there're some students, who belive that $$\frac10 = \infty $$

I need to teach them that this is not true and $\frac10 $ is undefined, mathematically and give a good picture (for their minds)

what is the proper way to teach them without to telling them to memorize this

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    $\begingroup$ Consider $1$ divided by $10^k$ as $k$ gets big. Draw the answer. Of course this shows that $1/x \to \infty$ as $x \to 0$ from above, which is why they think $1/0 = \infty$. Now do the same with $1$ divided by $-10^k$ as $k$ gets large. Finally, do the same with $(-10)^{k}$ as $k$ gets large. $\endgroup$
    – Dan Fox
    Commented Jan 13, 2018 at 9:11
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    $\begingroup$ That equation is true in the projective extended real number system. You should probably explain that you are using the real number system, as is the default, and in that system division by zero is undefined. $\endgroup$ Commented Jan 13, 2018 at 10:14
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    $\begingroup$ As an addendum to @RoryDaulton's point, and a countering view (not one I hold), many people are of the opinion that the extended real numbers are by default part of any calculus class. It's possible these students have been explicitly taught that in the past. See: matheducators.stackexchange.com/questions/13378/… $\endgroup$ Commented Jan 13, 2018 at 14:30
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    $\begingroup$ Possible duplicate of Dividing by zero $\endgroup$
    – mweiss
    Commented Jan 14, 2018 at 4:50
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    $\begingroup$ Do not teach: $\frac{1}{0} = \infty$ is wrong". At worst, teach: "$\frac{1}{0} = \infty$ is wrong in the real numbers". Later they may learn that: "$\frac{1}{0} = \infty$ is correct in the projective line." Instead of simply saying "wrong", specify what number system you mean. $\endgroup$ Commented Nov 22, 2018 at 11:06

6 Answers 6


What is $\frac 1 a$? It is the unique (real) number such that $a\cdot \frac 1 a=1$. Does there exist a real number that multiplied by $0$ gives $1$? No. Why is this? Because if $0\cdot b=0$ which ever is $b$. This is about not being defined. Still... why is $\frac 1 0=\infty$ not so completely wrong? Because they can see that the smaller is $a$ then the bigger is $\frac 1 a$ (and I would make them appreciate this on the real line, maybe letting $a=\frac 1 n$) and therefore... (vague explanation but your question is vague since you do not specify the level of your students).

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    $\begingroup$ To paraphrase: Disallow 1/0 when you are doing algebra, as it breaks algebra. Allow 1/0 when you are doing geometry, as $\infty$ is perfectly fine in geometry. $\endgroup$
    – Adam
    Commented Jan 13, 2018 at 13:49

A little late to the party, but I wanted to add my two cents.

The students should understand that they are looking at the value of $1/0$ as a limit. This is good but it's not the entire picture, and using that value in normal computation will break algebra.

See the following: $$ 0 = 0 \\ 0\cdot{1} = 0\cdot{2} \\ \frac{0\cdot 1}{0} = \frac{0\cdot 2}{0} \\ 1 = 2 $$

I know that there are many other issues with the "proof" beyond just dividing by zero but the point is to challenge students with the implication of this result to convince them that a statement like $1/0 = \infty$ is loaded.


I teach mostly physics, but have taught calculus a couple of times. From the physics end, I see things in almost the opposite way that you do. Here is a typical way that this plays out in my class.

We have a homework problem where a cable is stretched between two buildings, with a streetlight hanging from the middle, so that the cable makes the shape of a "V". The students calculate a formula for the tension $T$ in the table as a function of the height $h$ by which it sags in the middle. The problem asks them to do some interpretation of the result, including checking its units, the trend of $T$ as a function of $h$, and the special case of $h=0$. (Note that we automatically have $h\ge 0$ in this problem.)

A student comes to my office hours for help with the problem, and we get to the point where we're checking the special case $h=0$. They say, "Oh, it's undefined, because zero is in the denominator."

At this point they think they're done, and they need to be disabused of this notion that has been inculcated in them. They are missing out on the insight that is to be gained by realizing that the result is infinite.

If you like, you can dress this up in various language such as the language of limits, or saying that the result is undefined as a real number, but only because the result is infinite, and we don't have infinite real numbers. But these are issues of mathematical formalism that are way above the level of this type of student. To see the actual level that they're operating at, it may be useful to consider the following fragment of dialog, which I have had many times with many students:

Student: "Oh, it's undefined."

Me: "That's great that you've had such good mathematical training, and your math teachers have told you that dividing by zero is undefined. But suppose we really did divide by zero. What do you think you would really get?"

Student: "Zero?"

Me: "Well, hang on, what kind of result would you get if you divided by a really small number, like 1/0.001?"

Student: "It would come out really small?"

Me: "Let's try that on your calculator."

Student: "Oh, it's big. I see."

In other words, they're not at the level where they understand that the expression diverges but they just don't know how to formalize that statement. They're at the level where they lack the idea that it blows up at all. They need a first exposure to the idea that it blows up, and then on some later pass maybe they can learn about calculus, limits, the extended reals, and so on. The first exposure needs to happen in grade school, when they first learn about division. If they learn division at age 9, and calculus (possibly) at age 18, then IMO the right answer for them to give from age 9 to age 18 is that $1/0=\infty$ (or possibly $\pm\infty$, depending on context).

