0
$\begingroup$

This question already has an answer here:

apologize in advance if this is a duplicate, but I found a lot questions related to this but none answering this specific question.

My logic is: let's consider division the opposite of multiplication. So if I have

x / y = z. then that means that z * y = x

i understand that when y = 0 there can be so such z that would make the equation true for x if x is not zero. But if both x AND y = zero, then there is literally an infinite amount of numbers that can be z. e.g

1 * 0 = 0
2 * 0 = 0
3 * 0 = 0
4 * 0 = 0

So again, why is this undefined instead of infinity.

(s/n: after typing out this question i realize it may be my misconception of the true definition of infinity but alas I'm deciding to post this anyway)

$\endgroup$

marked as duplicate by David K, Mankind, mrf, user223391, user230715 Aug 27 '15 at 14:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Infinity is not a number. When we divide one number by another we must get, again, a number; say, real numbers. Since infinity is not a number, it does not make sense to say 0/0 = infinity. Think of a/b to be the number c such that a=bc. Now you are proposing that c = infinity is a solution. However, this not so simply because infinity is not a number. $\endgroup$ – User0112358 Aug 27 '15 at 13:39
  • 1
    $\begingroup$ In the answer math.stackexchange.com/a/26452 it was explained what the implications are of the fact that $1\times0=0$, $2\times0=0$, etc. If you can edit your question to say why that explanation did not answer your question then this will not be a duplicate. $\endgroup$ – David K Aug 27 '15 at 13:41
0
$\begingroup$

That's the whole point: if you were to say $0/0=1$ since $1×0=0$ then by the same logic we can say $0/0=2$ but then we would have to make $1=2$ (or any other number) which is absolute nonsense so the only logical option is to say that it is undefined.

Edit: If you introduce $\infty$ into our number system then you would also have to define what $\infty × 0$ is (should it be 0 or $\infty$ or something else?) and you run into more trouble.

$\endgroup$
0
$\begingroup$

In this context, infinity is not really defined, and there is no number such as $\infty$. What is defined is a sentence like ‘$f(x)$ tends to $\infty$ as $x$ tends to $0$’, but it's only a metaphor to say ‘$f(x)$ can be set a large as we want, as soon as we choose $x$ small enough’.

$\endgroup$
0
$\begingroup$

Can you definitely say that $0\over0$ is diverge towards infinity?

You can view the fact that you can't get a definite conclusion , so $0\over0$ is undefined . you can also view the fact that $\lim_{x\to0}\sin({1\over x})$ is undefined as we can't get a definite conclusion about this limit.

$\endgroup$
0
$\begingroup$

Matt B went into detail about why $\frac{0}0$ cannot be defined in the same sense that other division is defined since $a\cdot 0=0$ has infinitely many solutions that cause problems. This shows that it doesn't make sense for &\frac{0}0$ to have any finite value, but doesn't quite go into why it isn't infinity.

Now, to see that we need to talk about what infinity is. Essentially to say something is infinity, we need a way to estimate that thing, and we need a way to make that estimate better indefinitely. If in the process of refining our estimation, the estimate grows unbounded in a predictable way, we say this thing is infinity. This doesn't work here because we have no way of estimating $\frac{0}0$ or even if we did, there is no way to refine an estimate to make it "better".

$\endgroup$

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