Formula for an arithmetic series (Part 5)

In Precalculus, Discrete Mathematics or Real Analysis, an arithmetic series is often used as a student’s first example of a proof by mathematical induction. Recall, from Wikipedia:

Mathematical induction is a method of mathematical proof typically used to establish a given statement for all natural numbers. It is done in two steps. The first step, known as the base case, is to prove the given statement for the first natural number. The second step, known as the inductive step, is to prove that the given statement for any one natural number implies the given statement for the next natural number. From these two steps, mathematical induction is the rule from which we infer that the given statement is established for all natural numbers.

The simplest and most common form of mathematical induction infers that a statement involving a natural number n holds for all values of n. The proof consists of two steps:

The basis (base case): prove that the statement holds for the first natural number $n$ . Usually, $n=0$ or $n=1$ .

The inductive step: prove that, if the statement holds for some natural number $n$ , then the statement holds for $n+1$ .

The hypothesis in the inductive step that the statement holds for some $n$ is called the induction hypothesis (or inductive hypothesis). To perform the inductive step, one assumes the induction hypothesis and then uses this assumption to prove the statement for $n+1$ .

As an inference rule, mathematical induction can be justified as follows. Having proven the base case and the inductive step, then any value can be obtained by performing the inductive step repeatedly. It may be helpful to think of the domino effect. Consider a half line of dominoes each standing on end, and extending infinitely to the right. Suppose that:

The first domino falls right.

If a (fixed but arbitrary) domino falls right, then its next neighbor also falls right.

With these assumptions one can conclude (using mathematical induction) that all of the dominoes will fall right.

Mathematical induction… works because $n$ is used to represent an arbitrary natural number. Then, using the inductive hypothesis, i.e. that $P(n)$ is true, show $P(k+1)$ is also true. This allows us to “carry” the fact that $P(0)$ is true to the fact that $P(1)$ is also true, and carry $P(1)$ to $P(2)$ , etc., thus proving $P(n)$ holds for every natural number $n$ .

When students first encounter mathematical induction (in either Precalculus, Discrete Mathematics, or Real Analysis), the theorems that students are asked to prove usually fall into four categories:

Calculating a series (examples below).
Statements concerning divisibility (for example, proving that $4$ is always a factor of $5^n-1$ ).
Finding a closed-form expression for a recursively defined sequence (for example, if $a_1 = 4$ and $a_n = 3a_{n-1}$ if $n \ge 2$ , proving that $a_n = 4 \times 3^{n-1}$ 0
Statements concerning inequality (for example, proving that $n! > 4^n$ if $n \ge 9$ 0

Here’s a common first example of mathematical induction applied to an arithmetic series. Notice that the statement of the theorem matches the form $\displaystyle \frac{n}{2}(a_1 + a_n)$ seen earlier in this series (pardon the pun) of posts.

Theorem. $1 + 2 + \dots + (n-1) + n = \displaystyle \frac{n(n+1)}{2}$

Proof. Induction on $n$ .

$n = 1$ : The left-hand is simply $1$ , while the right-hand side is $\displaystyle \frac{(1)(2)}{2}$ , which is also equal to $1$ . So the base case works.

$n$ : Assume that the statement holds true for the integer $n$ .

$n+1$ . If I replace $n$ by $n+1$ in the statement of the theorem, then the right-hand side becomes

$\displaystyle \frac{(n+1)[(n+1)+1]}{2} = \displaystyle \frac{(n+1)(n+2)}{2}$

I find it helpful to describe this to students as my target. In other words, as I manipulate the left-hand side, my ultimate goal is to end up with this target. Once I have done that, then I have completed the proof.

If I replace $n$ by $n+1$ in the statement of the theorem, then the left-hand side will now end on $n+1$ instead of $n$ :

$1 + 2 + \dots + (n-1) + n + (n+1)$

Notice that we’ve seen almost all of this before, except for the extra term $n+1$ . So we will substitute using the induction hypothesis, carrying the extra $n+1$ along for the ride.

