# Factoring the time

True story: one way that I commit large numbers to (hopefully) short-term memory is by factoring. If I take the time to factor a big number, then I can usually remember it for a little while.

This approach has occasional disadvantages. For example, I now have stuck in my brain the completely useless information that, many years ago, my seat at a Texas Rangers ballgame was somewhere in Section 336 (which is $6 \times 7 \times 8$).

Source: http://www.xkcd.com/247/

# Engaging students: Introducing variables and expressions

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 Caitlin Kirk. Her topic, from Pre-Algebra: introducing variables and expressions.

To keep track of some of the coldest things in the universe, scientist use the Kelvin temperature scale that begins at 0 Kelvin, or Absolute Zero. Nothing can ever be colder than Absolute Zero because at this temperature, all motion stops. The table below shows some typical temperatures of different systems in the universe.

Table of Cold Places

 Temp.(K) Location 183 Vostok, Antarctica 160 Phobos- a moon of Mars 128 Europa in the summer 120 Moon at night 88 Miranda surface temp. 81 Enceladus in the summer 70 Mercury at night 55 Pluto in the summertime 50 Dwarf Planet Quaoar 33 Pluto in the wintertime 1 Boomerang Nebula 0 ABSOLUTE ZERO

You are probably already familiar with the Celsius (C) and Fahrenheit (F) temperature scales. The two formulas below show how to switch from degrees-C to degrees-F.

$C = \frac{5}{9} (F-32)$

$F = \frac{9}{5} C + 32$

Because the Kelvin scale is related to the Celsius scale, we can also convert from Celsius to Kelvin (K) using the equation:

$K = 273 + C$

Problems

Use these three equations to convert between the three temperature scales:

Problem 1: 212 F converted to K

Problem 2: 0 K converted to F

Problem 3: 100 C converted to K

Problem 4: Two scientists measure the daytime temperature of the moon using two different instruments. The first instrument gives a reading of +107 C while the second instrument gives +221 F.

a. What are the equivalent temperatures on the Kelvin scale?

b. What is the average daytime temperature on the Kelvin scale?

Problem 5: Humans can survive without protective clothing in temperatures ranging from 0 F to 130 F. In what, if any, locations from the table above can humans survive?

Solutions

Problem 1: First convert to C:  C = 5/9 (212-32) = +100 C. Then convert from C to K: K = 273 + 100 = 373 Kelvin.

Problem 2: First convert to Celsius:    0 = 273 + C so C = -273. Then convert from C to F: F = 9/5 (-273) + 32 = -459 Fahrenheit.

Problem 3: K = 273 – 100 = 173 Kelvin.

Problem 4:

a. 107 C becomes K = 273 + 107 = 380 Kelvin.  221 F becomes C = 5/9(221-32) = 105 C, and so K = 273 + 105 = 378 Kelvin.

b. (380 + 378)/2 = 379 Kelvin

Problem 5:

First convert 0 F and 130 F to Celsius so that the conversion to Kelvin is quicker. 0 F becomes C = 5/9(0-32) = -18 C (rounded to the nearest degree) and 130 F becomes C = 5/9 (130-32) = 54 C (rounded to the nearest degree).

Next, convert -18 C and 54 C to Kelvin. -18 C becomes K = 273-18 = 255 and 54 C becomes k = 273 + 54 = 327 K.

None of the locations on the table have temperatures between 255 K and 327 K, therefore humans could not survive in any of these space locations.

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

This topic is one of the first experiences students have with algebra. Since algebra is the point from which students dive into more advanced mathematics, this topic will be used in many different areas of future mathematics. After mastering the use of one variable, with the basic operations of addition, subtraction, multiplication, and division, students will be introduced to the use of more than one variable. They may be asked to calculate the area of a solid whose perimeter is given and whose side lengths are unknown variables. Or in a more advanced setting, they may be asked to calculate how much money will be in a bank account after five years of interest compounded continuously. In fact, the use of variables is present and important in every mathematics class from Algebra I through Calculus and beyond. There very well may never be a day in a mathematics students’ life where they will not see a variable after variables have been introduced.

