Musings on mathematics and teaching.

## Month: November, 2011

### On the determination of permutable primes

Edit: I have to retract this proof. It was pointed out to me that 10^k covers only 16 of the 17 congruence classes modulo 17, so the last step fails. The other steps are valid, and they show that a permutable prime greater than 991 must be a near-repdigit (all but one of the digits are equal) or a repunit.

A permutable prime (also called an absolute prime) is a prime number such that all of its digit permutations (in base 10) are prime. For example, 113 is a permutable prime, since each of the numbers 113, 131, and 311 are prime. Note that all permutable primes greater than 5 are composed of the digits 1, 3, 7, and 9. The following is a complete list of the permutable primes less than 1000.

2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, 97, 113, 131, 199, 311, 337, 373, 733, 919, 991

A repunit prime is a prime number that contains only the digit 1. Every repunit prime is permutable, and it is conjectured (but not proved) that there exist infinitely many repunit primes. I will prove that the above list contains every permutable prime, with the exception of repunit primes greater than 11.

The proof is based on the following observation. If N and m are integers greater than 1 such that the digit permutations of N cover all congruence classes modulo m, then every number containing the digits of N (as a multiset) is composite. We can find a finite number of pairs (N, m) that will cover every case; however, the number of cases is too large to check by hand. The complete proof was generated by a Python script, and the output is available here.

The first step is to find all permutable primes less than 106 by brute force search. We use the Miller-Rabin primality test (code is here). This is a probabilistic test, but we are not overly concerned with false positives, because we would run a separate deterministic test on any new candidate primes that were identified.

The second step is to verify that a permutable prime cannot contain all four of the digits 1, 3, 7, and 9, by showing that the permutations of 1379 cover all congruence classes modulo 7.

The third step is to verify that a permutable prime cannot contain exactly three of the four digits 1, 3, 7, and 9. There are two sub-cases: the number contains two digits that are repeated at least twice, or the number contains a digit that is repeated five or more times These sub-cases are denoted aabbc and aaaaabc, respectively. As in the previous case, we show that the permutations cover all congruence classes modulo 7. These two sub-cases cover all of the remaining possibilities, since we checked all numbers having fewer than seven digits in step 1.

The fourth step is to verify that a permutable prime cannot contain exactly two of the four digits 1, 3, 7, and 9, unless it is less than 106. There are two sub-cases: the number contains two digits that are repeated at least twice (such as 11133), or the number has all digits the same except for one (such as 11113). For the first sub-case, we consider permutations modulo 7 as before. Unfortunately, there is no proof of the second sub-case using modulo 7 arithmetic, so we must work harder.

Suppose that N is a positive integer with at least 17 digits, all of which are equal to a, except for the last digit which is equal to b. Then the following 17 numbers are digit permutations of N.

$N - (b-a) + (b-a) 10^k\quad (0 \le k \le 16)$

Since 10 is a primitive root modulo 17, these numbers cover all congruence classes. Therefore, N cannot be a permutable prime.

It remains to check that there do not exist any permutable primes of the form aaa…ab that have between 7 and 16 digits. We do this by checking each candidate with a Miller-Rabin primality test. This exhausts all of the possible cases, so the proof is complete.

Since 10 is a primitive root modulo 17, these numbers cover all congruence classes, except the one belonging to $N - (b-a)$. Therefore, N cannot be a permutable prime unless $N - (b-a) \equiv 0 \pmod{17}$, which happens when the number of digits of N is a multiple of 16.

### Generalized binomial coefficients

The reader is probably familiar with factorials and binomial coefficients. The factorial of a number n is the product of all positive integers between 1 and n, and it is denoted by n!. For example, $\displaystyle 6! = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 = 720$. We define 0! = 1.

Factorials are used to define the binomial coefficients. The symbol $\displaystyle \binom{n}{k}$ is defined by the equation $\displaystyle \binom{n}{k} = \frac{n!}{k!\,(n-k)!}$, provided that $0 \le k \le n$. It is not obvious that all binomial coefficients are integers, but this fact can be proved by induction on n using Pascal’s rule:

$\displaystyle \binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}\ .$

We can generalize both factorials and binomial coefficients by replacing the sequence of all positive integers {1, 2, 3, 4, 5, …} by an arbitrary sequence of positive integers. Let

$\displaystyle u = \{u_1, u_2, u_3, \ldots\}$

be any sequence of positive integers whatsoever, and define the u-factorial by the formula

$\displaystyle n!_u = u_1 u_2 u_3 \cdots u_n, \quad 0!_u = 1.$

We use the u-factorial function to define the u-binomial coefficient.

