4

$\dfrac 00=x$
$0x=0$
$x$ can be any value, therefore $\dfrac 00$ can be any value, and is indeterminate.

$\dfrac 10=x$
$0x=1$
There is no such $x$ that satisfies the above, therefore $\dfrac 10$ is undefined.

Is this a reasonable or naive thought process?
It seems too simple to be true.

helpme
  • 661
  • 1
    I would call it naive in the sense that when referring to "indeterminate forms in the form of $\frac{0}{0}$" we aren't referring to the actual explicit division of zero by zero, but rather we are talking about a limit of a ratio where both the numerator and denominator approach zero simultaneously. The explicit expression $\frac{0}{0}$ is also undefined. The limit $\lim\limits_{x\to 0} \frac{f(x)}{g(x)}$ where both $\lim\limits_{x\to 0}f(x)=0$ and $\lim\limits_{x\to 0}g(x)=0$ is indeterminate, the value depending on what $f$ and $g$ explicitly are. – JMoravitz Oct 28 '19 at 12:13

3 Answers3

5

Those expressions are about limits, not about numbers.

We say that $\frac00$ is an indeterminate form because a limit of that form can take any value:$$\lim_{y\to0}\frac{xy}y=x,$$for any real number $x$.

On the other hand, a limit of the type $\frac10$ cannot take any value. If it exists, it can only be $\infty$ or $-\infty$.

José Carlos Santos
  • 427,504
3

In the context of limits, $0/0$ is an indeterminate form (limit could be anything) while $1/0$ is not (limit either doesn't exist or is $\pm\infty$). This is a pretty reasonable way to think about why it is that $0/0$ is indeterminate and $1/0$ is not.

However, as algebraic expressions, neither is defined. Division requires multiplying by a multiplicative inverse, and $0$ doesn't have one.

eyeballfrog
  • 22,485
  • Why are they both undefined as algebraic expressions? – helpme Oct 28 '19 at 12:57
  • 1
    @helpme Division is really multiplying by the multiplicative inverse. $0$ has no multiplicative inverse, so division by it is undefined. – eyeballfrog Oct 28 '19 at 12:58
  • 1
    @helpme Or... you could formally invent a symbol called $"1/0"$, and then, you've just proven that $1/0$ would be equal to everything. So the cost of giving meaning to the expression $1/0$ is that suddenly $1=2$ and other statements that are clearly incompatible with standard arithmetic. – WoolierThanThou Oct 28 '19 at 13:00
  • Oh okay. So basically, 1/0 does not exist because if it does, then it wouldn't work with the math rules. Let τ=1/0. 0τ=1. x0τ=x. 0τ=x. τ=x/0. 1/0=x/0 which doesn't work (x represents any number). That means that 1/0, the multiplicative inverse of 0 does not exist. 0 multiplied by the multiplicative inverse of 0 does not make any sense and is undefined. Therefore both 1/0 and 0/0 are undefined. Is this correct? – helpme Oct 28 '19 at 13:11
0

The form $\frac{0}{0}$ at $x=x_0$ means that if you find $f(x_0\pm \delta)$ in the neighborhood of $x_0$, you may get a good number. All the following function are of the form $\frac{0}{0}$ at $x=1$; $f(x)=\frac{x-1}{x-1}, \frac{x^2-1}{x-1},\frac{x^3-1}{x-1},...$ but in the small neighborhood of $x=1$ they take different values as $1,~ 2\pm \epsilon(\delta),~ 3\pm \epsilon(\delta),...$, respectively. Here $epsilon(\delta)$ are as small as we may want and this is achieved by coosing very small values of $\delta$.

Z Ahmed
  • 43,235