101
$\begingroup$

Consider the following expression:

$$\frac{1}{2} \div \frac{4}{x}$$

Over here, one would state the restriction as $x \neq 0 $, as that would result in division by $0$.

But if we rearrange the expression, then:

$$\begin{align} \frac12\div\frac4x &= \frac{1}{2} \times \frac{x}{4} \\ &= \frac{x}{8} \end{align}$$

In this expression, there are no restrictions. If we substitute $x = 0$, then the answer is $\frac{0}{8} = 0$.

So, how come in the first unsimplified expression, when we substitute $ x =0$, we get undefined, whereas in the simplified expression we get $0$?

$\endgroup$
  • 60
    $\begingroup$ What is meant here, is that if you start from the false statement that $\frac 12 \div \frac 4x$ is defined for all $x$, you end up with the true statement that $\frac x8$ is defined for all $x$ after simplification. This is what is meant by false implying true. More generally, a false statement may very well imply a true statement, but one cannot deduce the truth of a proposition, from the truth of it's implications. $\endgroup$ – астон вілла олоф мэллбэрг Dec 2 '17 at 23:54
  • 2
    $\begingroup$ To give an example, let's start with the false statement : all multiples of $3$ are multiples of $2$. If you apply this to $6,12,18$ etc. you would get a true statement. But applying this to $3,9,15$ etc. would give you incorrect statements. $\endgroup$ – астон вілла олоф мэллбэрг Dec 2 '17 at 23:56
  • 6
    $\begingroup$ "But if we continue with the scenario" You can't "continue with the scenario" if x = 0. "So, how come in the first unsimplified expression when we substitite x=0, we get undefined, whereas in the simplified expression we get 0. " Because they are not the same expression. $\endgroup$ – fleablood Dec 2 '17 at 23:56
  • 14
    $\begingroup$ @Pedroski I'm showing you my zero apples right now! You can see all zero of them, right? (More seriously, I think you're quite wrong. Mathematicians in general are actually very good about how they use language; they know exactly what they mean when they say something is a number, and they know exactly why zero counts as such.) $\endgroup$ – Tim Pederick Dec 3 '17 at 13:20
  • 5
    $\begingroup$ @Pedroski: Tim is right when he is saying that he is showing you zero apple. The negation of "showing zero apple" is "showing at least one apple". Is he showing apples if he has an empty hand? Also, if you want to justify that $0$ is not a number, you should define "numbers" first, and this has probably nothing to do with apples. For example, for most people, $\pi$ is a real numbers, but I can hardly show $\pi$ apples. $\endgroup$ – Taladris Dec 4 '17 at 4:41
131
$\begingroup$

EDIT: I have made a small change in terminology based on the suggestion by @BPP in the comments below.


This is an excellent question!

The subtlety here is that in high-school mathematics we are taught to freely move between polynomials and polynomial functions. But if we do not make the distinction between the two carefully, then confusions like in your question will arise.

When I have the equality $$\frac{1/2}{4/x} = \frac{x}{8}$$ there is no problem when viewing this is an equality of rational fractions, because in this case, there is no question of evaluating $x$, and hence no question of what the domains are, simply because they are not functions.

On the other hand, when I look at the equality $$\frac{1/2}{4/x} = \frac{x}{8}$$ as an equality of rational functions, then it is necessary that the domain of both functions are the same. Since the function on the left-hand side is defined for all $x \neq 0$, I can take that to be the domain of the function, and so the domain of the function on the right-hand side is also $\mathbb{R} \setminus \{0\}$.

A lengthier explanation of this distinction follows below.


What is a polynomial?

Typically, it is taught at the high-school level that a polynomial (in one variable) is any expression of the form $$a_0 + a_1 x + \dots + a_n x^n$$ where $a_0,\dots,a_n$ are real numbers, and $n$ is a non-negative integer. We can denote this polynomial by $f(x)$. Here, $x$ is called a variable or indeterminate.

At the same time, we are also taught that we can substitute a real number, say $t$, for $x$ in order to evaluate the polynomial at $t$, and this is represented by $f(t)$. This is the point where the distinction between polynomials and polynomial functions should be made.

What is a polynomial function?

For our purposes, we will stick to functions from real numbers to real numbers. We define a polynomial function to be a function $f: \mathbb{R} \to \mathbb{R}$ such that there exist real numbers $a_0,\dots,a_n$ so that $f(t) = a_0 + a_1 t + \dots + a_n t^n$ for every $t \in \mathbb{R}$.

How are polynomials and polynomial functions the same?

