Today in class my teacher wrote something along the lines of:

$6^x = 0$

And proceed to heed a response from the class. A few people shouted undefined. So the teacher then writes:

no solution $\therefore$ undefined

Now my question: Is undefined the same thing as no solution? From what i understand about 'undefined' is not necessarily that the equation or expression has 'no solution' but rather that there can be infinitely many solutions. For example take $\frac{0}{0}$

Now if we have a line on the Cartesian plane, say, $ y=x$ then for every point on that line $\frac{y}{x} = 1$. But $(0;0)$ lies on the line so therefore $\frac{0}{0} = 1$

But now if we repeat that for the line $y = -x$ and follow the same argument we get $\frac{0}{0} = -1$ and so we can continue this argument for any line parsing through the origin with infinitely many gradients and thus infinitely many lines, and therefore we cannot 'define' $\frac{0}{0}$ by any one number.

Or at least that is what i was thinking when the question was written down.


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    $\begingroup$ Personally, I think your teacher is wrong to use the word undefined. The fact that the equation has no solution is just to say that $0$ is not in the range of the function $x \to 6^x$, but everything about the written equation is perfectly well defined. Undefined, to me means something more like your example $0/0$ where we literally have not defined what it means. $\endgroup$ – user38355 Feb 12 '13 at 18:42
  • $\begingroup$ ^ $0/0$ is undefined, but some teachers will insist on using the word "indeterminate" to refer to indeterminate forms like that one. $\endgroup$ – Joe Z. Feb 12 '13 at 18:46

To answer your question directly: The equation $6^x = 0$ has no solution for $x$, and therefore $\log_6 0$ is undefined.

To expand on the whole "no solution" vs. "undefined" thing, both "no solution" and "infinitely many solutions" (and in general anything other than "exactly one solution") mean that the expression representing the equation is undefined. For example, your example of $\frac 00$ can be represented as "the solution to $0x = 0$", which is any real or complex number (or in general any number in the field you're working in regardless of what that field is); therefore $\frac 00$ is undefined.

It is not always the case, however, that an undefined value will always stay that way. Take the case of the square roots. The equation $x^2 = 4$ has two roots, $2$ and $-2$, so technically $\sqrt{4}$ is "undefined". But because it suits our uses, we define the square root of a number to be its positive square root, and things work out.

In fact, the equation $x^2 + 1 = 0$ has no solution in the real numbers, so $\sqrt{-1}$ is also "undefined". But then we defined $i$, the imaginary unit, just to cover this case, and again, things worked out. Of course, by doing so we lost the property of ordering (Is $1$ or $i$ greater? There is no answer).

In general when you try and invent numbers to satisfy certain properties, you lose some properties that the previous system had. For example, extending the natural numbers to the integers to satisfy closure of subtraction means you lose well-ordering. Extending the integers to the rational numbers to satisfy division means you lose the existence of prime numbers, the division algorithm, and in general the possibility of numbers not being divisible by one another. Extending the rational numbers to the real numbers to satisfy the least-upper-bound property means that you lose finite representability (and maybe some other property that I don't know about), as some irrational numbers really do require an infinite, arbitrarily generated Cauchy sequence to define them.

In the case of your example of $\frac 00$, we could simply define $\frac 00$ to be $1$. However, by doing this, as you saw, we lose a lot of the properties that the real numbers give us, like multiplication being consistent. If we do this, then $1 = \frac 00 = \frac {2 \times 0}{0} = 2 \times \frac 00 = 2 \times 1 = 2$, meaning that things don't work out. In other words, by defining $\frac 00$, we lose too much other stuff for it to be useful. So we simply leave $\frac 00$ undefined and disallow it in most uses. (There are some cases in, say, calculus, where the form $\frac 00$ really is useful, but it's always in the context of limits and what value something takes as an expression approaches $\frac 00$.)

In short, the reason some things are undefined is simply because defining them causes trouble.

For more interesting reading on the subject of definitions and undefined numbers, try looking up the debate on whether $0^0 = 1$, or even the history of rational and irrational numbers. There have even been squabbles over whether negative numbers exist, and sometimes that makes for fun reading.

  • $\begingroup$ +1 I think this makes good sense. The equation has no solution. This is demonstrated since the normal operations for finding a solution are undefined given the arguments arising. $\endgroup$ – abnry Feb 12 '13 at 18:49
  • $\begingroup$ Don't you mean the other way around? $\endgroup$ – Mitch Feb 12 '13 at 20:09
  • $\begingroup$ $\log_6 0$ is not an equation, so it has no solution only because of that. The equation, however, is very well defined. $\endgroup$ – Joe Z. Feb 12 '13 at 20:23
  • $\begingroup$ A more popular question that goes into more detail about what I've been talking about here: math.stackexchange.com/questions/259584/… $\endgroup$ – Joe Z. Feb 15 '13 at 13:38
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    $\begingroup$ Yes, by an expression representing an equation I mean an expression that represent the solution of that equation using a term defined for that statement. $\endgroup$ – Joe Z. Apr 11 '14 at 18:03

I think the distinction is one of proper (English) usage.

An equation can have no solution(s). You do not say that an equation is undefined. Being undefined is not a property an equation can have.

For example, the equation $x^2 = -1$ has no (real) solutions. It (the equation) isn't "undefined".

