Why do cubic equations always have at least one real root, and why was it needed to introduce complex numbers?

Question

I am studying the history of complex numbers, and I don't understand the part on the screenshots. In particular, I don't understand why a cubic always has at least one real root.

I don't see why the reason is "since $y^3 − py − q$ is positive for sufficiently large positive $y$ and negative for sufficiently large negative $y$)."

Secondly, I don't see how this fact led to the need to introduce complex numbers. It seems like the book is hinting at this. But maybe I am wrong. I don't see why it was necessary to introduce complex numbers due to cubic equations if they were shown to have at least one real solution. Then the real solution could be enough, right?

Thank you for your explanation!

14.2 Quadratic Equations

The usual way to introduce complex numbers in a mathematics course is to point out that they are needed to solve certain quadratic equations, such as the equation $x^2 + 1 = 0$. However, this did not happen when quadratic equations first appeared, since at that time there was no need for all quadratic equations to have solutions. Many quadratic equations are implicit in Greek geometry, as one would expect when circles, parabolas, and the like are being investigated, but one does not demand that every geometric problem have a solution. If one asks whether a particular circle and line intersect, say, then the answer can be yes or no. If yes, the quadratic equation for the intersection has a solution; if no, it has no solution. An "imaginary solution" is uncalled for in this context.

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14.3 Cubic Equations

The del Ferro-Tartaglia-Cardano solution of the cubic equation $$ y^3 = py + q $$ is $$ y = \root{3}\of{ \frac{q}{2} + \sqrt{ \left(\frac{q}{2}\right)^2 - \left(\frac{p}{3}\right)^3 } } + \root{3}\of{ \frac{q}{2} - \sqrt{ \left(\frac{q}{2}\right)^2 - \left(\frac{p}{3}\right)^3 } } $$ as we saw in Section 6.5. The formula involves complex numbers when $(q/2)^2 - (p/3)^3 < 0$. However, it is not possible to dismiss this as a case with no solution, because a cubic always has at least one real root (since $y^3 - py - q$ is positive for sufficiently large positive $y$ and negative for sufficiently large negative $y$). Thus the Cardano formula raises the problem of reconciling a real value, found by inspection, say, with an expression of the form $$ \root{3}\of{ a + b\sqrt{ -1 } } + \root{3}\of{ a - b\sqrt{ -1 } } $$

The quote is from the Mathematics and its history book pp 276-277.

Answer 1

You ask in a comment

I am not sure, because if it was a real root, then I don't understand why this is used as an argument for introducing complex numbers? Having a real root feels unrelated to having a complex root, no? Maybe that's my real misunderstanding. Even if I saw why it always has a real root, I don't see how this leads to need for introducing complex numbers.

Even though the cubic has real coefficients the formula for the roots requires complex numbers.

That seemingly contradictory fact is probably what delayed discovering the formula. The book on the history of mathematics you quote from should discuss that.

Read about the casus irreducibilis on wikipedia.

Answer 2

I don't understand why a cubic always has at least one real root.

I don't see why the reason is "since $y^3 − py − q$ is positive for sufficiently large positive $y$ and negative for sufficiently large negative $y$)."

It is known that the graph of a cubic function is a single, unbroken curve (i.e. the function is continuous). If it is negative for some value and positive for another, then it must pass through zero. This is because any unbroken curve that goes from below the $x$-axis to above it must pass through it. Hence, every cubic of the form $y^3 - py - q$ has at least one real root. Since any cubic can be transformed into one of that form by a simple horizontal translation -- which doesn't add or remove any crossings of the axis -- we can extend that result to all cubics.

I don't see why it was necessary to introduce complex numbers due to cubic equations if they were shown to have at least one real solution. Then the real solution could be enough, right?

This is logical, but there's a step between putting in $p$ and $q$ and getting out that root. That step is evaluating the expression for the root given at the top of 14.3. The problem is that the evaluation of the inner square roots is not possible (exclusively in reals) for all $p$ and $q$, but as argued above, there is a root for all $p$ and $q$ and this expression should evaluate to that root in all cases. If you try some actual values of $p$ and $q$, using your knowledge of complex numbers, you will quickly see what's happening: the two terms of the expression are complex conjugates, and so their sum is always real. But if you don't have a theory of complex numbers, you have a gap in the evaluation of the expression and thus you cannot find the solution that you know is there.

