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Difference Between Root and Solution 📂Lemmas

Difference Between Root and Solution

Definitions

  1. A point of the domain that makes the function value of a given function become $0$ is called a root.
  2. Something that satisfies the conditions of a given problem is called a solution.

Explanation

In short, a root is something formal, and a solution is something conceptual. Many people not interested in mathematics get confused by these terms because in many cases, they are used interchangeably. $$ f(x) = 0 $$ For example, if we were to ask what satisfies $x$ for the equation above, the point that makes the function value of $f(x)$ become $0$ and also solves the equation problem at the same time. What is important here is the mention of ’equation problem’s solution’ not just ’equation’s solution’. The term solution is used in the context of a ‘problem’, whether it’s a partial differential equation problem or optimization problem, and the word ‘problem’ is just often omitted for convenience. On the other hand, a root isn’t about a problem but simply about a point on the blackboard that makes the function value become $0$. It might be sad not to solve a problem, but not being able to find a point where the function is $0$ isn’t something to be sad about, and they are distinctly different.

Cases Where Roots and Solutions Do Not Coincide

$$ f(x) = h $$ If we consider an equation problem like the above for $h \ne 0$, we can see that its solution is not a root of $f$. Conversely, from a root’s standpoint, there’s no way to satisfy $f(x) = h$, so it cannot be a solution. At first glance, it might seem that situations like the one described for $h \ne 0$ are much more common than those for $h = 0$, but in fact, if we introduce a new function as $g = f - h$, $$ g(x) = f(x) - h = 0 $$ the solution that satisfies this is also a root of $g$. By moving all the terms to one side, you can eventually make roots and solutions coincide, which is why ‘roots’ and ‘solutions’ can often become synonymous.

Why We Use Roots

At first glance, if algebraic manipulation can always make roots and solutions the same, it seems there’s no reason to consider a ‘root’ as a concept instead of a ‘solution’, which seems like a higher-level concept. The problem is that in the vast world of mathematics, the concept of a ‘problem’ can be so liberally defined that it’s not always easy to handle mathematically. For instance, consider the following two quadratic equation problems. $$ \begin{align*} x^{2} - 2 x + 1 =& 0 \\ x^{2} + 4 =& 4x \end{align*} $$ Of course, as objects of our intuition or contemplation, these can be distinctly identified. But consider what happens if you transfer $4x$ from the second equation to obtain the third equation below. $$ \begin{align*} x^{2} - 2 x + 1 =& 0 \\ x^{2} + 4 =& 4x \\ x^{2} - 4 x + 4 =& 0 \end{align*} $$ Can the second and third equations truly be considered the same? Strictly speaking, there’s not a single thing that’s the same between their left-hand and right-hand sides. Of course, intuitively we understand they’re ’effectively’ the same, so ’this time’ we might say they are. However, with more challenging problems, it’s unclear what might happen, but at least for quadratic equations, we could avoid this dilemma by establishing an equivalence relation that ‘if rearranging can make them look the same, then they are the same equation’. $$ \begin{align*} p_{1} (x) =& x^{2} - 2 x + 1 \\ p_{2} (x) =& x^{2} - 4 x + 4 \end{align*} $$ But then, wouldn’t it be better just to think about a set of functions like the above? Considering the roots of these functions has become exactly the same as considering the solution of equations, so there’s no need to forcibly create an ’equivalence relation’ to say ’two equations are effectively the same’. It’s much easier to navigate around by definition the ‘functions’ strictly defined in set theory when the domain and codomain are given than meticulously parsing through equations. I’m not exactly sure how to describe the entire set of equations mentioned in the example, but the entire set that contains ‘functions’ whose roots are also their solutions can be simply formalized as follows. $$ \left\{ p : \mathbb{R} \to \mathbb{R} \mid p(x) = a x^{2} + b x + c \text{ where } a,b,c \in \mathbb{R} \right\} $$

Etymology of Root

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I’m not quite sure, but if you draw the shape of the continuous function $f : \mathbb{R} \to \mathbb{R}$ on Cartesian coordinates as above, $0$ might seem like the root of a tree. Just as there was a time when mathematicians did not recognize complex numbers as solutions to problems, there was also a time when negative numbers were used in calculations but not recognized as ‘resulting values’ or solutions. At that time, the shapes of the functions could have all looked like they were sprouting upwards from $y = 0$, and even if negative numbers were accepted, there wouldn’t have been a big problem in calling the location $f(x) = 0$ a root. But anyway, that’s just one possibility.

Zeros

In abstract algebra, $0$ is specially treated, typically as the identity element of the group for addition in a given field. If it’s the field of polynomial functions, then naturally the constant function $0$ plays this role. The $0$ mentioned in the definition of a root doesn’t necessarily need to be an element of real numbers $0 \in \mathbb{R}$ anymore, and the phrase related to ‘roots’ from the above paragraphs could have been left as an expression of etymology alone.