I apologize for the outage on the site yesterday and today. Lamar University is in Beaumont Texas and Hurricane Laura came through here and caused a brief power outage at Lamar. Things should be up and running at this point and (hopefully) will stay that way, at least until the next hurricane comes through here which seems to happen about once every 10-15 years. Note that I wouldn't be too suprised if there are brief outages over the next couple of days as they work to get everything back up and running properly. I apologize for the inconvienence.

Paul

August 27, 2020

*i.e.*you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

## Calculus Errors

Many of the errors listed here are not really calculus errors, but errors that commonly occur in a calculus class and notational errors that are calculus related. If you haven’t had a calculus class then I would suggest that you not bother with this section as it probably won’t make a lot of sense to you.

If you are just starting a calculus class then I would also suggest that you be very careful with reading this. At some level this part is intended to be read by a student taking a calculus course as he/she is taking the course. In other words, after you’ve covered limits come back and look at the issues involving limits, then do the same after you’ve covered derivatives and then with integrals. Do not read this prior to the class and try to figure out how calculus works based on the few examples that I’ve given here! This will only cause you a great amount of grief down the road.

#### Derivatives and Integrals of Products/Quotients

Recall that while

\[{{\left( f\pm g \right)}^{\prime }}\left( x \right)={f}'\left( x \right)\pm {g}'\left( x \right) \hspace{0.5in} \int{f\left( x \right)\pm g\left( x \right)\,dx}=\int{f\left( x \right)dx}\pm \int{g\left( x \right)\,dx}\]are true, the same thing can’t be done for products and quotients. In other words,

\[\begin{align*} {{\left( fg \right)}^{\prime }}\left( x \right) & \ne {f}'\left( x \right){g}'\left( x \right) \hspace{0.5in} & \int{f\left( x \right)g\left( x \right)\,dx} & \ne \left( \int{f\left( x \right)dx} \right)\left( \int{g\left( x \right)\,dx} \right) \\ {{\left( \frac{f}{g} \right)}^{\prime }}\left( x \right) & \ne \frac{{f}'\left( x \right)}{{g}'\left( x \right)} \hspace{0.5in} & \int{\frac{f\left( x \right)}{g\left( x \right)}\,dx} & \ne \frac{\int{f\left( x \right)dx}}{\int{g\left( x \right)dx}} \\\end{align*}\]If you need convincing of this consider the example of \(f\left( x \right)={{x}^{4}}\) and \(g\left( x \right)={{x}^{10}}\).

\[\begin{align*} {{\left( fg \right)}^{\prime }}\left( x \right) & \ne {f}'\left( x \right){g}'\left( x \right) \\ {{\left( {{x}^{4}}{{x}^{10}} \right)}^{\prime }} & \ne {{\left( {{x}^{4}} \right)}^{\prime }}{{\left( {{x}^{10}} \right)}^{\prime }} \\ {{\left( {{x}^{14}} \right)}^{\prime }} & \ne \left( 4{{x}^{3}} \right)\left( 10{{x}^{9}} \right) \\ 14{{x}^{13}} & \ne 40{{x}^{12}}\end{align*}\]I only did the case of the derivative of a product, but clearly the two aren’t equal! I’ll leave it to you to check the remaining three cases if you’d like to.

Remember that in the case of derivatives we’ve got the product and quotient rule. In the case of integrals there are no such rules and when faced with an integral of a product or quotient they will have to be dealt with on a case by case basis.

#### Proper use of the formula for \(\int{{{x}^{n}}\,dx}\)

Many students forget that there is a restriction on this integration formula, so for the record here is the formula along with the restriction.

\[\int{{{x}^{n}}dx}=\frac{{{x}^{n+1}}}{n+1}+c, \hspace{0.5in} \text{provided }n\ne -1\]That restriction is incredibly important because if we allowed \(n=-1\) we would get division by zero in the formula! Here is what I see far too many students do when faced with this integral.

\[\int{{{x}^{-1}}dx}=\frac{{{x}^{0}}}{0}+c={{x}^{0}}+c=1+c\]THIS ISN’T TRUE!!!!!! There are all sorts of problems with this. First there’s the improper use of the formula, then there is the division by zero problem! This should NEVER be done this way.