$1 + 2 + \dots + (n-1) + n + (n+1) = \displaystyle \frac{n(n+1)}{2} + (n+1)$

Now our task is, by hook or by crook, using whatever algebraic tricks we can think of to convert this last expression into the target. Most students are completely comfortable doing this, although they typically multiply out the term $n(n+1)$ unnecessarily. Indeed, many early proofs by induction are simplified by factoring out terms whenever possible — in the example below, $(n+1)$ is factored on the last step — as opposed to multiplying them out. In my experience, proofs by induction often serve as a stringent test of students’ algebra skills as opposed to their skills in abstract reasoning.

In any event, here’s the end of the proof:

$1 + 2 + \dots + (n-1) + n + (n+1) = \displaystyle \frac{n(n+1)}{2} + (n+1)$

$1 + 2 + \dots + (n-1) + n + (n+1) = \displaystyle \frac{n(n+1) + 2(n+1)}{2}$

$1 + 2 + \dots + (n-1) + n + (n+1) = \displaystyle \frac{(n+1)(n + 2)}{2}$

Mathematical induction can be used to verify formulas for series which are not arithmetic, like

$1^2 + 2^2 + \dots + (n-1)^2 + n^2 = \displaystyle \frac{n(n+1)(2n+1)}{6}$

$1^3 + 2^3 + \dots + (n-1)^3 + n^3 = \displaystyle \frac{n^2(n+1)^2}{4}$

However, the downside of a proof by induction lies in the word verify, as it’s necessary to actually know what’s going to work before proceeding with the proof.

In the next post, I’ll describe a method of obtaining these series that does not require mathematical induction.

Formula for an arithmetic series (Part 4)

As I’ve said before, I’m not particularly a fan of memorizing formulas. Apparently, most college students aren’t fans either, because they often don’t have immediate recall of certain formulas from high school when they’re needed in the collegiate curriculum.

While I’m not a fan of making students memorize formulas, I am a fan of teaching students how to derive formulas. Speaking for myself, if I ever need to use a formula that I know exists but have long since forgotten, the ability to derive the formula allows me to get it again.

Which leads me to today’s post: the derivation of the formulas for the sum of an arithmetic series. This topic is commonly taught in Precalculus but, in my experience, is often forgotten by students years later when needed in later classes.

To get the idea across, consider the arithmetic series

$S = 16 + 19 + 22 + 25 + 28 + 31 + 34 + 37 + 40 + 43$

Now write the sum in reverse order. This doesn’t change the value of the sum, and so:

$S = 43 + 40 + 37 +34+ 31 + 28 + 25 + 22 + 19 + 16$

Now add these two lines vertically. Notice that $16 + 43 = 59$ , $19 + 40 = 59$ , and in fact each pair of numbers adds to $59$ . So

$2S = 59 + 59 + 59 + 59 + 59 + 59 + 59 + 59 + 59 + 59$

$2S = 59 \times 10 = 590$

$S = 295$

Naturally, this can be directly confirmed with a calculator by just adding the 10 numbers.

When I show this to my students, they often complain that there’s no way on earth that they would have thought of that for themselves. They wouldn’t have thought to set the sum equal to $S$ , and they certainly would not have thought to reverse the terms in the sum. To comfort them, I tell them my usual tongue-in-cheek story that this idea comes from the patented Bag of Tricks. Socrates gave the Bag of Tricks to Plato, Plato gave it to Aristotle, it passed down the generations, my teacher taught the Bag of Tricks to me, and I teach it to my students.

