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Hello everyone my question may be stupid but I really don't understand why $$0 \times +\infty$$ is undefined since o times any number is zero no matter how large the number may be

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    $\begingroup$ $\infty$ is not a number. Consider the following limits as $x \to 0$: $\lim x . \frac{1}{x}$, $\lim x . \frac{1}{x^2}$, $\lim x^2 . \frac{1}{x}$. They are all of the form $0 \times \infty$ but all of them have different answers. $\endgroup$ – Anurag A Jun 20 '17 at 19:00
  • $\begingroup$ Does this answer your question? $\endgroup$ – projectilemotion Jun 20 '17 at 19:00
  • $\begingroup$ Yeah it makes sense thanks very much projectile motion $\endgroup$ – arsene stein Jun 20 '17 at 19:03
  • $\begingroup$ @arsenestein There was a post long ago (probably the one projectilemotion linked), so I wrote below some of the facts I thought and learned through that question (and my personal knowledge). Take a look at it, it may help you ! Although, in the future try not to post duplicate question (try searching your question). $\endgroup$ – Rebellos Jun 20 '17 at 19:07
  • $\begingroup$ Note that the other question does not say a word about indeterminate forms so this is a different question. $\endgroup$ – Mikhail Katz Jun 21 '17 at 7:25
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The fact is, $+ \infty$ is not a number. The laws of multiplication hold for a number $x\in \mathbb C \supseteq \mathbb R \supseteq \mathbb Q \supseteq \mathbb Z \supseteq \mathbb N$, but infinity is not a natural number. If you start searching around your argument, you could observe that, for example :

$$\infty \times 0 = \lim_{x \to \infty} (x \times 1/x) = 1$$

$$\infty \times 0 = 0$$

But the above, are not statements that hold for infinity, because, as mentioned, it cannot be treated as a real/natural number.

Infinity must be defined, before working with and on it, but the problem is that there are many different ways to define it, exactly because it's not a normal number case.

For example, a different and classic case is that you could define infinity by $1/\infty = 0$ but in the same time, infinity can be defined while talking about infinite sets (cardinal numbers).

As it becomes clearer, you see that infinity is rather hard to define strictly and thus really hard to handle in such strict ways, as the multiplication law.

So, to sum up, any multiplication or summation law that holds for the natural numbers does not have any meaning for infinity and thus cannot be used to derive expressions and calculations, such as : $0 \times \infty = 0 $.

Also, take in account that an operation for a natural number $n$ such the one you mentioned, is :

$$n \times 0 = 0 + 0 + 0 +\cdots+ 0 = 0$$

which is clear that is a finite operation and not an operation regarding something infinite.

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  • $\begingroup$ "The laws of multiplication hold for natural numbers" —Actually the law not only holds for natural numbers, but also for integers, rational numbers, real numbers and complex numbers. All of which of course don't contain $\infty$. Just replace "natural numbers" with "numbers" and your answer is perfect. (If you are worried of the formulation because of ordinal numbers and cardinal numbers: Both contain infinite numbers, but neither contains a number $\infty$. And in both, multiplying an infinite number with zero gives, indeed, zero, so the laws of multiplication hold there as well). $\endgroup$ – celtschk Jun 21 '17 at 7:20
  • $\begingroup$ @celtschk Yeah, was a bit worried about cardinal and ordinal numbers to be honest. But I'll just change it to a set-style phrase. $\endgroup$ – Rebellos Jun 21 '17 at 18:13
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The easiest way of understanding this particular indeterminate form is as saying that an infinitesimal times an infinite number could have any desired order of magnitude: infinitesimal, appreciable, or infinite. Such a situation is summarized by the symbol $0\times\infty$ but in point of fact you are not actually multiplying zero by an infinite number but rather an infinitesimal by an infinite number.

For example, if $\epsilon>0$ is infinitesimal, then $H=\frac{1}{\epsilon}$ is infinite. Then the products $\epsilon^2\times H$, $\quad \epsilon\times H$, and $\epsilon\times H^2$ are respectively infinitesimal, appreciable, and infinite.

See this textbook for details.

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