$\displaystyle\binom{n}{k}_u = \frac{n!_u}{k!_u\,(n-k)!_u}$

These definitions are very general, and they do not guarantee that our binomial coefficients are integers. To remedy this deficiency, we will assume that u is a strong divisibility sequence, which means that $d = \gcd(m,n)$ implies $u_d = \gcd(u_m, u_n)$. Here are some examples of strong divisibility sequences.

1. The identity sequence $u_n = n$.
2. The Fibonacci numbers.
3. The q-bracket $\displaystyle u_n = [n]_q = \frac{q^n - 1}{q - 1}$ where $q\ge2$ is an integer.

I claim that if u is a strong divisibility sequence then the u-binomial coefficients are always integers, and I will explain how I discovered a proof of this fact. The key idea is to search for a generalization of Pascal’s rule:

$\displaystyle \binom{n}{k}_u = r \binom{n-1}{k-1}_u + s \binom{n-1}{k}_u$

We write this equation out in full, and then simplify using the fact that $m!_u = u_m \cdot (m-1)!_u$.

It remains to find r and s. Since u is a strong divisibility sequence, $\gcd(u_k, u_{n-k}) = u_d$ where $d = \gcd(k, n-k)$. But ud divides un, since d divides n. Therefore, Bézout’s identity implies that the integers r and s exist. This allows us to prove by induction on n that all u-binomial coefficients are integers.

If we start with the Fibonacci numbers, then the numbers defined by this process are called Fibonomial coefficients. We can also define the q-binomial coefficients by starting with the q-brackets.

The Catalan numbers are ubiquitous in combinatorics. The nth Catalan number is defined by

$\displaystyle \frac{1}{n+1} \binom{2n}{n} = \frac{(2n)!}{n!\,(n+1)!}.$

Alexander Bogomolny has given a delightfully simple proof that this quantity is always an integer. I will leave it to the reader to check that the proof remains valid if the factorials are replaced with u-factorials. This produces integer sequences that are analogous to the Catalan numbers, such as the Fibonomial Catalan numbers when un is the nth Fibonacci number, and the q-Catalan numbers when $u_n = (q^n - 1)/(q - 1)$.

### Is mathematics useful in everyday life?

To be honest, I don’t know the answer to this question. Teachers are supposed to teach students how they will use mathematics in daily life, but I find myself unconvinced by many of the supposed practical applications of mathematics.

Here is a typical example of a practical math lesson, from Yummy Math. The students are asked to adjust a recipe for mashed potatoes to accommodate various numbers of guests for Thanksgiving dinner. The lesson appears to be well-designed, and I have no objections to any of it, except that the questions seem rather dull.

But is it really necessary to be able to convert a recipe by hand? If you use any of the major websites for recipes, then they will convert the recipes for you with a click of a button.There are also websites that will allow you to enter your own recipes and convert them to different quantities. The average person no longer needs to know how to convert a recipe.

The same is true for computing interest payments on a mortgage, or any kind of routine calculation. If it is a problem that people typically encounter in everyday life, then there’s an app for that. Of course, the people who write the apps need to know the math behind them, but for most people this is unnecessary.

So, if we don’t need math to solve routine problems, then what is left? Non-routine problems! We learn math (in part) because we hope to solve problems that are unique to our situations; because we have questions that nobody has ever thought to ask, let alone answer; because we want to find better ways to do things. Also, understanding mathematics allows us to formulate questions that could not be asked (or even thought) without mathematics.

The paradox is that we can’t tell students how they will use math, even though it is tremendously useful. As soon as we name an application of math that our students are likely to encounter, we know that other people have encountered the same problem and solved it before us, and there is little need to solve it again. The goal of a math teacher should be to prepare students to answer questions that we cannot even conceive.

Or maybe not. I’m just rambling. Thanks for reading this far.

### Interesting integer sequence, or a crisis in rationalism?

What is the next number in this sequence? 1, 12, 144, …

You might think that these are powers of 12, and that the next number is 1728. But what if I told you that the sequence continues like this?

1, 12, 144, 1750, 23420, 303240, 3641100, 46113200, 575360400, 7346545000, …

I will explain this sequence after the jump.