Recall that the correct integral of \({{x}^{-1}}\) is,

\[\int{{{x}^{-1}}dx}=\int{\frac{1}{x}\,dx}=\ln |x|+c\]This leads us to the next error.

#### Dropping the absolute value when integrating \[\int{\frac{1}{x}\,dx}\]

Recall that in the formula

\[\int{\frac{1}{x}\,dx}=\ln |x|+c\]the absolute value bars on the argument are required! It is certainly true that on occasion they can be dropped after the integration is done, but they are required in most cases. For instance, contrast the two integrals,

\[\begin{align*} \int{\frac{2x}{{{x}^{2}}+10}\,dx} & =\ln |{{x}^{2}}+10|+C=\ln \left( {{x}^{2}}+10 \right)+c \\ \int{\frac{2x}{{{x}^{2}}-10}dx} & =\ln |{{x}^{2}}-10|+c \\\end{align*}\]In the first case the \({{x}^{2}}\)is positive and adding 10 on will not change that fact so since \({{x}^{2}}+10>0\) we can drop the absolute value bars. In the second case however, since we don’t know what the value of \(x\) is, there is no way to know the sign of \({{x}^{2}}-10\) and so the absolute value bars are required.

#### Improper use of the formula \(\int{\frac{1}{x}\,dx}=\ln |x|+c\)

Gotten the impression yet that there are more than a few mistakes made by students when integrating \(\frac{1}{x}\)? I hope so, because many students lose huge amounts of points on these mistakes. This is the last one that I’ll be covering however.

In this case, students seem to make the mistake of assuming that if \(\frac{1}{x}\) integrates to \(\ln |x|\) then so must one over anything! The following table gives some examples of incorrect uses of this formula.

Integral | Incorrect Answer | Correct Answer |
---|---|---|

\(\displaystyle \int{\frac{1}{{{x}^{2}}+1}\,dx}\) | \(\ln \left( {{x}^{2}}+1 \right)+c\) | \({{\tan }^{-1}}\left( x \right)+c\) |

\(\displaystyle \int{\frac{1}{{{x}^{2}}}\,dx}\) | \(\ln \left( {{x}^{2}} \right)+c\) | \(-{{x}^{-1}}+c=-\frac{1}{x}+c\) |

\(\displaystyle \int{\frac{1}{\cos x}dx}\) | \(\ln |\cos x|+c\) | \(\ln |\sec x+\tan x|+c\) |

So, be careful when attempting to use this formula. This formula can only be used when the integral is of the form \(\int{\frac{1}{x}\,dx}\). Often, an integral can be written in this form with an appropriate \(u\)-substitution (the two integrals from previous example for instance), but if it can’t be then the integral will NOT use this formula so don’t try to.

#### Improper use of Integration formulas in general

This one is really the same issue as the previous one, but so many students have trouble with logarithms that I wanted to treat that example separately to make the point.

So, as with the previous issue students tend to try and use “simple” formulas that they know to be true on integrals that, on the surface, kind of look the same. So, for instance we’ve got the following two formulas,

\[\begin{align*} \int{\sqrt{u}\,du} & =\frac{2}{3}{{u}^{\frac{3}{2}}}+C \\ \int{{{u}^{2}}\,du} &=\frac{1}{3}{{u}^{3}}+C \\\end{align*}\]The mistake here is to assume that if these are true then the following must also be true.

\[\begin{align*} \int{\sqrt{\text{anything}}\,du} & =\frac{2}{3}{{\left( \text{anything} \right)}^{\frac{3}{2}}}+C \\ \int{{{\left( \text{anything} \right)}^{2}}\,du} &=\frac{1}{3}{{\left( \text{anything} \right)}^{3}}+C \\\end{align*}\]This just isn’t true! The first set of formulas work because it is the square root of a single variable or a single variable squared. If there is anything other than a single u under the square root or being squared then those formulas are worthless. On occasion these will hold for things other than a single u, but in general they won’t hold so be careful!

Here’s another table with a couple of examples of these formulas not being used correctly.