An expression can be undefined. You do not say that an expression has no solution.

For example, $\lim_{x \to \infty} \sin x$ is undefined. You do not say that this limit has "no solutions". (Nor should you write, as I sometimes see, ``$\lim_{x \to \infty} \sin x = \emptyset$''.)

  • $\begingroup$ Well certainly the solution set to $\lim_{x \to \infty} \sin x$ is empty. $\endgroup$ – Joe Z. Feb 12 '13 at 20:31
  • $\begingroup$ @Joe Zeng, you're falling into the same trap that Bruce is explaining. The limit expression is an expression. It does not make sense to talk about its "solution set". $\endgroup$ – Hammerite Feb 12 '13 at 21:52
  • $\begingroup$ Sorry, I meant the solution set to the corresponding equation. I imagine you could say "the set of values $L$ such that for every $\varepsilon > 0$, there exist an $N$ big enough that if $x > N$, $|\sin x - L| < \varepsilon$" to be the corresponding "equation" that the expression solves. $\endgroup$ – Joe Z. Feb 12 '13 at 23:41
  • $\begingroup$ ^ where $L$ is the value being solved for, not $x$, of course. $\endgroup$ – Joe Z. Feb 13 '13 at 0:26
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    $\begingroup$ @MatthewLeingang : Well, anything operated on with a DNE becomes a DNE, kind of like NaN in any programming language that implements it. $\endgroup$ – Joe Z. Feb 13 '13 at 3:37

This problem has no solution. The word ``undefined'' is being misused.


An equation's solution is a set. That set can be empty, or it can consist of a single unique solution, some bounded number of solutions, or an infinite number of solutions.

An empty set is not undefined in any sense. The solution for an equation which has no solutions is perfectly well defined: it is the empty set object: $\emptyset$.

If an equation is formed using this pattern:

$$f(x) = C$$

and if that equation has no solutions, then it means that the inverse relation $f^{-1}(C)$ is undefined at $C$. This is a different concern. (Note that the inverse relation is is not necessarily a function.)

The equation not having solutions means that there exists no value of $x$ such that $f(x)$ makes $C$. We can say that $C$ lies outside of the range of the function $f$. And since $C$ lies outside of the range of $f$, it means that $C$ lies outside of the domain of the inverse relation $f^{-1}$. A relation is only defined for values inside its domain by, ahem, the definition of what it means to be defined.


For the sake of variety....

The actual practice of mathematics tends to implicitly use more flexible semantics than is actually explicitly described. One interpretation is that semantics not based on functions but on partial functions.

For example, if you've ever really thought through questions like

What is the domain of the function $f(x) = \sqrt{1 - x^2}$?

you'd see that they are very poor questions when taken at face value, when interpreted in the usual way. There are two problems:

  • The domain of $f$ is, by definition, the set of values that the variable $x$ ranges over. So there's nothing to prove.
  • $1-x^2$ isn't an expression known to be limited to the domain of $\sqrt{}$, so the equation is nonsense anyways.

However, when interpreted in terms of partial functions, the question both makes sense and is meaningful. The function $f$ so defined assigns a value to every $x$ such that $1-x^2$ is in the domain of $\sqrt{}$, and it does not assign a value to any other $x$.

This is a special case of composition of relations.

To resolve ambiguity, when considering a partial function $f:X \to Y$, I will call $X$ and $Y$ the "source" and "target" respectively. The "domain" of $f$ is defined to be the subset of $X$ of values that $f$ assigns a value to.

So if $x$ is a real variable, then the function $f(x) = \sqrt{1-x^2}$ has source $\mathbb{R}$ (the set of real numbers), target $\mathbb{R}$, and domain $[-1,1]$.

Let $I$ denote a set with one element. Recall the notion of "element of a set $S$" is equivalent to the notion of "function $I \to S$". Keeping in the spirit that notation should be more along the lines of partial functions than functions, this means that there should be an idea of a "partial element".

The set of all partial functions from $I \to S$ consists of all ordinary functions and one new partial function: the one whose whose domain is empty.

Switching back to the notion of element, it is not unreasonable to call the partial element of $S$ corresponding to the partial function with empty domain "the undefined element of $S$".

And this matches conventional usage. Recall that if $a$ is a partial element of $S$, $i$ is the corresponding partial function, and $f$ is a partial function $S \to T$, then $f(a)$ is the element of $T$ corresponding to the composite $f \circ i$.

So continuing our example of $f(x) = \sqrt{1-x^2}$, if we follow the above rule to plug in the value $2$, we get that $f(2)$ is the undefined real number.

One awkwardness of this notation is that there's an ambiguity of what "$=$" should mean. Specifically, while $f(2)$ is the undefined real number, it would be reasonable to insist that $f(2) = undefined$ is not a true statement; instead, it is the undefined truth value. I haven't thought through the specifics of this notation enough to have decided how to treat "$=$".

Another feature that comes up in practice is generalized elements, e.g. in the form of indeterminate variables and expressions involving them.

If I define the variable $x$ to be the general real solution to the equation $6^x = 0$, this definition makes sense in terms of generalized elements; it is more or less equivalent to the "undefined real number" described above. And interpreted in the context of generalized elements, it does make sense to say $x \in \emptyset$.


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