The derivation of the general formula proceeds using the same idea. If $a_1, \dots, a_n$ are the first $n$ terms of an arithmetic sequence, let

$S = a_1 + a_2 + \dots + a_{n-1} + a_n$

Recalling the formula for an arithmetic sequence, we know that

$a_2 = a_1 + d$

$\vdots$

$a_{n-1} = a_1 + (n-2)d$

$a_n = a_1 + (n-1)d$

Substituting, we find

$S = a_1 + [a_1 + d] + \dots + [a_1 + (n-2)d] + [a_1 + (n-1)d]$

As above, we now return the order…

$S = [a_1 + (n-1)d] + [a_1 + (n-2)d] + \dots + [a_1 + d] + a_1$

… and add the two equations:

$2S = [2a_1 + (n-1)d] + [2a_1 + d+(n-2)d] + \dots + [2a_1 +(n-2)d+ d] + [2a_1+(n-1)d]$

$2S = [2a_1 + (n-1)d] + [2a_1 + (n-1)d] + \dots + [2a_1 +(n-1)d] + [2a_1+(n-1)d]$

$2S = n[2a_1 + (n-1)d]$

$S = \displaystyle \frac{n}{2} [2a_1 + (n-1)d]$

We also note that the formula may be rewritten as

$S = \displaystyle \frac{n}{2} [a_1 + \{a_1 + (n-1)d\} ]$

$S = \displaystyle \frac{n}{2} [a_1 + a_n]$

This latter form isn’t too difficult to state as a sentence: the sum of a series with $n$ is the average of the first and last terms, multiplied by the number of terms.

Indeed, I have seen textbooks offer proofs of this formula by using the same logic that young Gauss used to find the sum $1 + 2 + \dots + 99 + 100$ . The “proof” goes like this: Take the terms in pairs. The first term plus the last term is $a_1 + a_n$ . The second term plus the second-to-last term is $a_2 + a_{n-1} = a_1 + d + a_n - d = a_1 + a_n$ . And so on. So each pair adds to $a_1 + a_n$ . Since there are $n$ terms, there are $n/2$ pairs, and so we derive the above formula for $S$ .

You’ll notice I put “proof” in quotation marks. There’s a slight catch with the above logic: it only works if $n$ is an even number. If $n$ is odd, the result is still correct, but the logic to get the result is slightly different. That’s why I don’t particularly recommend using the above paragraph to prove this formula for students, even though it fits nicely with the almost unforgettable Gauss story.

That said, for talented students looking for a challenge, I would recommend showing this idea, then point out the flaw in the argument, and then ask the students to come up with an alternate proof for handling odd values of $n$ .

Calculation of a famous arithmetic series (Part 3)

In this post, we’ll consider the calculation of a very famous arithmetic series… not because the series is particularly important, but because it’s part of a legendary story about one of the greatest mathematicians who ever lived. My frank opinion is that every math teacher should know this story. While I’m not 100% certain about small details of the story — like whether young Gauss was 9 or 10 years old when the following event happened — I’m just going to go with the story as told by the website http://www.math.wichita.edu/history/men/gauss.html.

Carl Friedrich Gauss (1777-1855) is considered to be the greatest German mathematician of the nineteenth century. His discoveries and writings influenced and left a lasting mark in the areas of number theory, astronomy, geodesy, and physics, particularly the study of electromagnetism.

Gauss was born in Brunswick, Germany, on April 30, 1777, to poor, working-class parents. His father labored as a gardner and brick-layer and was regarded as an upright, honest man. However, he was a harsh parent who discouraged his young son from attending school, with expectations that he would follow one of the family trades. Luckily, Gauss’ mother and uncle, Friedrich, recognized Carl’s genius early on and knew that he must develop this gifted intelligence with education.

While in arithmetic class, at the age of ten, Gauss exhibited his skills as a math prodigy when the stern schoolmaster gave the following assignment: “Write down all the whole numbers from $1$ to $100$ and add up their sum.” When each student finished, he was to bring his slate forward and place it on the schoolmaster’s desk, one on top of the other. The teacher expected the beginner’s class to take a good while to finish this exercise. But in a few seconds, to his teacher’s surprise, Carl proceeded to the front of the room and placed his slate on the desk. Much later the other students handed in their slates.