At first glance, there may not appear to be much of a difference between polynomials and polynomial functions, for the following reason: given a polynomial $f(x) = a_0 + a_1 x + \dots + a_n x^n$ I can define a polynomial function $\tilde{f} : \mathbb{R} \to \mathbb{R}$ by $\tilde{f}(t) = a_0 + a_1 t + \dots + a_n t^n$ for each $t \in \mathbb{R}$. And, given a polynomial function $\tilde{g} : \mathbb{R} \to \mathbb{R}$ such that $\tilde{g}(t) = b_0 + b_1 t + \dots + b_m t^m$ for each $t \in \mathbb{R}$, I can define a polynomial $g$ by $g(x) = b_0 + b_1 x + \dots + b_m x^m$. This back and forth between polynomials and polynomial functions is compatible in the sense that the map $F$ from the space of polynomials (over $\mathbb{R}$ in one variable) to the space of polynomial functions (from $\mathbb{R}$ to $\mathbb{R}$) that sends the polynomial $f$ to the polynomial function $\tilde{f}$ by the above rule, preserves the structure of both spaces. More precisely, it satisfies the conditions: $$ \begin{gather*} f + g \mapsto \tilde{f} + \tilde{g}\\ cf \mapsto c\tilde{f}\\ fg \mapsto \tilde{f} \tilde{g}\\ 1 \mapsto \operatorname{id} \end{gather*} $$ where $\operatorname{id}$ is the constant function $\operatorname{id}(t) = 1$ for all $t \in \mathbb{R}$. (As an aside, all this is often expressed by saying that the map $F$ given by $F(f) = \tilde{f}$ is an isomorphism of algebras.)

The fact that the function $F$ has these properties essentially means that polynomials and polynomial functions are not all that different.

How are polynomials and polynomial functions different?

Despite the previous section, it is important to make a distinction between polynomials and polynomial functions for the following reason. When we divide a polynomial $f(x)$ by another polynomial $g(x)$ we get the rational fraction $f(x)/g(x)$. (Of course, $g(x)$ could divide $f(x)$ without leaving a remainder, in which case $f(x)/g(x)$ is in fact a polynomial, so every polynomial can also be considered a rational fraction.) Two rational fractions $f_1(x)/g_1(x)$ and $f_2(x)/g_2(x)$ will be considered equal if, after cancelling their common factors, we end up with the same expression in both cases. Or equivalently, the two rational fractions will be considered equal if $f_1(x) g_2(x) = f_2(x) g_1(x)$. Note that this is completely analogous to the definitions we make when talking about rational numbers.

Note that a rational fraction is not a function (since polynomials are not 'really' functions). However, we can define a rational function $h$ to be a function such that there exist real numbers $a_0,\dots,a_n$ and $b_0,\dots,b_m$ so that for every $t$ in the domain of $h$, we have $$h(t) = \frac{a_0 + a_1 t + \dots + a_n t^n}{b_0 + b_1 t + \dots b_m t^m}.$$

Clearly, I can repeat the same process as earlier: to every rational fraction I can associate a rational function, and to every rational function I can associate a rational fraction, in the obvious way. So, where does the problem arise?

The problem arises in the fact that two different representations of the same rational fraction can give rise to two different rational functions. For example, let $f(x) = x^2 - 4$, $g(x) = x-2$, $h(x) = x+2$ and $k(x) = 1$. Clearly, $f(x)/g(x) = h(x)/k(x)$. However, the function $u$ defined by $u(t) = f(t)/g(t)$ and the function $v$ defined by $v(t) = h(t)/k(t)$ are not the same function, because their domains are different! (I am assuming that all the functions have their domain as the maximal set on which they are defined. So, the domain of $u$ is $\mathbb{R} \setminus \{ 2 \}$ and domain of $v$ is $\mathbb{R}$.)

So, let's say we started with the rational function $v$. We construct the 'natural' rational fraction using $v$, namely $h(x)/k(x)$. We write $h(x)/k(x) = f(x)/g(x)$. Then we construct the rational function $u$ from the rational fraction $f(x)/g(x)$. But we end up with a different function! So, the realisation is that the 'evaluation map', which is basically 'replace $x$ by the real number $t$', is not an 'isomorphism' between the set of all rational fractions and the set of rational functions. Hence, we cannot move back and forth freely between the two notions unless we sort out how we want to define the evaluation map compatibly.

How does all this answer the question?

When I have the equality $$\frac{1/2}{4/x} = \frac{x}{8},$$ this is certainly true when I view both sides as rational fractions. In this case, there is no question of evaluating $x$, and hence no question of what the domains are, simply because they are not functions.

On the other hand, when I look at the equality $$\frac{1/2}{4/x} = \frac{x}{8}$$ as an equality of rational functions, then it is necessary that the domain of both functions are the same. Since the function on the left hand side is defined for all $x \neq 0$, I can take that to be the domain of the function, and so the domain of the function on the right-hand side is also $\mathbb{R} \setminus \{0\}$.

