arrows in the plane'. In particular, the complex number i means 'an arrow of size 1, pointed up' [and -i means 'an arrow of size 1, pointed down'].

When we do this, we now ask 'what do addition and multiplication look like, in terms of arrows?' Addition is easy: it is what we call 'vector addition'. Multiplication is trickier. It turns out that, by interpreting 'arrow' in polar coordinates, and translating back to cartesian coordinates using sin() and cos() in the usual way, and applying the usual complex multiplication rule to that, you end up with the following simple rule: the product of two 'arrows' is an 'arrow' which, in polar coordinates, has length the product of the two given lengths, and angle the sum of the two given angles!

The 'square root' connection can now be explained: Since i is 'an arrow of length 1 and angle 90-degrees', its square - the product i*i - must be 'an arrow of length 1*1 and angle 90+90 degrees', i.e 'an arrow of length 1 and angle 180 degrees', i.e -1.

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