At the end of the classtime, the results were examined, with most of them wrong. But when the schoolmaster looked at Carl’s slate, he was astounded to see only one number: $5050$ . Carl then had to explain to his teacher that he found the result because he could see that, $1+100=101$ , $2+99=101$ , $3+98=101$ , so that he could find $50$ pairs of numbers that each add up to $101$ . Thus, $50$ times $101$ will equal $5050$ .

Pedagogical thoughts about sequences and series (Part 2)

After yesterday’s post about arithmetic and geometric sequences, I’d like to contribute some thoughts about teaching this topic, based on my own experience over the years.

1. Some students really resist the subscript notation $a_n$ when encountering it for the first time. To allay these concerns, I usually ask my students, “Why can’t we just label the terms in the sequence as $a$ , $b$ , $c$ , and so on?” They usually can answer: what if there are more than 26 terms? That’s the right answer, and so the $a_n$ is used so that we’re not limited to just the letters of the English alphabet.

Another way of selling the $a_n$ notation to students is by telling them that it’s completely analogous to the $f(x)$ notation used more commonly in Algebra II and Precalculus. For a “regular” function $f(x)$ , the number $x$ is chosen from the domain of real numbers. For a sequence $a_n$ , the number $n$ is chosen from the domain of positive (or nonnegative) integers.

2. The formulas in Part 1 of this series (pardon the pun) only apply to arithmetic and geometric sequences, respectively. In other words, if the sequence is neither arithmetic nor geometric, then the above formulas should not be used.

While this is easy to state, my observation is that some students panic a bit when working with sequences and tend to use these formulas on homework and test questions even when the sequence is specified to be something else besides these two types of sequences. For example, consider the following problem:

Find the 10th term of the sequence $1, 4, 9, 16, \dots$

I’ve known pretty bright students who immediately saw that the first term was $1$ and the difference between the first and second terms was $3$ , and so they answered that the tenth term is $1 + (10-1)\times 3 = 28$ … even though the sequence was never claimed to be arithmetic.

I’m guessing that these arithmetic and geometric sequences are emphasized so much in class that some students are conditioned to expect that every series is either arithmetic or geometric, forgetting (especially on tests) that there are sequences other than these two.

3. Regarding arithmetic sequences, sometimes it helps by giving students a visual picture by explicitly make the connection between the terms of an arithmetic sequence and the points of a line. For example, consider the arithmetic sequence which begins

$13, 16, 19, 22, \dots$

The first term is $13$ , the second term is $16$ , and so on. Now imagine plotting the points $(1,13)$ , $(2,16)$ , $(3,19)$ , and $(4,22)$ on the coordinate plane. Clearly the points lie on a straight line. This is not surprising since there’s a common difference between terms. Moreover, the slope of the line is $3$ . This matches the common difference of the arithmetic sequence.

4. In ordinary English, the words sequence and series are virtually synonymous. For example, if someone says either, “a sequence of unusual events” or “a series of unusual events,” the speaker means pretty much the same thing

However, in mathematics, the words sequence and series have different meanings. In mathematics, an example of an arithmetic sequence are the terms

$1, 3, 5, 7, 9, \dots, 99$

However, an example of an arithmetic series would be

$1 + 3 + 5 + 7 + 9 + \dots + 99$

In other words, a sequence provides the individual terms, while a series is a sum of the terms.

When teaching this topic, I make sure to take a minute to emphasize that the words sequence and series will mean something different in my class, even though they basically mean the same thing in ordinary English.

Formulas for arithmetic and geometric sequences (Part 1)

I’m not particularly a fan of memorizing formulas. Apparently, most college students aren’t fans either, because they often don’t have immediate recall of certain formulas from high school when they’re needed in the collegiate curriculum.

Which leads me to today’s post: the derivation of the formulas for the $n$ th term of an arithmetic sequence and of a geometric sequence. This topic is commonly taught in Precalculus but, in my experience, is often forgotten by students years later when needed in later classes.