B.  How does this topic extend what your students should have learned in previous courses?

In basic arithmetic, probably in elementary or early middle school math classes, students learn how to do calculations with numbers using the four basic operations of addition, subtraction, multiplication and division. They also learn simple applications of these basic operations by calculating the area and perimeter of a rectangle, for example. Introducing variables and expressions is a continuation of those same ideas except that one or more of the numbers is now an unknown variable. Students can rely on the arithmetic skills they already possess when learning this introduction to algebra with variables and expressions.

Students are familiar with calculating the area and perimeter of figures like the one on the left before they are introduced to variables. Later, they may see the same figure with the addition of a variable, as shown on the right. The addition of the variable will come with new instructions as well.

The difficulty of problems using variables is determined by the information given in the problems. For instance, the problem on the right can be a one step equation if an area and perimeter are given so that students only need to solve for w. The difficulty can be increased by giving only a perimeter so that students must solve for w and then for the area.

# Engaging students: Computing the determinant of a matrix

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 Caitlin Kirk. Her topic: computing the determinant of a matrix.

B. Curriculum: How does this topic extend what your students should have learned in previous courses?

Students learn early in their mathematical careers how to calculate the area of simple polygons such as triangles and parallelograms. They learn by memorizing formulas and plugging given values into the formulas. Matrices, and more specifically the determinant of a matrix, can be used to do the same thing.

For example, consider a triangle with vertices $(1,2)$, $(3, -4)$, and $(-2,3)$. The traditional method for finding the area of this circle would be to use the distance formula to find the length of each side and the height before plugging and chugging with the formula $A = \frac{1}{2} bh$. Matrices can be used to compute the same area in fewer steps using the fact that the area of a triangle the absolute value of one-half times the determinant of a matrix containing the vertices of the triangle as shown below.

First, put the vertices of the triangle into a matrix using the x-values as the first column and the corresponding y-values as the second column. Then fill the third column with 1’s as shown:

Next, compute the determinant of the matrix and multiply it by ½ (because the traditional area formula for a triangle calls for multiplying by ½ to account for the fact that a triangle is half of a rectangle, it is necessary to keep the ½ here also) as shown:

Obviously, the area of a triangle cannot be negative. Therefore it is necessary to take the absolute value of the final answer. In this case $|-8| = 8$, making the area positive eight instead of negative eight.

The same idea can be applied to extend students knowledge of the area of other polygons such as a parallelogram, rectangle, or square. Determinants of matrices are a great extension of the basic mathematical concept of area that students will have learned in previous courses.

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

The history of matrices can be traced to four different cultures. First, Babylonians as early as 300 BC began attempting to solve simultaneous linear equations like the following:

There are two fields whose total area is eighteen hundred square yards. One produces grain at the rate of two-thirds of a bushel per square yard while the other produces grain at the rate of on-half a bushel per square yard. If the total yield is eleven hundred bushels, what is the size of each field?

While the Babylonians at this time did not actually set up matrices or calculate any determinants, they laid the framework for later cultures to do so by creating systems of linear equations.

The Chinese, between 200 BC and 100 BC, worked with similar systems and began to solve them using columns of numbers that resemble matrices. One such problem that they worked with is given below:

There are three types of corn, of which three bundles of the first, two of the second, and one of the third make 39 measures. Two of the first, three of the second and one of the third make 34 measures. And one of the first, two of the second and three of the third make 26 measures. How many measures of corn are contained of one bundle of each type?

Unlike the Babylonians, the Chinese answered this question using their version of matrices, called a counting board. The counting board functions the same way as modern matrices but is turned on its side. Modern matrices write a single equation in a row and the next equation in the next row and so forth. Chinese counting boards write the equations in columns. The counting board below corresponds to the question above:

1   2   3

2   3   2

3   1   1

26  34  39

They then used what we know as Gaussian elimination and back substitution to solve the system by performing operations on the columns until all but the bottom row contains only zeros and ones. Gaussian elimination with back substitution did not become a well known method until the early 19th century, however.

Next, in 1683, the Japanese and Europeans simultaneously saw the discovery and use of a determinant, though the Japanese published it first. Seki, in Japan, wrote Method of Solving the Dissimulated Problems which contains tables written in the same manner as the Chinese counting board. Without having a word to correspond to his calculations, Seki calculated the determinant and introduced a general method for calculating it based on examples. Using his methods, Seki was able to find the determinants of 2×2, 3×3, 4×4, and 5×5 matrices.