Integral | Incorrect Answer | Correct Answer |
---|---|---|

\(\displaystyle \int{\sqrt{{{x}^{2}}+1}\,dx}\) | \(\displaystyle \frac{2}{3}{{\left( {{x}^{2}}+1 \right)}^{\frac{3}{2}}}+C\) | \(\displaystyle \frac{1}{2}\left( x\sqrt{{{x}^{2}}+1}+\ln |x+\sqrt{{{x}^{2}}+1}| \right)+C\) |

\(\displaystyle \int{{{\cos }^{2}}xdx}\) | \(\displaystyle \frac{1}{3}{{\cos }^{3}}x+C\) | \(\displaystyle \frac{x}{2}+\frac{1}{4}\sin \left( 2x \right)+C\) |

If you aren’t convinced that the incorrect answers really aren’t correct then remember that you can always check you answers to indefinite integrals by differentiating the answer. If you did everything correctly you should get the function you originally integrated, although in each case it will take some simplification to get the answers to be the same.

Also, if you don’t see how to get the correct answer for these they typically show up in a Calculus II class. The second however, you could do with only Calculus I under your belt if you can remember an appropriate trig formula.

#### Dropping limit notation

The remainder of the errors in this document consists mostly of notational errors that students tend to make.

I’ll start with limits. Students tend to get lazy and start dropping limit notation after the first step. For example, an incorrectly worked problem is

\[\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{2}}-9}{x-3}=\frac{\left( x-3 \right)\left( x+3 \right)}{x-3}=x+3=6\]There are several things wrong with this. First, when you drop the limit symbol you are saying that you’ve in fact taken the limit. So, in the first equality,

\[\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{2}}-9}{x-3}=\frac{\left( x-3 \right)\left( x+3 \right)}{x-3}\]you are saying that the value of the limit is

\[\frac{\left( x-3 \right)\left( x+3 \right)}{x-3}\]and this is clearly not the case. Also, in the final equality,

\[x+3=6\]you are making the claim that each side is the same, but this is only true provided \(x=3\) and what you really are trying to say is

\[\underset{x\to 3}{\mathop{\lim }}\,x+3=6\]You may know what you mean, but someone else will have a very hard time deciphering your work. Also, your instructor will not know what you mean by this and won’t know if you understand that the limit symbols are required in every step until you actually take the limit. If you are one of my students, I won’t even try to read your mind and I will assume that you didn’t understand and take points off accordingly.

So, while you may feel that it is silly and unnecessary to write limits down at every step it is proper notation and in my class I expect you to use proper notation. The correct way to work this limit is.

\[\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{2}}-9}{x-3}=\underset{x\to 3}{\mathop{\lim }}\,\frac{\left( x-3 \right)\left( x+3 \right)}{x-3}=\underset{x\to 3}{\mathop{\lim }}\,x+3=6\]The limit is required at every step until you actually take the limit, at which point the limit must be dropped as I have done above.

#### Improper derivative notation

When asked to differentiate \(f\left( x \right)=x\left( {{x}^{3}}-2 \right)\) I will get the following for an answer on occasion.

\[\left( x \right)=x\left( {{x}^{3}}-2 \right)={{x}^{4}}-2x=4{{x}^{3}}-2\]This is again a situation where you may know what you’re intending to say here, but anyone else who reads this will come away with the idea that \({{x}^{4}}-2x=4{{x}^{3}}-2\) and that is clearly NOT what you are trying to say. However, it IS what you are saying when you write it this way.

The proper notation is

\[ \begin{align*} f\left( x \right) & = x\left( {{x}^{3}}-2 \right)={{x}^{4}}-2x \\ {f}'\left( x \right) & =4{{x}^{3}}-2 \end{align*} \]#### Loss of integration notation

There are many dropped notation errors that occur with integrals. Let’s start with this example.

\[\int{x\left( 3x-2 \right)dx}=3{{x}^{2}}-2x={{x}^{3}}-{{x}^{2}}+c\]As with the derivative example above, both of these equalities are incorrect. The minute you drop the integral sign you are saying that you’ve done the integral! So, this means that the first equality is saying that the value of the integral is \(3{{x}^{2}}-2x\), when in reality all you’re doing is simplifying the function. Likewise, the last equality says that the two functions, \(3{{x}^{2}}-2x\) and \({{x}^{3}}-{{x}^{2}}+c\) are equal, when they are not! Here is the correct way to work this problem.