### Fibonacci Pigeons

Here is a funny picture that has been circulating the Internet since September 2010. I think that analyzing this picture would be an interesting project for a high school math class. My own analysis is included below.

This picture is amusing, but I wondered if it was real. Some people claim that the picture was Photoshopped. I don’t know how to tell if it was faked, but I do know how to count pixels.

I copied the picture into Microsoft Paint, and I recorded the x-coordinates of the arrow tips. Magnifying the picture to 800% made this task much easier. I entered these coordinates into Microsoft Excel, and I also calculated the first differences.

If the pigeons are truly spaced according to the Fibonacci numbers, then the first differences should be roughly proportional to 1.618x. To test this hypothesis, I made a scatter plot, and I fitted an exponential trend line to the data.

The fit is good, but not spectacular (R2 = 0.9357). However, the base of exponents is not even close to 1.618. According to Excel, the exponential function of best fit is y = 5.9093 * exp(0.2453*x), which can also be written as y = 5.9093 * 1.278^x. If we omit the last two data points as outliers, then the correlation improves to R2 = 0.986, but the base of exponents is even smaller (b = 1.216). The spacing between pigeons is simply not increasing as rapidly as the Fibonacci numbers.

Conclusion: the picture is funny, and perhaps one should not over-analyze a good joke, but it does not show any evidence that pigeons arrange themselves according to the Fibonacci numbers.

### Sums of consecutive integers

Some numbers can be written as the sum of two or more consecutive positive integers, and some cannot. For example, 15 can be expressed in three different ways: 7+8, 4+5+6, and 1+2+3+4+5. But it is not possible to express 8 in this way. This raises two interesting questions:

1. Which numbers can be expressed as the sum of two or more consecutive positive integers?
2. In how many ways can a given number be expressed as the sum of two or more consecutive positive integers?

These are excellent questions for students to investigate, and there have been many articles written about them. The NRICH website has a guide for using this problem in the classroom. I would like to offer a fresh perspective on the problem.

Generalization is one of the most powerful tools that are available to a mathematician. We can generalize the problem by allowing the integers in the sum to be negative or zero, and also by allowing the sum to consist of a single term. With this generalization, there are now 8 ways to write 15 as a sum of consecutive integers.

 15 −14 + … + 14 + 15 7 + 8 −6 + …  + 6 + 7 + 8 4 + 5 + 6 −3 + … + 3 + 4 + 5 + 6 1 + 2 + 3 + 4 + 5 0 + 1 + 2 + 3 + 4 + 5

Let us examine this table closely. The first column lists the solutions whose terms are all positive, and the second column lists the solutions that have at least one non-positive term. These columns have the same length; in fact, there is a one-to-one correspondence between the two sets of solutions.

A + … + B   ↔   (−A+1) + … + B

This transformation has another important property: it changes an expression having an even number of terms into an expression having an odd number of terms, and vice versa. This implies that half of the solutions have an even number of terms, and half of the solutions have an odd number of terms.

But if half of the solutions have an odd number of terms, and half of the solutions have only positive terms, then it follows that the number of solutions with positive terms is equal to the number of solutions with an odd number of terms.

It remains to count the solutions with an odd number of terms. If N is the sum of d consecutive integers, where d is odd, then the middle term must be N/d, hence N/d is an integer. Conversely, if d is odd and N/d is an integer, then the sum of d consecutive integers centered at N/d is equal to N. Therefore, the number of ways to write N as the sum of an odd number of consecutive integers is equal to the number of odd positive divisors of N.

Consider the prime factorization of N

where pn denotes the nth prime (so p1 = 2). Every odd divisor d of N can be written as

where 0 ≤ ai ≤ ei for all i between 2 and k. Therefore, the number of odd divisors of N is equal to

since there are (1 + ei) choices for each exponent ai in the prime factorization of d.

Example 1: In how many ways can 1024 be expressed as the sum of two or more consecutive positive integers?

Solution: Since 1024 is a power of two, it has only one odd divisor (namely 1). Therefore, it is not possible to write 1024 as the sum of two or more consecutive positive integers.

Example 2 : In how many ways can 600 be expressed as the sum of two or more consecutive positive integers?

Solution: By the preceding discussion, this is equal to the number of odd divisors of 600, minus one (to eliminate the trivial solution with one term.) The prime factorization of 600 is

and the number of odd divisors of 600 is (1+1)*(1+2) = 6. Therefore, there are 5 ways to write 600 as the sum of two or more consecutive positive integers.