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    $\begingroup$ I didn't vote and I really like your physical example, but "knowing" $1/0 = \infty$ for 9 years leads to a messy understanding of $\infty$ and has probably no benefit. $\endgroup$
    – Jasper
    Commented Nov 21, 2018 at 20:53
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    $\begingroup$ After many re-readings, to quote Pauli, this feels like it is "not even wrong". $\endgroup$ Commented Nov 23, 2018 at 4:58
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    $\begingroup$ Can you actually have something that's an infinite weight (or tension)? If not then (other number) / 0 doesn't give you a valid, physical answer which sounds to me like the physics equivalent of a mathematician saying, "It's undefined." $\endgroup$
    – G. Allen
    Commented Nov 23, 2018 at 22:58
  • $\begingroup$ @G.Allen (I'm late to the party but) that's an interesting point - you can't have an infinite tension in a real system, but when using a physical/mathematical model of that real system, you might legitimately get an infinite or undefined result from the model. That tends to mean you have encountered a situation where the model doesn't apply, and you need a better model. Of course, that's a physics insight, probably not something to bring up in the context of pure math. $\endgroup$
    – David Z
    Commented Feb 17, 2021 at 17:53

If the students can think about graphs, you can graph y=1/x. So if 1/0 = ∞, this should approach ∞ as x -> 0. It does on one side. But on the other, it approaches -∞. Since it doesn't approach ∞ from both sides, we must say it's undefined.

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    $\begingroup$ One caveat would be this opens up a counterargument of inspecting $1/|x|$, say. Second caveat is when I see extended reals defined (i.e., $\infty$ as a usable number), it actually is defined that $x/0 = sgn(x) \cdot \infty$ (assuming $x \ne 0$; Ray, Real Analysis, Appendix A). $\endgroup$ Commented Jan 13, 2018 at 23:35
  • $\begingroup$ @DanielR.Collins: That's an unusual convention, since it would mean division is not continuous on its domain; usually $1/0$ is left undefined, for the reason of this answer. Although one can conceive of a "positive zero", so that they can write $x/0^+ = \infty$. $\endgroup$
    – user797
    Commented Jan 15, 2018 at 16:08
  • $\begingroup$ @Hurkyl: Extended reals aren't a field, so any way you cut it something gets lost (pick your poison, basically). The alternative is to not have division be onto the whole set under discussion. $\endgroup$ Commented Jan 15, 2018 at 18:06
  • $\begingroup$ @DanielR.Collins: Yes, something has to be lost, but losing continuity is a poor poison to pick, since it sours one of the major applications of the number system: they simplify the computation of limits by continuously extending the operations involved. I'm not sure what you were getting at with your last sentence, but note that $\infty / 1 = \infty$ and there are a number of other values we need to leave undefined as well. $\endgroup$
    – user797
    Commented Jan 15, 2018 at 19:25
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    $\begingroup$ The other option is to use the projective reals instead of the extended reals, because then you have $\infty = - \infty$ and $y=\frac{1}{x}$ is continuous. $\endgroup$
    – Kevin
    Commented Feb 11, 2018 at 16:55

It is possible for an expression to be undefined. If you work in $\mathbb{N}$, you can consider $2 - 3$ to be undefined. We just have meaningless expressions. Similarly when working in $\mathbb{R}$, we can also consider the expression $\frac{1}{0}$ to be a meaningless expression. $\frac{1}{0}$ is considred undefined because $\frac{1}{0}$ means the number that if you multiply by 0 you get 1 but there is no such number. If that doesn't work, here's an alternate explanation.

Actually, we could define $\mathbb{R}$ with addition, multiplication, and division in such a way to satisfy the desirable properties. Suppose we defined $\frac{1}{0}$ to actually represent a number. Then we find that $\frac{1}{0} \times 0 = 0$ but then the property that $\frac{x}{y} \times y = x$ when ever $\frac{x}{y}$ is defined would be lost. Then maybe we could instead define that $\frac{1}{0} \times 0 = 1$ inventing an infinite number and defining $\frac{1}{0}$ to equal that number. Then we would find that $\frac{1}{0} \times 0 = 1 \neq 2 = (\frac{1}{0} \times 0) + (\frac{1}{0} \times 0) = \frac{1}{0} \times (0 + 0)$ so the distributivity of multiplication over addition would be lost.

If after explaining all that, the person who was wondering how $\frac{1}{0}$ could be undefined still knows how to derive the contradiction but cannot make any sense of it and is all confused, they may eventually figure out the answer to their own question. They may realize that it is a matter of how you define the number system and its operations and in order to define it in a way that avoids other problems, you have to define the system in such a way that multiplication by 0 always gets you 0 and saying $\frac{1}{0}$ is another way of saying the number that if you multiply by 0 gets you 1 but there is no such number so $\frac{1}{0}$ is a meaningless expression according to that definition.


If the students have sufficient algebraic skills, point out that division is defined in terms of multiplication as follows:

For any real numbers $x$, $y$ and $z$, we define $x/y = z$ iff $y\neq 0$ and $x=zy$.

Why the condition $y\neq 0$? When $y=0$ then $x=z\times 0$ (the equation on the RHS) would have no unique solution for $z$.

Example 1: If $x=0$ and $y=0$, then the equation on the RHS becomes $0=z\times 0$. Then any value would work for $z$.

Example 2: If $x=1$ and $y=0$, then the equation on the RHS becomes $1=z\times 0$. Then no value would work for $z$.


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