\[\int{x\left( 3x-2 \right)dx}=\int{3{{x}^{2}}-2xdx}={{x}^{3}}-{{x}^{2}}+c\]Another big problem in dropped notation is students dropping the \(dx\) at the end of the integrals. For instance,

\[\int{3{{x}^{2}}-2x}\]The problem with this is that the \(dx\) tells us where the integral stops! So, this can mean a couple of different things.

\[\begin{align*} \int{3{{x}^{2}}-2xdx} & ={{x}^{3}}-{{x}^{2}}+c \hspace{0.5in} \text{OR} \\ \int{3{{x}^{2}}dx}-2x & ={{x}^{3}}-2x+c \\\end{align*}\]Without the dx a reader is left to try and intuit where exactly the integral ends! The best way to think of this is that parenthesis always come in pairs “(” and “)”. You don’t open a set of parenthesis without closing it. Likewise, \(\int{{}}\) is always paired up with a \(dx\). You can always think of \(\int{{}}\) as the opening parenthesis and the \(dx\) as the closing parenthesis.

Another dropped notation error that I see on a regular basis is with definite integrals. Students tend to drop the limits of integration after the first step and do the rest of the problem with *implied* limits of integration as follows.

Again, the first equality here just doesn’t make sense! The answer to a definite integral is a number, while the answer to an indefinite integral is a function. When written as above you are saying the answer to the definite integral and the answer to the indefinite integral are the same when they clearly aren’t!

Likewise, the second to last equality just doesn’t make sense. Here you are saying that the function, \({{x}^{3}}-{{x}^{2}}\) is equal to \(8-4-\left( 1-1 \right)=4\) and again, this just isn’t true! Here is the correct way to work this problem.

\[\int_{\,1}^{\,2}{x\left( 3x-2 \right)dx}=\int_{\,1}^{\,2}{3{{x}^{2}}-2xdx}=\left. \left( {{x}^{3}}-{{x}^{2}} \right) \right|_{1}^{2}=8-4-\left( 1-1 \right)=4\]#### Loss of notation in general

The previous three topics that I’ve discussed have all been examples of dropped notation errors that students first learning calculus tend to make on a regular basis. Be careful with these kinds of errors. You may know what you’re trying to say, but improper notation may imply something totally different.

Remember that in many ways written mathematics is like a language. If you mean to say to someone

You wouldn’t drop words that you considered extraneous to the message and just say

This is meaningless and the person that you were talking to may get the idea that you are thirsty and wanted to drink something. They would definitely not get the idea that you wanted water to drink or that you were asking them to get it for you. You would know that is what you wanted, but those two words would not convey that to anyone else.

This may seem like a silly example to you, because you would never do something like this. You would give the whole sentence and not just two words because you are fully aware of how confusing simply saying those two words would be. That, however, is exactly the point of the example.

You know better than to skip important words in spoken language, so you shouldn’t skip important notation (i.e. words) in writing down the language of mathematics. You may feel that they aren’t important parts to the message, but they are. Anyone else reading the message you wrote down would not necessarily know that you neglected to write down those important pieces of notation and would very likely misread the message you were trying to impart.

So, be careful with proper notation. In my class, I grade the “message” you write down not the “message” that you meant to impart. I can’t read your mind so I don’t even try to. If the “message” that I read in grading your homework or exam is wrong, I will grade it appropriately.

#### Dropped constant of integration

Dropping the constant of integration on indefinite integrals (the \(+c\) part) is one of the biggest errors that students make in integration. There are actually two errors here that students make. Some students just don’t put it in at all, and others drop it from intermediate steps and then just tack it onto the final answer.

Those that don’t include it at all tend to be the students that don’t remember (or never really understood) that the indefinite integrals give the most general possible function that we could differentiate to get the integrand (the function we integrated). Because it is the most general possible function we’ve got to include the constant, since constants differentiate to zero.

For those that drop it from all intermediate steps and just tack it on at the end there are other issues. I suppose that the problem is these (in fact it’s probably most) students just don’t see why it’s important to include the constant of integration. This is partially a problem with the class itself. Calculus classes just don’t really have good examples of why the constant of integration is so important or how it comes into play in later steps.