Example 3: What is the smallest number that can be written as the sum of two or more consecutive integers in exactly 1000 ways?

Solution: This is left as a challenge to the reader.

Note: Nick Hobson arrived at a similar solution independently.

### A Simple Mathematical Model of Economic Inequality

Most people acknowledge that economic inequality has been increasing in the United States. Statistics bear this out. In 2007, the top 1% of families received 24% of the nation’s income, but their share during the 1960s and 1970s was about 9%.

Liberals and conservatives have vastly different explanations for economic inequality. Liberals generally believe that the rules of the game favor the wealthy, while the poor face discrimination and diminished opportunities. Conservatives generally believe that people are wealthy due to superior ability or effort, and that inequality is the natural result of giving people the opportunity to achieve their potential.

Both of these explanations for inequality are deterministic; they discount the role of blind luck. But I claim that inequality can arise purely by chance, and I will explain how this occurs.

Let us begin with a simple thought experiment. Imagine a society in which all citizens begin with equal wealth of 100 units. Every year, half of the population enjoys a 20% increase in wealth, and the other half suffers a 20% decrease in wealth. We suppose that the process is completely random, and it does not favor the rich over the poor. In this imaginary society, the (expected) average wealth remains constant at 100 units, but the median wealth will gradually decline.

 Years Median wealth 0 100.0 10 81.5 20 66.5 30 54.2 40 44.2 50 36.0

The reason for this decline is that the median person’s wealth will experience the same number of up years and down years; but an increase of 20% followed by a decline of 20% results in a decline of 4%, since 1.20 × 0.80 = 0.96. Nevertheless, the society’s total wealth remains the same, because the increases and decreases in wealth cancel each other out. Consequently, the majority of the society’s wealth is concentrated in fewer hands each year.

Let’s generalize this model. We suppose that the distribution of wealth in a society is specified by a random variable X. (This does not mean that wealth is distributed randomly, only that it can vary between individuals.)  We need a mathematical measure of the inequality of a distribution, and we choose to define it by the following formula.

I(X) = E[X2]/(E[X])2

What does this formula mean? E[X] denotes the expected value (or average) of X, so we are dividing the average value of X2 by the square of the average wealth. The quantity I(X) is equal to 1 in the case of perfect equality, and it is greater than 1 otherwise. Larger values of I(X) indicate greater income inequality. An equivalent formula is

I(X) = 1 + (σ/μ)2

where σ is the standard deviation and μ is the mean.

Now, we suppose that each person’s wealth increases or decreases by a random percentage. This is equivalent to multiplying X by a positive random variable Y. We assume that the percentage of increase in wealth is independent of a person’s current wealth; that is, X and Y are independent. But if X and Y are independent, then X2 and Y2 are also independent, which implies that

I(XY) = E[X2Y2]/(E[XY])2 = E[X2]/(E[X])2 × E[Y2]/(E[Y])2 = I(X) I(Y).

But I(Y) > 1, and so I(XY) > I(X), which means that social inequality has increased by a factor of I(Y).

There are many objections that could be raised to this analysis. The model is too simple to capture the complexity of a national economy, and I make no claims to the contrary. If the model were literally true, then we would expect the distribution of wealth to follow a log-normal distribution; but as Ben Goldacre observed, the true distribution more closely resembles a power law. But I do argue that this simple model demonstrates that, in the absence of other factors, random chance will inexorably lead to unequal distribution of wealth.

A more sophisticated treatment of this idea is discussed in the article Entrepreneurs, Chance, and the Deterministic Concentration of Wealth by Joseph E. Fargione, Clarence Lehman, and Stephen Polasky. See this article for a non-technical summary.

### Octagon puzzle

Dr. Gordon Hamilton of MathPickle.Com posed a very interesting question about regular octagons. In this note, I will explain how I solved the problem, and I will generalize to other regular polygons.

Here is the problem. Given a regular octagon in the plane, suppose that we join the vertices using line segments to form a continuous loop that visits each vertex exactly once. Is it necessarily the case that two or more of the line segments are parallel?

I claim that every such circuit contains a pair of parallel line segments. Suppose to the contrary that there exists a circuit which has no parallel segments. In order to analyze the problem mathematically, we number the vertices consecutively from 0 to 7. We also label each edge with the sum of its vertices; for example, the edge from vertex 3 to vertex 7 is labeled with 10.