Answer 3

(this is just a simple historical explanation, without delving in the theory of cubic equations)

The trouble with the solution of the cubic equation is that when the equation has three real solutions the formula found by Tartaglia required to take square roots of negative numbers. Eventually he noticed that if he supposed that such number existed and the usual rules for sum and multiplication could be applied, at the end of the calculation those number disappeared and an actual (real) root was found, so Tartaglia did not worry too much. Of course, this was not a sound reasoning, but at that time it was enough as long as it worked.

Answer 4

First you were asking about: I don't see why the reason is "since $y^3 − py − q$ is positive for sufficiently large positive $y$ and negative for sufficiently large negative $y$)."

Answer for this question: If $f(x) = x^3 +ax^2+bx+c$, then it is a continuous function and observe that $\displaystyle \lim_{x \to \infty} f(x) = \infty$ and $\displaystyle \lim_{x \to -\infty} f(x) = -\infty$. As it is continuous, it can't always be positive or negative. So there exists $x_1,x_2 \in \Bbb R$ such that $f(x_1) < 0$ and $f(x_2) > 0$, so by IVP there exists $c$ between $x_1$ and $x_2$ such that $f(c) = 0$.

Also, by the Fundamental Theorem of Algebra, we know that any $3$ degree polynomial has exactly $3$ roots. Also, if $z \in \Bbb C$ is a root of $f(x)$, then $\bar{z}$ is also a root of $f(x)$(as $f(\bar{z}) = \bar{f}(z)$). These two situations tell us

1. $f(x)$ can have one real and two complex roots.
2. $f(x)$ have three real roots.

Your second query was: Secondly, I don't see how this fact led to the need to introduce complex numbers.

Answer: Each polynomial in $\Bbb R[x]$(set of all polynomials with real coefficients) doesn't necessarily have a real root. A great example is the polynomial $x^2+1$, which has no real roots. So, we have to extend the field where we can find the roots of this polynomial. This field extension is also called a splitting field for a polynomial, which can be split into linear factors.

Edit: You are right because extension fields were made to introduce the subject for studying the roots of polynomials. A cubic equation always has at least one real solution. Cardan discovered that to obtain real solutions to a cubic equation, you need to make sense of the square roots of negative numbers. So they faced a dilemma: you either say that square roots of negative numbers do not exist and then lose these legitimate real solutions for cubics, or else introduce a theory for square roots of negative numbers so that we can keep these legitimate real solutions. So, solving polynomials is the actual motivation. Galois made this generalization. See this.

Answer 5

With tools from Calculus the intermediate value theorem would do the trick.

If calculus is not part of the tool box, the Descartes' sign change criteria works well for real polynomials of the form $x^3-px-p$.

Answer 6

A cubic function $y = Ax^3 + Bx^2 + Cx$ can have two turning values. If you plot it out, it can have a maximum and a minimum. Some cubic functions such as $y = x^3 + x$ have no turning values.

A graphical solution for $D = Ax^3 + Bx^2 + Cx$ would be the points where $y = Ax^3 + Bx^2 + Cx$ crosses the horizontal line $y=D$. If D lies between the two turning values then the line will cross the curve three times. If D is not between the two turning values, or there are no turning values then you will get only one intersection. If you have two intersections then your curve has a single turning value, and so it is a quadratic, and not a cubic.

You do not have to use complex numbers. You don't have to use negative numbers, or zero if it comes to that. But it does help understanding.

Answer 7

You already had good answers, so here is just a hint to grasp the idea:

I don't see why the reason is "since $y^3−py−q$ is positive for sufficiently large positive $y$ and negative for sufficiently large negative $y$."

In a polynomial, the order of magnitude when the variable (here, $y$) approaches $+\infty$ or $-\infty$ is given by the term of higher degree, here $y^3$. This is a general rule (you have to remember it, it's useful!).

Said in another way, when $y$ is a very big number, the other terms ($p y - q$) are negligible, and so, as $y \rightarrow \pm \infty$,

$$y^3−py−q \sim y^3$$

So your function looks like:

Thus, there exists $a$ such that $f(a) < 0$, and there exists $b$ such that $f(b) > 0$. Since the function is continuous, this means that there exists $c$ such that $f(c)=0$.

You got it, there is always a real root!