The first place where constants of integration play a major role is a first course in differential equations. Here the constant of integration will show up in the middle of the problem. If it’s dropped there and then just added back in on the final answer or not put in at all, the answer will be very wrong. The answer won’t be wrong because the instructor said that it was wrong without the constant or because it was only added in at the last step. The answer will be wrong because the function you get without dropping it will be totally different from the function you get if you do drop it!

#### Misconceptions about \(\frac{1}{0}\) and \(\frac{1}{\infty }\)

This is not so much about an actual error that students make, but instead a misconception that can, on occasion, lead to errors. This is also a misconception that is often encouraged by laziness on the part of the instructor.

So, just what is this misconception? Often, we will write \(\frac{1}{\infty }=0\) and \(\frac{1}{0}=\infty \). The problem is that neither of these are technically correct and in fact the second, depending on the situation, can actually be \(\frac{1}{0}=-\infty \). All three of these are really limits and we just short hand them. What we really should write is

\[\begin{align*} \underset{x\to \infty }{\mathop{\lim }}\,\frac{1}{x} & =0 \\ \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{1}{x} & =\infty \\ \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\frac{1}{x} & =-\infty \\\end{align*}\]In the first case 1 over something increasingly large is increasingly small and so **in the limit** we get zero. In the last two cases note that we’ve got to use one-sided limits as \(\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{x}\) doesn’t even exist! In these two cases, 1 over something increasingly small is increasingly large and will have the sign of the denominator and so in the limit it goes to either \(\infty \) or \(-\infty \).

#### Indeterminate forms

This is actually a generalization of the previous topic. The two operations above, \(\infty -\infty \) and \(\frac{\infty }{\infty }\) are called *indeterminate forms* because there is no one single value for them. Depending on the situation they have a very wide range of possible answers.

There are many more indeterminate forms that you need to look out for. As with the previous discussion there is no way to determine their value without taking the situation into consideration. Here are a few of the more common indeterminate forms.

\[\begin{array}{cccc} \infty -\infty \hspace{0.25in} & \displaystyle \frac{\infty }{\infty } \hspace{0.25in} & \displaystyle \frac{0}{0} \hspace{0.25in} & 0\cdot \infty \\ {{0}^{0}} \hspace{0.25in} & {{1}^{\infty }} \hspace{0.25in} & {{\infty }^{0}} \hspace{0.25in} & \\ \end{array}\]Let’s just take a brief look at \({{0}^{0}}\) to see the potential problems. Here we really have two separate rules that are at odds with each other. Typically we have \({{0}^{n}}=0\) (provided \(n\) is positive) and \({{a}^{0}}=1\). Each of these rules implies that we could get different answers. Depending on the situation we could get either 0 or 1 as an answer here. In fact, it’s also possible to get something totally different from 0 or 1 as an answer here as well.

All the others listed here have similar problems. So, when dealing with indeterminate forms you need to be careful and not jump to conclusions about the value.

#### Treating infinity as a number

In the following discussion I’m going to be working exclusively with real numbers (things can be different with say complex numbers). I’m also going to think of infinity (\(\infty \)) as a really, really large number. This is not technically accurate as infinity is really a concept to denote a state of endlessness or a state of no limits in any direction. In terms of a number line infinity (\(\infty \)) denotes moving in the positive direction without ever stopping. Likewise, negative infinity (\(-\infty \)) on a number line denotes moving in the negative direction without ever stopping.

The problem with the conceptual definition of infinity is that many students have a hard time dealing with arithmetic involving infinity when they think if in it terms of its conceptual definition. However, if we simply call it a really, really large number it seems to help a little so that’s how I’m going to think of it for the purposes of this discussion.