An arithmetic sequence is specified by two numbers: the first term and the common difference between terms. For example, if the first term is $16$ and the common difference is $3$ , then the sequence begins as

$16, 19, 22, 25, 28, 31, 34, \dots$

If the first term is $29$ and the common difference is $-4$ , then the sequence begins as

$29, 25, 21, 17, 13, 9, 5, 1, -3, \dots$

For those of us old enough to remember, our favorite arithmetic sequences came from Schoolhouse Rock:

Let’s discuss the first arithmetic sequence, whose first seven terms are:

$16, 19, 22, 25, 28, 31, 34, \dots$

How do we get the $8$ th term? That’s easy: we just add $3$ to $34$ to get $37$ .

How to we get the $100$ th term. That’s easy: we just add $3$ to the $99$ th term.

Oops. We don’t know the $99$ th term. To get the $99th$ term, we need the $98$ th term, which in turn requires the $97$ th term. Et cetera, et cetera, et cetera.

The trouble (so far) is that an arithmetic sequence is recursively defined: to get one term, I add something to the previous term. Mathematically, the arithmetic sequence is defined by

$a_n = a_{n-1} + d$ ,

where $d$ is the common difference. This can be very intimidating to students when seeing it for the first time. So, to make this formula less intimidating, I usually read this equation as “Each next term in the sequence is equal to the previous term in the sequence plus the common difference.”

It would be far better to have a closed-form formula, where I could just plug in $100$ to get the $100$ th term, without first figuring out the previous $99$ terms.

To this end, we notice the following pattern:

Second term: $19 = 16 + 3$
Third term: $22 = 19 + 3 = 16 + 3 + 3 = 16 + 2 \times 3$
Fourth term: $25 = 22+ 3 = 16 + (2 \times 3) + 3 = 16 + 3 \times 3$
Fifth term: $28 = 25+ 3 = 16 + (3 \times 3) + 3 = 16 + 4 \times 3$
Sixth term: $31 = 28+ 3 = 16+ (4 \times 3) + 3 = 16 + 5 \times 3$
Seventh term: $34 = 31 + 3 = 16 + (5 \times 3) + 3 = 16 + 6 \times 3$

It looks like we have a pattern, so we can guess that:

One hundredth term = $16 + (100-1) \times 3 = 313$

In general, we have justified the closed-form formula

$a_n = a_1 + (n-1)d$ ,

where $a_1$ is the first term, and $d$ is the common difference. In words: to get the $n$ th term of an arithmetic sequence, we add $d$ to the first term $n-1$ times. (This may be formally proven using mathematical induction, though I won’t do so here.)

A closed-form formula for a geometric sequence is similarly obtained. In a geometric sequence, each term is equal to the previous term multiplied by a common ratio. Mathematically, the geometric sequence is recursively defined by

$a_n = a_{n-1}r$ ,

where $r$ is the common ratio. For example, if the first term is $3$ and the common ratio is $2$ , then the first few terms of the sequence are

$3, 6, 12, 24, 48, dots$

By the same logic used above, to get the $n$ th term of an geometric sequence, we multiply $r$ to the first term $n-1$ times. Thus justifies the formula

$a_n = a_1 r^{n-1}$ ,

which may be formally proven using mathematical induction.

Dating pools

Source: http://www.xkcd.com/314/

Engaging students: Deriving the Pythagorean theorem

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission again comes from my former student Maranda Edmonson. Her topic, from Geometry: deriving the Pythagorean theorem.

D. History: What are the contributions of various cultures to this topic?

Legend has it that Pythagoras was so happy about the discovery of his most famous theorem that he offered a sacrifice of oxen. His theorem states that “the area of the square built upon the hypotenuse of a right triangle is equal to the sum of the areas of the squares upon the remaining sides.” It is likely, though, that the ancient Babylonians and Egyptians knew the result much earlier than Pythagoras, but it is uncertain how they originally demonstrated the proof. As for the Greeks, it is likely that methods similar to Euclid’s Elements were used. Also, though there are many proofs of the Pythagorean Theorem, one came from the contemporary Chinese civilization found in the Arithmetic Classic of the Gnoman and the Circular Paths of Heaven, a Chinese text containing formal mathematical theories.