In the same year in Europe, Leibniz wrote that the system of equations below:

$10+11x+12y=0$

$20+21x+22y=0$

$30+31x+32y=0$

has a solution because

$(10 \times 21 \times 32)+(11 \times 22 \times 30)+(12 \times 20 \times 31)=(10 \times 22 \times 31)+(11 \times 20 \times 32)+(12 \times 21 \times 30)$.

This is the exact condition under which the matrix representing the system has a determinant of zero. Leibniz was the first to apply the determinant to finding a solution to a linear system. Later, other European mathematicians such as Cramer, Bezout, Vandermond, and Maclaurin, refined the use of determinants and published rules for how and when to use them.

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

Calculating the determinant is used in many lessons in future mathematics courses, mainly in algebra II and pre-calculus. The determinant is the basis for Cramer’s rule that allows a student to solve a system of linear equations. This leads to other methods of solving linear systems using matrices such as Gaussian elimination and back substitution.  It can also be used in determining the invertibility of matrices.  A matrix whose determinant is zero does not have an inverse. Invertibility of matrices determines what other properties of matrix theory a given matrix will follow. If students were to continue pursuing math after high school, understanding determinants is essential to linear algebra.

# Collaborative Mathematics: Challenge 05

My colleague Jason Ermer is back from summer hiatus and has posted his fifth challenge video, shown below.

Video responses can be posted to his website, http://www.collaborativemathematics.org. In the words of his website, this is a unique forum for connecting a worldwide community of mathematical problem-solvers, and I think these unorthodox but simply stated problems are a fun way for engaging students with the mathematical curriculum.

# Thoughts on 1/7 and other rational numbers (Part 10)

In the previous post, I showed a quick way of obtaining a full decimal representation using a calculator that only displays ten digits at a time. To review: here’s what a TI-83 Plus returns as the (approximate) value of $8/17$:

Using this result and the Euler totient function, we concluded that the repeating block had length $16$. So we multiply twice by $10^8$ (since $10^8 \times 10^8 = 10^{16}$) to deduce the decimal representation, concluding that

$\displaystyle \frac{8}{17} = 0.\overline{4705882352941176}$

Though this is essentially multi-digit long division, most students are still a little suspicious of this result on first exposure. So here’s a second way of confirming that we did indeed get the right answer. The calculators show that

$8 \times 10^8 = 17 \times 47058823 + 9$ and $9 \times 10^8 = 17 \times 52941176 + 8$

Therefore,

$8 \times 10^{16} = 17 \times (47058823 \times 10^8) + 9 \times 10^8$ and $9 \times 10^8 = 17 \times 52941176 + 8$

so that

$8 \times 10^{16} = 17 \times (47058823 \times 10^8) + 17 \times 52941176 + 8$

$8 \times 10^{16} = 17 \times 4705882352941176 + 8$

$8 \times 10^{16} - 8 = 17 \times 4705882352941176$

$8 (10^{16}-1) = 17 \times 4705882352941176$

$\displaystyle \frac{8}{17} = \displaystyle \frac{4705882352941176}{10^{16}-1}$

Using the rule for dividing by $10^k -1$, we conclude that

$\displaystyle \frac{8}{17} = 0.\overline{4705882352941176}$

# Thoughts on 1/7 and other rational numbers (Part 9)

Let’s now consider the decimal representation of $\displaystyle \frac{8}{17}$.

There’s no obvious repeating pattern. But we know that, since 17 has neither 2 nor 5 as a factor, that there has to be a repeating decimal pattern.

So… what is it?

When I ask this question to my students, I can see their stomachs churning a slow dance of death. They figure that the calculator didn’t give the answer, and so they have to settle for long division by hand.

That’s partially correct.

However, using the ideas presented below, we can perform the long division extracting multiple digits at once. Through clever use of the calculator, we can quickly obtain the full decimal representation even though the calculator can only give ten digits at a time.

Let’s now return to where this series began… the decimal representation of $\displaystyle \frac{1}{7}$ using long division. As shown below, the repeating block has length $6$, which can be found in a few minutes with enough patience. By the end of this post, we’ll consider a modification of ordinary long division that facilitates the computation of really long repeating blocks.

Because we arrived at a repeated remainder, we know that we have found the repeating block. So we can conclude that $\displaystyle \frac{1}{7} = 0.\overline{142857}$.