Most students have run across infinity at some point in time prior to a calculus class. However, when they have dealt with it, it was just a symbol used to represent a really, really large positive or negative number and that was the extent of it. Once they get into a calculus class students are asked to do some basic algebra with infinity and this is where they get into trouble. Infinity is NOT a number and for the most part doesn’t behave like a number. When you add two non-zero numbers you get a new number. For example, \(4+7=11\). With infinity this is not true. With infinity you have the following.

\[\begin{align*} \infty +a & =\infty \hspace{0.5in} \text{where }a\ne -\infty \\ \infty +\infty & =\infty \end{align*}\]In other words, a really, really large positive number (\(\infty \)) plus any positive number, regardless of the size, is still a really, really large positive number. Likewise, you can add a negative number (*i.e.* \(a<0\)) to a really, really large positive number and stay really, really large and positive. So, addition involving infinity can be dealt with in an intuitive way if you’re careful. Note as well that the \(a\) must NOT be negative infinity. If it is, there are some serious issues that we need to deal with.

Subtraction with negative infinity can also be dealt with in an intuitive way. A really, really large negative number minus any positive number, regardless of its size, is still a really, really large negative number. Subtracting a negative number (i.e. \(a<0\)) from a really, really large negative number will still be a really, really large negative number. Or,

\[\begin{align*} -\infty -a & =-\infty \hspace{0.5in} \text{where }a\ne -\infty \\ -\infty -\infty & =-\infty \end{align*}\]Again, \(a\) must not be negative infinity to avoid some potentially serious difficulties.

Multiplication can also be dealt with fairly intuitively. A really, really large number (positive, or negative) times any number, regardless of size, is still a really, really large number. In the case of multiplication we have

\[\begin{align*} \left( a \right)\left( \infty \right) & =\infty & \hspace{0.25in} & \text{if }a>0 \\ \left( a \right)\left( \infty \right) & =-\infty & \hspace{0.25in} & \text{if }a<0 \\ \left( \infty \right)\left( \infty \right)&=\infty & & \\ \left( -\infty \right)\left( -\infty \right)& =\infty & & \\ \left( -\infty \right)\left( \infty \right) & =-\infty & & \end{align*}\]What you know about products of positive and negative numbers is still true.

Some forms of division can be dealt with intuitively as well. A really, really large number divided by a number that isn’t too large is still a really, really large number.

\[\begin{align*} \frac{\infty }{a} & =\infty & \hspace{0.25in} & \text{if }a>0 \\ \frac{\infty }{a} & =-\infty & \hspace{0.25in} & \text{if }a<0 \\ \frac{-\infty }{a} & =-\infty & \hspace{0.25in} & \text{ if }a>0 \\ \frac{-\infty }{a} & = \infty & \hspace{0.25in} & \text{ if }a<0 \end{align*}\]Division of a number by infinity is somewhat intuitive, but there are a couple of subtleties that you need to be aware of. I go into this in more detail in the section about **Misconceptions about \(\frac{1}{0}\) and \(\frac{1}{\infty }\)** above, but one way to think of it is the following. A number that isn’t too large divided by infinity (a really, really large number) is a very, very, very small number. In other words,

So, I’ve dealt with almost every basic algebraic operation involving infinity. There are two cases that that I haven’t dealt with yet. These are

\[\begin{align*} & \infty -\infty =\text{?} \\ & \frac{\pm \,\infty }{\pm \,\infty }=? \\\end{align*}\]The problem with these two is that intuition doesn’t really help here. A really, really large number minus a really, really large number can be anything (\(-\infty \), a constant, or \(\infty \)). Likewise, a really, really large number divided by a really, really large number can also be anything (\(\pm \,\infty \) - this depends on sign issues, 0, or a non-zero constant).

What you’ve got to remember here is that there are really, really large numbers and then there are really, really, really large numbers. In other words, some infinities are larger than other infinities. With addition, multiplication and the first sets of division I worked this isn’t an issue. The general size of the infinity just doesn’t affect the answer. However, with the subtraction and division I listed above, it does matter as you will see.

Here is one way to think of this idea that some infinities are larger than others. This is a fairly dry and technical way to think of this and your calculus problems will probably never use this stuff, but this it is a nice way of looking at this. Also, please note that I’m not trying to give a precise proof of anything here. I’m just trying to give you a little insight into the problems with infinity and how some infinities can be thought of as larger than others. For a much better (and definitely more precise) discussion see,

http://www.math.vanderbilt.edu/~schectex/courses/infinity.pdfLet’s start by looking at how many integers there are. Clearly, I hope, there are an infinite number of them, but let’s try to get a better grasp on the “size” of this infinity. So, pick any two integers completely at random. Start at the smaller of the two and list, in increasing order, all the integers that come after that. Eventually we will reach the larger of the two integers that you picked.