http://jwilson.coe.uga.edu/emt669/student.folders/morris.stephanie/emt.669/essay.1/pythagorean.html

E. Technology: How can technology be used to effectively engage students with this topic?

The following link is for a video that not only engages students from the very beginning by playing the Mission: Impossible theme and giving students a mission – “should they choose to accept it” – but that has great information. It begins with a short engagement, as stated before, and goes into a little bit of history about Pythagoras and the Pythagoreans. It then briefly describes what the Pythagorean Theorem is before the commentator says, “Does it have applications in our lives today?” At this point (2:43 in the video), it would be beneficial to stop the video and let students discuss where they could use the theorem. The rest of the video simply shows some examples of how the Pythagorean Theorem is used on sailboats, inclined planes, and televisions. It would be up to the teacher whether or not to show the last five minutes of the video to show students these examples, but they could take notes on these examples as they are worked out on the screen.

http://digitalstorytelling.coe.uh.edu/movie_mathematics_02.html

B. Applications: How can this topic be used in your students’ future courses in mathematics or science?

After students learn the Pythagorean Theorem in their Geometry classes, they will use it throughout their mathematical careers. They will use it specifically in Pre-Calculus when they are learning about the unit circle. The theorem is fundamental to proving the basic identities in Trigonometry. It is also used in some of the trigonometric identities, aptly named the Pythagorean Identities based on the nature of their derivation.

In Physics, the kinetic energy of an object is

$\displaystyle \frac{1}{2} (\hbox{mass})(\hbox{velocity})^2$ .

But, in terms of energy, energy at 500 mph = energy at 300 mph + energy at 400 mph. This equation means that, with the energy used to accelerate something at 500 mph, two other objects could use that same energy to be accelerated to 300 mph and 400 mph. Looks like a Pythagorean triple, right? The theorem is also used in Computer Science with processing time. Other examples are found in the link below.

http://betterexplained.com/articles/surprising-uses-of-the-pythagorean-theorem/

Why does 0.999… = 1? (Part 5)

Here’s one more way of convincing students that $0.\overline{9} = 1$ . Here’s the idea: how far apart are the two numbers?

First off, since $1 \ge 0.\overline{9}$ , we know that $1 - \overline{9} \ge 0$ .

Of course, we know that $1-0.9 = 0.1$ . Since $0.\overline{9}$ must lie between $0.9$ and $1$ , we know that $1 - 0.\overline{9}$ must be less than $0.1$ .

Second, we know that $1-0.99 = 0.01$ . Since $0.\overline{9}$ must lie between $0.99$ and $1$ , we know that $1 - 0.\overline{9}$ must be less than $0.01$ .

Third, we know that $1-0.999 = 0.001$ . Since $0.\overline{9}$ must lie between $0.999$ and $1$ , we know that $1 - 0.\overline{9}$ must be less than $0.001$ .

By the same reasoning, we conclude that

$0 \le 1 - 0.\overline{9} < \displaystyle \frac{1}{10^n}$

for every integer $n$ . What’s the only number that’s greater than or equal to $0$ and less than every decimal of the form $0.00\dots001$ ? Clearly, the only such number is $0$ . Therefore,

$1 - 0.\overline{9} = 0$ , or $0.\overline{9} = 1$ .

I like this approach because it really gets at the heart of the difference between integers $\mathbb{Z}$ and real numbers $\mathbb{R}$ . For integers, there is always an integer to the immediate left and to the immediate right. In other words, if you give me any integer (say, $15$ ), I can tell you the largest integer that’s less than your number (in our example, $14$ ) and the smallest integer that’s bigger than your number ( $16$ ).

Real numbers, however, do not have this property. There is no real number to the immediate right of $0$ . This is easy to prove by contradiction. Suppose $x > 0$ is the real number to the immediate left of $0$ . That means that there are no real numbers between $0$ and $x$ . However, $x/2$ is bigger than $0$ and less than $x$ , providing the contradiction.