Students are taught long division in elementary school and are so familiar with the procedure that not much thought is given to the logic behind the procedure. The underlying theorem behind long division is typically called the division algorithm. From Wikipedia:

Given two integers $a$ and $b$, with $b \ne 0$, there exist unique integers $q$ and $r$ such that $a = bq+r$ and $0 \le r < |b|$,  where $|b|$ denotes the absolute value of $b$.

The number $q$ is typically called the quotient, while the number $r$ is called the remainder.

Repeated application of this theorem is the basis for long division. For the example above:

Step 1.

$10 = 1 \times 7 + 3$. Dividing by $10$, $1 = 0.1 \times 7 + 0.3$

Step 2.

$30 = 4 \times 7 + 2$. Dividing by $100$, $0.3 = 0.04 \times 7 + 0.02$

Returning to the end of Step 1, we see that

$1 = 0.1 \times 7 + 0.3 = 0.1 \times 7 + 0.04 \times 7 + 0.02 = 0.14 \times 7 + 0.02$

Step 3.

$20 = 2 \times 7 + 6$. Dividing by $1000$, $0.02 = 0.002 \times 7 + 0.006$

Returning to the end of Step 2, we see that

$1 = 0.14 \times 7 + 0.02 = 0.14 \times 7 + 0.0002 \times 7 + 0.006 = 0.142 \times 7 + 0.006$

And so on.

By adding an extra zero and using the division algorithm, the digits in the decimal representation are found one at a time. That said, it is possible (with a calculator) to find multiple digits in a single step by adding extra zeroes. For example:

Alternate Step 1.

$1000 = 142 \times 7 + 6$. Dividing by $1000$, $1 = 0.142 \times 7 + 0.006$

Alternate Step 2.

$6000 = 587 \times 7 + 1$. Dividing by $100000$, $0.006 = 0.000587 \times 7 + 0.000001$

Returning to the end of Alternate Step 1, we see that

$1 = 0.142 \times 7 + 0.006= 0.142 \times 7 + 0.000587\times 7 + 0.000001 = 0.142857 \times 7 + 0.000001$

So, with these two alternate steps, we arrive at a remainder of $1$ and have found the length of the repeating block.

The big catch is that, if $a = 1000$ or $a = 6000$ and $b = 7$, the appropriate values of $q$ and $r$ have to be found. This can be facilitated with a calculator. The integer part of $1000/7$ and $6000/7$ are the two quotients needed above, and subtraction is used to find the remainders (which must be less than $7$, of course).

At first blush, it seems silly to use a calculator to find these values of $q$ and $r$ when a calculator could have been used to just find the decimal representation of $1/7$ in the first place. However, the advantage of this method becomes clear when we consider fractions who repeating blocks are longer than 10 digits.

Let’s now return to the question posed at the top of this post: finding the decimal representation of $\displaystyle \frac{8}{17}$. As noted in Part 6 of this series, the length of the repeating block must be a factor of $\phi(17)$, where $\phi$ is the Euler toitent function, or the number of integers less than $17$ that are relatively prime with $17$. Since $17$ is prime, we clearly see that $\phi(17) = 16$. So we can conclude that the length of the repeating block is a factor of $16$, or either $1$, $2$, $4$, $8$, or $16$.

Here’s the result of the calculator again:

We clearly see from the calculator that the repeating block doesn’t have a length less than or equal to $8$. By process of elimination, the repeating block must have a length of $16$ digits.

Now we perform the division algorithm to obtain these digits, as before. This can be done in two steps by multiplying by $10^8 = 100,000,000$.

So, by the same logic used above, we can conclude that

$\displaystyle \frac{8}{17} = 0.\overline{4705882352941176}$

In other words, through clever use of the calculator, the full decimal representation can be quickly found even if the calculator itself returns only ten digits at a time… and had rounded the final $2941176$ of the repeating block up to $3$.

# Thoughts on 1/7 and other rational numbers (Part 8)

In Part 6 of this series, I mentioned the following fact concerning the decimal representation of $\displaystyle \frac{a}{b}$: if neither $2$ nor $5$ is a factor of $b$, then the repeating block in the decimal representation of $\displaystyle \frac{a}{b}$ has a length $k$ that must be a factor of $\phi(b)$. This function is the Euler toitent function or the number of integers less than $b$ that are relatively prime with $b$.