Depending on the relative size of the two integers it might take a very, very long time to list all the integers between them and there isn’t really a purpose to doing it. But, it could be done if we wanted to and that’s the important part.

Because we could list all these integers between two randomly chosen integers we say that the integers are *countably infinite*. Again, there is no real reason to actually do this, it is simply something that can be done if we should chose to do so.

In general, a set of numbers is called countably infinite if we can find a way to list them all out. In a more precise mathematical setting this is generally done with a special kind of function called a *bijection* that associates each number in the set with exactly one of the positive integers. To see some more details of this see the pdf given above.

It can also be shown that the set of all fractions are also countably infinite, although this is a little harder to show and is not really the purpose of this discussion. To see a proof of this see the pdf given above. It has a very nice proof of this fact.

Let’s contrast this by trying to figure out how many numbers there are in the interval \( (0,1) \). By numbers, I mean all possible fractions that lie between zero and one as well as all possible decimals (that aren’t fractions) that lie between zero and one. The following is similar to the proof given in the pdf above, but was nice enough and easy enough (I hope) that I wanted to include it here.

To start let’s assume that all the numbers in the interval \( (0,1) \) are countably infinite. This means that there should be a way to list all of them out. We could have something like the following,

\[\begin{align*} {{x}_{1}}& = 0.692096\cdots \\ {{x}_{2}}& = 0.171034\cdots \\ {{x}_{3}}& = 0.993671\cdots \\ {{x}_{4}}& = 0.045908\cdots \\ \vdots & \hspace{0.65in} \vdots \end{align*}\]Now, select the \(i^{th}\) decimal out of \({{x}_{i}}\) as shown below

\[\begin{align*} {{x}_{1}} & = 0.\underline{6}92096\cdots \\ {{x}_{2}} & = 0.1\underline{7}1034\cdots \\ {{x}_{3}} & = 0.99\underline{3}671\cdots \\ {{x}_{4}} & = 0.045\underline{9}08\cdots \\ \vdots & \hspace{0.65in} \vdots \end{align*}\]and form a new number with these digits. So, for our example we would have the number

\[x=0.6739\cdots \]In this new decimal replace all the 3’s with a 1 and then replace every other number with a 3. In the case of our example this would yield the new number

\[\overline{x}=0.3313\cdots \]Notice that this number is in the interval \( (0,1) \) and also notice that given how we choose the digits of the number this number will not be equal to the first number in our list, \({{x}_{1}}\), because the first digit of each is guaranteed to not be the same. Likewise, this new number will not get the same number as the second in our list, \({{x}_{2}}\), because the second digit of each is guaranteed to not be the same. Continuing in this manner we can see that this new number we constructed, \(\overline{x}\), is guaranteed to not be in our listing. But this contradicts the initial assumption that we could list out all the numbers in the interval \( (0,1) \). Hence, it must not be possible to list out all the numbers in the interval \( (0,1) \).

Sets of numbers, such as all the numbers in (0,1), that we can’t write down in a list are called *uncountably infinite*.

The reason for going over this is the following. An infinity that is uncountably infinite is significantly larger than an infinity that is only countably infinite. So, if we take the difference of two infinities we have a couple of possibilities.

\[\begin{align*} \infty \left( \text{uncountable} \right)-\infty \left( \text{countable} \right) & =\infty \\ \infty \left( \text{countable} \right)-\infty \left( \text{uncountable} \right)& =-\infty \end{align*}\]Notice that we didn’t put down a difference of two infinities of the same type. Depending upon the context there might still have some ambiguity about just what the answer would be in this case, but that is a whole different topic.

We could also do something similar for quotients of infinities.

\[\begin{align*} \frac{\infty \left( \text{countable} \right)}{\infty \left( \text{uncountable} \right)} & =0 \\ \frac{\infty \left( \text{uncountable} \right)}{\infty \left( \text{countable} \right)} & =\infty \end{align*}\]Again, we avoided a quotient of two infinities of the same type since, again depending upon the context, there might still be ambiguities about its value.