Now we observe that two line segments are parallel if and only if their labels or the same, or they differ by 8. For example, the line segment between vertex 0 and vertex 4 is parallel to the line segment between vertex 1 and vertex 3, and it is also parallel to the line segment between vertex 5 and vertex 7.

Since there are 8 possible sums, there are also 8 possible directions for the line segments, and a circuit with no parallel segments must have exactly one segment in each direction. Therefore, the sum of the edge labels is 1+2+3+4+5+6+7+8+8k = 36+8k for some integer k.

But this sum must also equal 2*(0 + 1 + 2 + 3 + 4 + 5 + 6 + 7) = 56, because the sum of the edge labels includes every vertex twice.  Therefore, 36 + 8k = 56, or k = 2.5.

This is a contradiction, since k is required to be an integer. Therefore, every circuit joining the vertices of a regular octagon must contain a pair of parallel edges.

Exercise: Show that a similar argument works for any regular polygon having an even number of sides.  What happens if the polygon has an odd number of sides?

### What is a negative number?

Negative numbers can be a difficult concept to understand. In fact, the widespread acceptance of negative numbers is a relatively recent event in the history of mathematics. There are many ways to understand and explain negative numbers; and some of these interpretations are listed below.

1. The opposite of a positive number
2. A number that is less than zero
3. A number that is to the left of zero on a number line
4. A value on a scale that extends beyond zero
5. The amount of a loss or absence
6. A directed quantity
7. A comparison between two quantities
8. The result of subtracting a larger number from a smaller number
9. An equivalence class of ordered pairs of natural numbers

I will discuss my thoughts about these interpretations below. My purpose is not to tell the reader how to think about negative numbers, but to encourage the reader to think deeply about the matter and reach his or her own conclusions. My remarks are intended for other math teachers, so they might be too advanced for beginning learners.

Interpretation 1: The opposite of a positive number.

We know that the opposite of hot is cold, and the opposite of love is hate, but the opposite of a number might be an unfamiliar concept. Two numbers are opposites if their sum is zero. For example, the opposite of 6 is –6, because 6 + (–6) = 0. The same addition also shows that the opposite of –6 is 6.

This interpretation can be modeled using black chips and red chips, where black chips are used to show positive numbers and red chips are used to show negative numbers. We adopt a rule that we can add or remove chips, as long as we add or remove equal numbers of black chips and red chips. For example, a pile containing 4 black chips and 7 red chips represents the number –3, because we may remove 4 black chips and 4 red chips from the pile, leaving 3 red chips.

This interpretation is unsatisfying on its own, because it does not show how negative numbers are used in real life.

Interpretations 2, 3, and 4: A number that is less than zero.

This is self-explanatory, although it is not obvious how a number could be less than zero. Indeed, the concept makes no sense for many kinds of quantities. You can’t have less than zero enemies, eat less than zero servings of vegetables, or walk less than zero miles. A successful explanation of negative numbers has to explain why some quantities can be negative and some quantities cannot.

We can show the natural numbers on a number line, and we interpret “less than” and “greater than” to mean “to the left of” and “to the right of”, respectively. The number line continues to the right without end. Negative numbers arise when we extend the line to the left of zero.

A thermometer is a real-life example of a number line. The zero point is 0º on the Fahrenheit or Celsius scale. On a cold winter day, the temperature can go below zero on either scale. Students should be able to come up with other examples of scales that extend below zero. For example, elevations are defined in reference to sea level, and they can be negative. The lowest point in North America is the Badwater Basin in Death Valley, California. Its elevation is –86 meters, which means that it is 86 meters below sea level.

What these scales have in common is that each has a reference point which is designated as zero, and the quantity can be greater or less than this reference point. In the Celsius scale, the reference point is the freezing temperature of water. When measuring elevation, the reference point is sea level. When the reference point represents an absolute minimum value (such as absolute zero temperature) then negative numbers have no meaning.

Interpretation 5: Loss or absence.

Negative numbers often indicate a loss or decrease in a quantity. A decrease in a quantity may be thought of as a negative change or negative increase. We often see negative numbers in financial news stories when a company loses money or a market index declines in value. An American football team may gain or lose yards on a play, and this can be represented by a positive or negative number.

Negative numbers can also indicate the absence of something, such as a debt or deficit. In accounting, credits and debits are represented by positive and negative numbers respectively.

Interpretation 6: A directed quantity.