In this post, I’d like to provide a justification for this theorem.

As discussed earlier, $k$ is the least integer so that $b$ is a factor of $10^k - 1$. In the language of congruence, $k$ is the least integer so that

$10^k \equiv 1 (\mod b)$

In other words, let $G_b$ be the multiplicative group of numbers less than $b$ that are relatively prime with $b$. By assumption $10 \in G_b$. Then $k$ is the order of $10$ in $G_b$, and there’s a theorem that states that the order of an element of a group must be a factor of the order of the group, or the number of elements in the group. In our case, the order of $G_b$ is the number of integers less than $b$ that are relatively prime with $b$, or $\phi(b)$.

In other words, using these ideas from group theory, we can prove that $k \mid \phi(b)$.

Naturally, we don’t expect middle school students seeing long division for the first time to appreciate this property of decimal representations. Still, my main purpose in writing this post was to give a concrete example of how ideas from higher-level mathematics — like group theory — actually can shed insight into ideas that are first seen in school — even middle school. In other words, there’s a reason why UNT (and other universities) requires that college students who want to earn mathematics teaching certification with their degrees must have a major in mathematics.

# Thoughts on 1/7 and other rational numbers (Part 7)

In a previous post concerning roundoff error, I mentioned that the number $1/10$ equals

$\displaystyle \frac{1}{2^4} + \frac{1}{2^5} +\frac{1}{2^8} + \frac{1}{2^9} + \frac{1}{2^{12}} + \frac{1}{2^{13}} + \dots$

In other words, the binary expansion of $1/10$ is

$0.0001100110011001100110011001100....$

That’s the expansion of the fraction in base $2$, as opposed to base $10$.

In the previous post, I verified that the above infinite series actually converges to $1/10$:

$S = \displaystyle \left(\frac{1}{2^4} + \frac{1}{2^5}\right) +\left(\frac{1}{2^8} + \frac{1}{2^9}\right) + \left(\frac{1}{2^{12}} + \frac{1}{2^{13}}\right) + \dots$

$S = \displaystyle \frac{3}{2^5} + \frac{3}{2^9} + \frac{3}{2^{13}} + \dots$

$S = \displaystyle \frac{\displaystyle \frac{3}{2^5}}{\quad \displaystyle 1 - \frac{1}{2^4} \quad}$

$S = \displaystyle \frac{\displaystyle \frac{3}{32}}{\quad \displaystyle \frac{15}{16} \quad}$

$S = \displaystyle \frac{3}{32} \times \frac{16}{15}$

$S = \displaystyle \frac{1}{10}$

Still, a curious student may wonder how one earth one could directly convert $1/10$ into binary without knowing the above series ahead of time.

This can be addressed by using the principles that we’ve gleaned in this study of decimal representations, except translating this work into the language of base $2$. In the following, I will use the subscripts $\hbox{ten}$ and $\hbox{two}$ so that I’m clear about when I’m using decimal and binary, respectively.

To begin, we note that $10_{\hbox{\scriptsize ten}} = 1010_{\hbox{\scriptsize two}} = 10_{\hbox{\scriptsize two}} \times 101_{\hbox{\scriptsize two}}$. (In other words, ten is equal to two times five.) So, following Case 3 of the previous post, we will attempt to write the denominator in the form

$10_{\hbox{\scriptsize two}}^d \left(10_{\hbox{\scriptsize two}}^k - 1 \right)$, or $2_{\hbox{\scriptsize ten}}^d \left(2_{\hbox{\scriptsize ten}}^k - 1 \right)$

• If $k = 1_{\hbox{\scriptsize ten}}$, then $2_{\hbox{\scriptsize ten}}^1 - 1 = 1_{\hbox{\scriptsize ten}}$, but $1_{\hbox{\scriptsize ten}} \div 5_{\hbox{\scriptsize ten}}$  is not an integer.
• If $k = 2_{\hbox{\scriptsize ten}}$, then $2_{\hbox{\scriptsize ten}}^2 - 1 = 3_{\hbox{\scriptsize ten}}$, but $3_{\hbox{\scriptsize ten}} \div 5_{\hbox{\scriptsize ten}}$  is not an integer.
• If $k = 3_{\hbox{\scriptsize ten}}$, then $2_{\hbox{\scriptsize ten}}^3 - 1 = 7_{\hbox{\scriptsize ten}}$, but $7_{\hbox{\scriptsize ten}} \div 5_{\hbox{\scriptsize ten}}$  is not an integer.
• If $k = 4_{\hbox{\scriptsize ten}}$, then $2_{\hbox{\scriptsize ten}}^4 - 1 = 15_{\hbox{\scriptsize ten}}$. This time, $15_{\hbox{\scriptsize ten}} \div 5_{\hbox{\scriptsize ten}} = 3_{\hbox{\scriptsize ten}}$. Written in binary,