The question, “Why am I able to make sense of the right-hand side at $x = 0$ and not able to make sense of the left-hand side at $x = 0$?” arises because we expect the evaluation map to be 'compatible', in some sense. That this is not so, is the content of the discussion above.

What have I glossed over?

I have certainly been handwaving a bit when talking about isomorphisms, but this can be made fully precise without much trouble. All these ideas are discussed in any senior undergraduate course in algebra.

Also, as @AlexProvost commented below, I have defined a rational fraction to be a ratio of polynomials, and yet I am calling $\frac{1/2}{4/x}$ a rational fraction, despite $4/x$ not being a polynomial. Here, I have glossed over the fact that dividing a rational fraction by a rational fraction gives one a rational fraction. This can be shown quite easily, but I had still skipped this point.


Explanation of the edit: Earlier I had called ‘rational fractions’ as ‘rational functions’, and ‘rational functions’ as ‘functions that are rational’. I think the new terminology is cleaner, and I am not sure whether there is any standard terminology.

$\endgroup$
  • 3
    $\begingroup$ Maybe it's better to call them rational fractions instead of rational function since they aren't functions. $\endgroup$ – Paracosmiste Dec 3 '17 at 15:50
  • 1
    $\begingroup$ One does not have to restrict the domain of the RHS, one can use the alternative condition: "on the intersection of the domains of the RHS and LHS, equality holds". This is equivalent to what I was taught: "identities only have to hold where both sides are defined." $\endgroup$ – Eric Towers Dec 3 '17 at 22:57
  • 1
    $\begingroup$ Related: math.stackexchange.com/questions/2185587/… $\endgroup$ – Ethan Bolker Dec 4 '17 at 14:15
  • 3
    $\begingroup$ I have had polynomial called a formal polynomial, to make it extremely clear we where not defining a function. $\endgroup$ – Yakk Dec 5 '17 at 19:10
  • 2
    $\begingroup$ How exactly is $\frac{1/2}{4/x}$ a rational fraction? The denominator $4/x$ is not a polynomial. $\endgroup$ – Alex Provost Dec 5 '17 at 22:54
78
$\begingroup$

Your computation assumes that dividing by $\frac4x$ is equivalent to multiplying by $\frac x4$, but in fact they're equivalent only as long as $x\neq0$. The general theorem that justifies this sort of computation step says: If neither $a$ nor $b$ is zero (and you're working with real numbers or, more generally with a field), then $x\div\frac ab=x\cdot\frac ba$. Before applying the conclusion of this (or any) theorem, you need to check that the hypotheses are satisfied.

$\endgroup$
31
$\begingroup$

When we make a simplification like that, we are implicitly assuming that $x\neq 0$. Really, one should write that $$ \frac{1}{2} \div \frac{4}{x}=\begin{cases} x/8&\text{if $x\neq 0$}\\ \text{undefined}&\text{if $x=0$}. \end{cases} $$

$\endgroup$
  • $\begingroup$ Now you're defining a function $f:\mathbb R \to (\mathbb R \cup \{\text{undefined}\})$ which might be okay, but this undefined element is a bit weird. It should be much nicer to just simply define $f:\mathbb R\setminus\{0\} \to \mathbb R$. Then we would find $f(0)$ to make no sense, and we would say that $f$ is not defined at $0$, thus it does not actually yield a result of "undefined", rather it simply was never defined for zero, just like it has not been defined for complex numbers, strings, or any other abstract object that is not a real number. $\endgroup$ – Benji Altman Dec 6 '17 at 16:40
17
$\begingroup$

The simplification you did is correct. The only issue is that you must keep the domain restrictions when simplifying, otherwise the simplified function and the original function are not equivalent, as they have different domains.

Take, for example, the function $f(x) = \frac{x(x-2)}{x-2}$.

This function simplifies to $f(x) = x$. However, when $x=2$, the original function does not exist while the simplified one equals $2$. So $f(x) = x$ and $f(x) = \frac{x(x-2)}{x-2}$ are not equivalent functions unless you keep the domain restrictions of $x \not = 2$.

To summarize: the manipulations you made are only equivalent to what you started with if you keep the domain restrictions. So you can't "continue with the scenario" assuming that it's defined for $x=0$, because the initial expression implicitly implies that it's not.

$\endgroup$
14
$\begingroup$

The function, as first written, may be rewritten: \begin{align} f : \; \; \left(\mathbb R-\{0\}\right) & \rightarrow \mathbb R \\ x & \mapsto \frac x 8 \end{align} ...in other words, $x/8$ where $x\neq0$, and undefined otherwise.