(For what it’s worth, the above proof doesn’t apply to the set of integers $\mathbb{Z}$ since $x/2$ doesn’t have to be an integer.)

By the same logic — visually, you can imagine reflecting the number line across the point $x = 0.5$ — there is no number to the immediate left of $1$ . So while $0.\overline{9}$ would appear to be to the immediate left of $1$ , they are in reality the same point.

Why does 0.999… = 1? (Part 4)

In this series, I discuss some ways of convincing students that $0.999\dots = 1$ and that, more generally, a real number may have more than one decimal representation even though a decimal representation corresponds to only one real number. This can be a major conceptual barrier for even bright students to overcome. I have met a few math majors within a semester of graduating — that is, they weren’t dummies — who could recite all of these ways and were perhaps logically convinced but remained psychologically unconvinced.

Method #5. This is a proof by contradiction; however, I think it should be convincing to a middle-school student who’s comfortable with decimal representations. Also, perhaps unlike Methods #1-4, this argument really gets to the heart of the matter: there can’t be a number in between $0.999\dots$ and $1$ , and so the two numbers have to be equal.

In the proof below, I’m deliberating avoiding the explicit use of algebra (say, letting $x$ be the midpoint) to make the proof accessible to pre-algebra students.

Suppose that $0.999\dots < 1$ . Then the midpoint of $0.999\dots$ and $1$ has to be strictly greater than $0.999\dots$ , since

$\displaystyle \frac{0.999\dots + 1}{2} > \displaystyle \frac{0.999\dots + 0.999\dots}{2} = 0.999\dots$

Similarly, the midpoint is strictly less than $1$ :

$\displaystyle \frac{0.999\dots + 1}{2} < \displaystyle \frac{1 +1}{2} =1$

(For the sake of convincing middle-school students, a number line with three tick marks — for $0.999\dots$ , $1$ , and the midpoint — might be more believable than the above inequalities.)

So what is the decimal representation of the midpoint? Since the midpoint is less than $1$ , the decimal representation has to be $0.\hbox{something}$ Furthermore, the midpoint does not equal $0.999\dots$ . That means, somewhere in the decimal representation of the midpoint, there’s a digit that’s not equal to $9$ . In other words, the midpoint has to have one of the following 9 forms:

midpoint = $0.999\dots 990 \, \_ \, \_ \dots$

midpoint = $0.999\dots 991 \, \_ \, \_ \dots$

midpoint = $0.999\dots 992 \, \_ \, \_ \dots$

midpoint = $0.999\dots 993 \, \_ \, \_ \dots$

midpoint = $0.999\dots 994 \, \_ \, \_ \dots$

midpoint = $0.999\dots 995 \, \_ \, \_ \dots$

midpoint = $0.999\dots 996 \, \_ \, \_ \dots$

midpoint = $0.999\dots 997 \, \_ \, \_ \dots$

midpoint = $0.999\dots 998 \, \_ \, \_ \dots$

In any event, $9$ is the largest digit. That means that, no matter what, the midpoint is less than $0.999\dots$ , contradicting the fact that the midpoint is larger than $0.999\dots$ (if $0.999\dots < 1$ ).

Why does 0.999… = 1? (Part 3)

Method #4. This is a direct method using the formula for an infinite geometric series… and hence will only be convincing to students if they’re comfortable with using this formula. By definition,

$0.999\dots = \displaystyle \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \dots$

This is an infinite geometric series. Its first term is $\displaystyle \frac{9}{10}$ , and the common ratio needed to go from one term to the next term is $\displaystyle \frac{1}{10}$ . Therefore,

$0.999\dots = \displaystyle \frac{ \displaystyle \frac{9}{10}}{ \quad \displaystyle 1 - \frac{1}{10} \quad}$

$0.999\dots = \displaystyle \frac{ \displaystyle \frac{9}{10}}{ \quad \displaystyle \frac{9}{10} \quad}$

$0.999\dots = 1$