$101_{\hbox{\scriptsize two}} \times 11_{\hbox{\scriptsize two}} = 1111_{\hbox{\scriptsize two}}$

We now return to the binary representation of $1/10_{\hbox{\scriptsize ten}} = 1/1010_{\hbox{\scriptsize two}}$.

$\displaystyle \frac{1}{1010_{\hbox{\scriptsize two}}} = \displaystyle \frac{1}{1010_{\hbox{\scriptsize two}}} \times \frac{11_{\hbox{\scriptsize two}}}{11_{\hbox{\scriptsize two}}}$

$\displaystyle \frac{1}{1010_{\hbox{\scriptsize two}}} = \frac{11_{\hbox{\scriptsize two}}}{11110_{\hbox{\scriptsize two}}}$

Therefore, the binary representation has a delay of one digit and a repeating block of four digits:

$\displaystyle \frac{1}{1010_{\hbox{\scriptsize two}}} = 0.0\overline{0011}$

Naturally, this matches the binary representation given earlier.

# Thoughts on 1/7 and other rational numbers (Part 6)

In Part 5 of this series, I showed that fractions of the form $\displaystyle \frac{M}{10^d}$, $\displaystyle \frac{M}{10^k - 1}$, and $\displaystyle \frac{M}{10^d (10^k-1)}$ can be converted into their decimal representations without using long division and without using a calculator.

The amazing thing is that every rational number $\displaystyle \frac{a}{b}$ can be written in one of these three forms. Therefore, after this conversion is made, then the decimal expansion can be found without a calculator.

Case 1. If the denominator $b$ has a prime factorization of the form $2^m 5^n$, then $\displaystyle \frac{a}{b}$ can be rewritten in the form $\displaystyle \frac{M}{10^d}$, where $d = \max(m,n)$.

For example,

$\displaystyle \frac{3}{160} = \displaystyle \frac{3}{2^5 \times 5}$

$\displaystyle \frac{3}{160} = \displaystyle \frac{3}{2^5 \times 5} \times \frac{5^4}{5^4}$

$\displaystyle \frac{3}{160} = \displaystyle \frac{3 \times 5^4}{2^5 \times 5^5}$

$\displaystyle \frac{3}{160} = \displaystyle \frac{1875}{10^5}$

$\displaystyle \frac{3}{160} = 0.01875$

The step of multiplying both sides by $\displaystyle \frac{5^4}{5^4}$ is perhaps unusual, since we’re so accustomed to converting fractions into lowest terms and not making the numerators and denominators larger. This particular form of $1$ was chosen in order to get a power of $10$ in the denominator, thus facilitating the construction of the decimal expansion.

Case 2. If the denominator $b$ is neither a multiple of $2$ nor $5$, then  $\displaystyle \frac{a}{b}$ can be rewritten in the form $\displaystyle \frac{M}{10^k - 1}$.

For example,

$\displaystyle \frac{3}{11} = \displaystyle \frac{3}{11} \times \frac{9}{9}$

$\displaystyle \frac{3}{11} = \displaystyle \frac{27}{99}$

$\displaystyle \frac{3}{11} = 0.\overline{27}$

This example wasn’t too difficult since we knew that $9 \times 11 = 99$. However, finding the smallest value of $k$ that works can be a difficult task requiring laborious trial and error.

However, we do have a couple of theorems that can assist in finding $k$. First, since $k$ is the length of the repeating block, we are guaranteed that $k$ must be less than the denominator $b$ since, using ordinary long division, the length of the repeating block is determined by how many steps are required until we get a remainder that was seen before.