However, if we take a look at this graph, we notice that there's probably a very natural value to assign to $f(0)$...

enter image description here

...that value being $f(0)=0$. Technically speaking, what we're doing is extending the function $f$ like so: \begin{align} \bar f : x & \mapsto \lim_{y\rightarrow x} f(y) \quad \quad \text{where the limit exists} \end{align}

We may perform this extension wherever the original function is missing only discrete points from its domain, and the limits approaching these points from either side, $\lim_{y\rightarrow x^+}$ and $\lim_{y \rightarrow x^-}$, are well-defined and equal.

This new, extended function is: \begin{align} \bar f : \; \; \mathbb R & \rightarrow \mathbb R \\ x & \mapsto \frac x 8 \end{align} ...and is well-defined for its entire domain.

enter image description here

$\endgroup$
7
$\begingroup$

Your first phrase has a restriction $x\neq 0$ and it is equal to $\frac{x}8$ this is OK.

You can use $\frac{x}8$ where ever you like. But the condition $x\neq 0$ is still in effect. You cannot remove it. If you allow $x=0$, then $\frac4x$ was not defined. But, it raises the question why you have used it before? It invalidates the rest of your work.

$\endgroup$
4
$\begingroup$

A more computationally oriented point of view provides a better explanation. We use awful ambiguous notation in the standard language. Switch to a notation that is more oriented to performing this with a computer; a constructivist notation.

Declare the existence of some variables like this $$ a:Int, b:Int, c:Int $$ If I declare a new variable, I need to declare a type for it.

Declare an operator called equals, such that it is commutative, associative, and everything left distributes and right distributes over it, etc. Use $"\to"$ to mean a one-way reduction step:

Right distribute $$ ((a = b) * c) \to ((a*c) = (b*c)) $$ Right factor $$ ((a*c) = (b*c)) \to ((a=b) * c) $$ Those examples don't have any side-conditions. But our first example of one that accumulates a side-condition, or equivalently has type constraints, happens with division. We define a division operator such that you can only form a division expression with the precondition that $d$ is of type $NonZeroInt$, which would be defined in the language something like $NonZeroInt(x): Int(x : (not(x=0)))$. But in any case, $(a:Int / b:Int)$ is not actually defined as a valid construction. The int b must be converted to NonZeroInt first.

$$ (a/(d:NonZeroInt)) $$

This is called a dependent type. What this means is that if you have a bunch of expressions, the variables in the expressions have types, and there are explicitly given (and named) reduction rules that can be applied to perform a reduction step. As a data structure, performing steps will produce a proof object that is a tree of reductions, like:

$$ A \to B, B \to C, B \to D, A \to E $$

You can turn this idea inside out, and just perform the algebraic operations. If your reduction did not invoke an $(a/b)$ step, then $b$ is not constrained to be $NonZeroInt$. This is kind of like "duck typed" algebra, where you can apply a bunch of steps and have constraints on what the type could be.

$(a/(b/0)) \to 0$ can be defined explicitly, because you cannot explicitly perform a step $(b/0)$.

$\endgroup$
4
$\begingroup$

You should distinguish between functions and formulas which are used to describe the former. $x/8$ is (being very rigorous here) no function, but a formula which you can use to define a function. A function can be more seen as the process which takes in a number (or something else) and spits out another number (or something else). There are other ways to give functions than by using formulas (e.g. graphs, tables, algorithms, ...)

So you have to decide what kind of function you want to express. Do you want the function which is defined in zero, or the one which is not. Once you have decided this, you know the domain: either $\Bbb R$ or $\Bbb R-\{0\}$.

The formula $x/8$ can be used to describe the bahavior of both these functions on their domain. While $x/8$ is defined for $x=0$, we can always ignore this value. On the other hand, your other expression

$$\frac12\div\frac 4x$$

can only be used to describe the function with domain $\Bbb R-\{0\}$ because this specific formula does not provide a value in the case $x=0$. It cannot be used to define a function on all of $\Bbb R$.

The fact that you can rearrange the first formula into the second does only show that the usual rearrangements procedures do not preserve the "range of validity" of those expressions. A line like

$$[\text{formula 1}]=[\text{formula 2}]$$

expresses that both sides agree on the value (of the function they define) if and only if both sides represent a value in the first place.

You should look at formulas as tools rather than the ground truth. They are not the function, but a way to denote them. And there are more appropriate ones, and less apropriate ones for specific cases.

$\endgroup$
3
$\begingroup$

Consider the function f(x)=x^3/x^4 here if you apply limit and put x=0 then definitely one will get 1/0 form which is undefined. So make use of the L'Hospitals rule here that is take derivative of numerator and denominator term separately that is, Applying limit when x approaches to zero and then taking limit

$\endgroup$

protected by Parcly Taxel Dec 6 '17 at 5:09

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.