Negative numbers are useful for representing quantities that have two opposing directions. The directions can be literal (e.g. up or down, East or West) or metaphorical (e.g. profit or loss, winning or losing). In physics, one usually assumes that up is positive and down is negative. A falling object near Earth’s surface undergoes an acceleration of –9.8 meters per second per second; the acceleration is negative because the object is being pulled downwards by gravity. Latitude and longitude are measured with respect to the equator and the prime meridian. A location that is south of the equator has a negative latitude, and a location that is west of the prime meridian has a negative longitude.

Interpretations 7 and 8:  A comparison between two quantities.

Negative numbers usually arise when we are comparing two measurements of the same type. This is seen to be a generalization of the earlier interpretations. We may be comparing a quantity with an earlier value of the same quantity, which leads to an increase or decrease. Or we may be comparing the quantity to some reference point, such as the prime meridian or the freezing point of water.

The operation of subtraction expresses the difference between two quantities. We may usually think of subtraction as “taking away”, but we also use subtraction to answer the question “how many more?”  The subtraction fact 100 – 86 = 14 tells us that if we take 86 away from 100 then 14 remain, but it also tells us that 100 is 14 more than 86. This allows us to make sense of subtracting a larger number from a smaller number.  We can’t take 5 from 2, but we can say that 2 is 3 less than 5, and this is expressed by the subtraction fact 2 – 5 = –3.

Interpretation 9: An equivalence class of ordered pairs of natural numbers.

In formal mathematics, we construct the integers by defining an equivalence relation on ordered pairs of natural numbers. This approach is much too abstract for beginning learners, but it is a valuable perspective for math teachers.

The set of natural numbers is {0, 1, 2, 3, 4, …}. (Some people exclude 0.) Each integer is represented by an infinite number of different ordered pairs of natural numbers. Two ordered pairs (a,b) and (c,d) represent the same integer if and only if a+d = b+c. For example, the following ordered pairs all represent the same integer:

(0,3), (1,4), (2,5), (3,6), (4,7), …

The idea can be seen more clearly if we change our notation. If we write the ordered pair (a,b) as (a – b) instead, then we have

(0 – 3) = (1 – 4) = (2 – 5) = (3 – 6) = (4 – 7) = …

The standard notation for this integer is –3. What you should take away from this is that it is possible to define an integer as the result of subtracting two natural numbers.

### Graph Transformations and Daylight Saving Time

Saturday was the last day of Daylight Saving Time in the United States. On this day, most of set our clocks back an hour, and we enjoyed an extra hour of sleep. I say “most of us” because Arizona, Puerto Rico, Hawaii, U.S. Virgin Islands and American Samoa do not observe Daylight Saving Time.

Some people find Daylight Saving Time to be confusing, or at least hard to remember. The phrase “spring forward, fall back” is meant to remind us to set our clocks ahead one hour in the spring, and set it one hour back in the fall. One possible reason for the confusion is that there are two different ways to think about Daylight Saving Time. The usual way to express it is that we set the clocks ahead in the spring, and set them back in the fall. We might call this the clock’s point of view. The other point of view is that we must wake up an hour earlier in the spring, but we are permitted to sleep an hour later in the fall.

We encounter a similar concept when we transform the graph of a function by shifting it to the left or right. In algebra, we learn that replacing x with x−1 in an equation will shift the graph one unit to the right, and replacing x with x+1 will shift the graph one unit to the left. This is very perplexing. Surely, subtracting 1 should be a move to the left, and adding 1 should be a move to the right. Why is everything backwards when it comes to graph transformations?

But just as with Daylight Saving Time, there are two ways to think about graph transformations. When we replace x with x−1, we usually say that the graph moves one unit to the right. But an alternative interpretation is that the coordinate system moves one unit to the left, and the graph stays still! To put it another way, suppose you start with the graph of y = f(x), but then you subtract one to each of the number labels on the x-axis. After some thought, you will see that the same graph now describes the equation y = f(x−1).

These two ways of viewing a transformation are called “alibi” and “alias”. An alibi transformation moves the points (“alibi” is Latin for “in another place”.) An alias transformation does not move the points, but only renames them. The difference between an alias and an alibi is illustrated by the hilarious train scene gag in the movie Top Secret (1984).

Sleeping an hour later is an alibi transformation — we are shifting our “sleep curves” one hour to the right. Setting the clock an hour back is an alias transformation — the time coordinate is shifted to the left.