However, we can do even better than that. Using ideas from number theory, it can be proven that $k$ must be a factor of $\phi(b)$, which is the Euler toitent function or the number of integers less than $b$ that are relatively prime with $b$. In the example above, the denominator was $11$, and clearly, if $1 \le n < 11$, then $\gcd(n,11) = 1$. Since there are $10$ such numbers, we know that $k$ must be a factor of $10$. In other words, $k$ must be either $1$, $2$, $5$, or $10$, thus considerably reducing the amount of guessing and checking that has to be done. (Of course, for the example above, $k=2$ was the least value of $k$ that worked.)

In general, if $n = p_1^{a_1} p_2^{a_2} \dots p_r^{a_r}$ is the prime factorization of $n$, then

$\phi(n) = n \left( \displaystyle 1 - \frac{1}{p_1} \right) \left( \displaystyle 1 - \frac{1}{p_2} \right) \dots \left( \displaystyle 1 - \frac{1}{p_r} \right)$

For the example above, since $11$ was prime, we have $\phi(11) = 11 \left( \displaystyle 1 - \frac{1}{11} \right) = 10$.

Case 3. Suppose the prime factorization of the denominator $b$ both (1) contains $2$ and/or $5$ and also (2) another prime other than $2$ and $5$. This is a mixture of Cases 1 and 2, and the fraction $\displaystyle \frac{a}{b}$ can be rewritten in the form $\displaystyle \frac{M}{10^d (10^k-1)}$.

For example, consider

$\displaystyle \frac{11}{74} = \displaystyle \frac{11}{2 \times 37}$

Following the rule for Case 1, we should multiply by $\displaystyle \frac{5}{5}$ to get a $10$ in the denominator:

$\displaystyle \frac{11}{74} = \displaystyle \frac{11}{2 \times 37} \times \frac{5}{5} = \frac{55}{10 \times 37}$

Next, we need to multiply $37$ by something to get a number of the form $10^k - 1$. Since $37$ is prime, every number less than $37$ is relatively prime with $37$, so $\phi(37) = 36$. Therefore, $k$ must be a factor of $36$. So, $k$ must be one of $1$, $2$, $3$, $4$, $6$, $9$, $12$, $18$, and $36$.

(Parenthetically, while we’ve still got some work to do, it’s still pretty impressive that — without doing any real work — we can reduce the choices of $k$ to these nine numbers. In that sense, the use of $\phi(n)$ parallels how the Rational Root Test is used to determine possible roots of polynomials with integer coefficients.)

So let’s try to find the least value of $k$ that works.

• If $k = 1$, then $10^1 - 1 = 9$, but $9 \div 37$ is not an integer.
• If $k = 2$, then $10^2 - 1 = 99$, but $99 \div 37$ is not an integer.
• If $k = 3$, then $10^3 - 1 = 999$, and it turns out that $999 \div 37 = 27$, an integer.

Therefore,

$\displaystyle \frac{11}{74} = \frac{55}{10 \times 37} \times \frac{27}{27}$

$\displaystyle \frac{11}{74} = \frac{1485}{10 \times 999}$

$\displaystyle \frac{11}{74} = \frac{999 + 486}{10 \times 999}$

$\displaystyle \frac{11}{74} = \frac{999}{10 \times 999}+ \frac{486}{10 \times 999}$

$\displaystyle \frac{11}{74} = \frac{1}{10}+ \frac{486}{10 \times 999}$

$\displaystyle \frac{11}{74} = 0.1 + 0.0\overline{486}$

$\displaystyle \frac{11}{74} = 0.1\overline{486}$

# Thoughts on 1/7 and other rational numbers (Part 5)

Students are quite accustomed to obtaining the decimal expansion of a fraction by using a calculator. Here’s an (uncommonly, I think) taught technique for converting certain fractions into a decimal expansion without using long division and without using a calculator. I’ve taught this technique to college students who want to be future high school teachers for several years, and it never fails to surprise.

First off, it’s easy to divide any number by a power of $10$, or $10^k$. For example,

$\displaystyle \frac{4312}{1000} = 4.312$ and $\displaystyle \frac{71}{10000} = 0.00071$

What’s less commonly known is that it’s also easy to divide by $10^k - 1$, or $99\dots 9$, a numeral with $k$ consecutive $9$s. (This number can be used to prove the divisibility rules for 3 and 9 and is also the subject of one of my best math jokes.) The rule can be illustrated with a calculator:

In other words, if $M < 10^k - 1$, then the decimal expansion of $\displaystyle \frac{M}{10^k-1}$ is a repeating block of $k$ digits containing the numeral $M$, possibly adding enough zeroes to fill all $k$ digits.

To prove that this actually works, we notice that

$\displaystyle \frac{M}{10^k - 1} = M \times \frac{ \displaystyle \frac{1}{10^k}}{\quad \displaystyle 1 - \frac{1}{10^k} \quad}$

$\displaystyle \frac{M}{10^k - 1} = M \times \left(\displaystyle \frac{1}{10^k} + \frac{1}{10^{2k}} + \frac{1}{10^{3k}} + \dots \right)$

$\displaystyle \frac{M}{10^k-1} = M \times 0.\overline{00\dots01}$

The first line is obtained by multiplying the numerator and denominator by $\displaystyle \frac{1}{10^k}$. The second line is obtained by using the formula for an infinite geometric series in reverse, so that the first term is $\displaystyle \frac{1}{10^k}$ and the common ratio is also $\displaystyle \frac{1}{10^k}$. The third line is obtained by converting the series — including only powers of $10$ — into a decimal expansion.

If $M > 10^k - 1$, then the division algorithm must be used to get a numerator that is less than $10^k-1$. Fortunately, dividing big numbers by $10^k-1$ is quite easy and can be done without a calculator. For example, let’s find the decimal expansion of $\displaystyle \frac{123456}{9999}$ without a calculator. First,

$123456 = 12(10000) + 3456$

$123456 = 12(9999 + 1) + 3456$

$123456 = 12(9999) + 12(1) + 3456$

$123456 = 12(9999) + 3468$

Therefore,

$\displaystyle \frac{123456}{9999} = \displaystyle \frac{12(9999) + 3468}{9999}$

$\displaystyle \frac{123456}{9999} = \displaystyle \frac{12(9999)}{9999} + \frac{3468}{9999}$

$\displaystyle \frac{123456}{9999} = \displaystyle 12 + \frac{3468}{9999}$

$\displaystyle \frac{123456}{9999} = \displaystyle 12.\overline{3468}$

This can be confirmed with a calculator. Notice that the repeating block doesn’t quite match the digits of the numerator because of the intermediate step of applying the division algorithm.

In the same vein, it’s also straightforward to find the decimal expansion of fractions of the form $\displaystyle \frac{M}{10^d (10^k-1)}$, so that the denominator has the form $99\dots9900\dots00$. This is especially easy if $M < 10^k -1$. For example,

$\displaystyle \frac{123}{99900} = \frac{1}{100} \times \frac{123}{999} = \frac{1}{100} \times 0.\overline{123} = 0.00\overline{123}$

On the other hand, if $M > 10^k-1$, then the division algorithm must be applied as before. For example, let’s find the decimal expansion of $\displaystyle \frac{51237}{99000}$. To begin, we need to divide the numerator by $99$, as before. Notice that, for this example, an extra iteration of the division algorithm is needed to get a remainder less than $99$.

$51237 = 512(100) + 37$

$51237 = 512(99 + 1) + 37$

$51237 = 512(99) + 512 + 37$

$51237 = 512(99) + 549$

$51237= 512(99) + 5(100) + 49$

$51237 = 512(99) + 5(99 + 1) + 49$

$51237 = 512(99) + 5(99) + 5 + 49$

$51237 = 517(99) + 54$

Therefore,

$\displaystyle \frac{51237}{99000} = \displaystyle \frac{517(99) + 54}{99000}$

$\displaystyle \frac{51237}{99000} = \displaystyle \frac{517(99)}{99000} + \frac{54}{99000}$

$\displaystyle \frac{51237}{99000} = \displaystyle \frac{517}{1000} + \frac{54}{99000}$

$\displaystyle \frac{51237}{99000} = 0.517 + 0.000\overline{54}$

$\displaystyle \frac{51237}{99000} = 0.517\overline{54}$

In particular, notice that the three $0$s in the denominator correspond to a delay of length 3 (the digits $517$), while the $99 = 10^2 - 1$ in the denominator corresponds to the repeating block of length $2$.

These can be confirmed for students who may be reluctant to believe that decimal expansions can be found without a calculator.