Understanding School Mathematics in Terms of Linear Measure and Discrete Real Additive Groups

Abstract

In this chapter, I describe a capstone course developed for secondary teachers. Its content represents my mathematical perspective on school mathematics, aimed at conceptual coherence. It centrally features the real number line, with its geometric and arithmetic structures. It starts with linear measurement, expressed through division with remainder (DwR), which leads directly to place value and modular congruence. Abstract algebra enters through the study of discrete additive groups of real numbers, from which multiplicative arithmetic and commensurability (irrationality) naturally emerge; DwR is the foundation of this development. Brief treatments of polynomial algebra and combinatorics then culminate in Discrete Calculus, the natural generalization of the “pattern generalization” activities in school mathematics. Finally, I present and discuss some problem-solving designs, which are intended to cultivate important mathematical practices in the course.

Appendices

Appendix 1: Analysis of the Extended Usiskin Problem Set

Problems Ar1, Ar2, Ar3, Ar4, in order, lead directly to the following Diophantine equations (“Diophantine” because one seeks (positive) integer solutions):

$$ 1/n+1/m=1/2 $$

(7.1)

$$ 2\left(n+m\right)= nm $$

(7.2)

$$ 2n=m\left(n-2\right) $$

(7.3)

$$ \mathrm{For}\kern0.17em \mathrm{which}\;n>1\;\mathrm{does}\;n-2\;\mathrm{divide}\;2n $$

(7.4)

Version (7.4) is essentially a verbal expression of the Eq. (7.3). Moreover, it is not difficult to see how Eqs. (7.1–7.3) are algebraically equivalent. For example, multiply (7.1) by 2nm to get (7.2); and subtract 2m from (7.2) to get (7.3). Hence, solving any one of them provides solutions to the others.

My students generally preferred to use (7.3) to express m in terms of n:

$$ m=2n/(n-2). $$

(7.5)

They then did numerical experiments to find those n for which 2n/(n − 2) is an integer. (Some students even graphed m in (7.5) as a function of n > 0, and highlighted the integer points on the graph.) The solutions they found were:

$$ \left(n,m\right)=\left(4,4\right),\left(3,6\right),\mathrm{or}\;\left(6,3\right). $$

(7.6)

None of the students tried to work directly with (7.1), which is my preferred approach. Using the symmetric roles of m and n, we can assume that n ≤ m. Then n ≥ 3; otherwise 1/n ≥ 1/2. Also n ≤ 4; otherwise 1/n + 1/m < 1/2. Thus either n = 3 (and so m = 6) or n = 4 (and so m = 4).

Problem R1 corresponds to the equation,

$$ 1/4=\left(1/2\right)\left(1/n+1/m\right), $$

(7.7)

which is (7.1) multiplied by 1/2.

For Problem R3: If one travels distance d at speed v in time t, then: d = vt and t = d/v. Now suppose that one travels distance d at speed v₁ in time t₁, and then returns at speed v₂ in time t₂. What is the average speed for the whole trip? It is

$$ {\displaystyle \begin{array}{c}{v}_{ave}=\left(\mathrm{total}\kern0.17em \mathrm{distance}\right)/\left(\mathrm{total}\kern0.17em \mathrm{time}\right)\\ {}=2d/\left({t}_1+{t}_2\right)\\ {}=\frac{2d}{\frac{d}{v_1}+\frac{d}{v_1}}=\frac{2}{\frac{1}{v_1}+\frac{1}{v_2}}\end{array}} $$

Thus,

$$ \frac{1}{v_{\mathrm{ave}}}=\frac{1}{2}\left(\frac{1}{v_1}+\frac{1}{v_2}\right). $$

In other words, v_ave is the harmonic mean of v₁ and v₂. In problem R2, d would be the work of painting the house, and n and m are the rates at which Nan and her Mom do that job. The rate of doing it together (analogous to average speed) is the harmonic mean of the two rates.

The geometry problems are less obviously related, but they too lead to the same Diophantine equations. In Problem G1, let α(n) denote the (equal) interior angle(s) of a regular n-gon: then it is known that $\alpha (n)=\frac{n-2}{n}\cdot {180}^{{}^{\circ}}$. For some number, say m, of these regular n-gons to fit together to cover the area around a point P, we would need: $m\cdot \frac{n-2}{n}\cdot {180}^{{}^{\circ}}={360}^{{}^{\circ}}$, i.e.,

$$ m\left(n-2\right)=2n,\mathrm{as}\ \mathrm{in}\ \mathbf{Ar3}. $$

(7.8)

For G2 (also framed as the “crossing ladders problem”), consider the diagram:

Using similar triangles we have:

$$ \left(a+b\right)/n=a/2,\mathrm{and} $$

$$ \left(a+b\right)/m=b/2 $$

Adding these equations, and then dividing by a + b, gives

$$ 1/n+1/m=1/2,\mathrm{as}\ \mathrm{in}\ \mathbf{Ar1}. $$

(7.9)

For G3, consider the diagram:

The big triangle and the one above the square are similar (corresponding sides are parallel), and so h/b = (h − 2)/2, whence, multiplying this by 2b, the equation

$$ 2h=b\left(h-2\right),\mathrm{as}\ \mathrm{in}\ \mathbf{Ar3}. $$

(7.10)

In Al1, if we formally factor p:

p(x) = x ² − sx + 2s = (x − n)(x − m)…(n, m being integers)

we find that

$$ n+m=s,\mathrm{and} $$

$$ nm=2s $$

whence n and m are positive, since s is, and so we have the equation

$$ nm=2\left(n+m\right),\mathrm{as}\ \mathrm{in}\ \mathbf{Ar2}. $$

(7.11)

Then s (=n + m) = 8 (=4 + 4) or 9 (=3 + 6).

In Al2, the mathematics is mainly happening in the exponents: (uv)² = u ⁿ = v ^m

We first get, from (uv)² = v ^m, that u ² = v ^m − 2, so

$$ v={u}^{2/\left(m-2\right)}. $$

Then, substituting for v in (uv)² = u ⁿ gives: (u ∙ u ^{2/(m − 2)}) = u ⁿ

Equating exponents then gives:

$$ n=2\left[1+2/\left(m-2\right)\right]=2m/\left(m\hbox{--} 2\right), $$

whence, again, equation

$$ 2m=n\left(m-2\right). $$

(7.12)

In Al3, the conditions on (r, b) are that, r + b = rb/2. Dividing this by rb gives,

$$ 1/b+1/r=1/2, \mathrm{as}\ \mathrm{in}\ \mathbf{Ar1}. $$

(7.13)

Appendix 2: Group Theoretic Derivation of Properties of gcd and lcm

Note that all of what follows precedes, and does not depend on, prime factorization.

(DM0): Let a and b be real numbers. Then we have proved that:

   ℤa + ℤb is discrete ⇔ a and b are commensurable

In what follows we shall assume that a and b are commensurable, unless the contrary is indicated. In this case we define

d =  gcd (a, b) ≥ 0 and m =  lcm (a, b) ≥ 0

by

ℤa + ℤb = ℤd and ℤa ∩ ℤb = ℤm

Note that a and b are integers if and only if d is an integer. When d = 1, we say that a and b are “relatively prime.” We say that a fraction a/b is “reduced” if gcd(a, b) = 1.

To ease writing we shall here abbreviate:

(a, b) =  gcd (a, b) and [a, b] =  lcm (a, b)

We now record some basic properties. We omit proofs if they follow easily from the definitions. We have bolded the items that are especially important and/or useful.

(DM1): There exist integers r, s such that d(=(a, b)) = ra + sb.

(DM2): (d) d | a and d | b. If d ^′ | a and d ^′ | b, then d ^′ | d

   “d is the greatest common divisor of a and b”

   (m) a | m and b | m. If a | m ^′ and b | m ^′, then m | m ^′

   “m is the least common multiple of a and b”

(DM3): Suppose that a | A and b | B. Then (a, b) | (A, B) and [a, b] | [A, B].

Put d = (a, b) and D = (A, B). Then d | a and a | A, so d | A. Similarly d | B, and so d | D, by (DM3(a)). In similar fashion one shows that [a, b] | [A, B].

(DM4): (a, b) = (b, a) = (| a| , | b| ), and

  [a, b] = [b, a] = [| a| , | b| ] “Absolute Symmetry”

(DM5): (a, 0) =  | a| and [a, 0] = 0.

(DM6): For any real number c, (ac, bc) = (a, b) · | c|, and

   [ ac,bc] = [a, b] ·  | c| “Multiplicative scaling”

This follows from the easily verified relations:

$$ \mathbb{Z} ac+\mathbb{Z} bc=\left(\mathbb{Z}a+\mathbb{Z}b\right)\cdot \mathrm{c},\mathrm{and} $$

$$ \mathbb{Z} ac\cap \mathbb{Z} bc=\left(\mathbb{Z}a\cap \mathbb{Z}b\right)\cdot c $$

Multiplicative scaling is a very useful property. For example, when a and b are commensurable, we know that there is a nonzero number c such that ca and cb are integers. Then, for example, (a, b) =  | c ⁻¹ | (ca, cb), so this reduces the calculation of (a, b) to the case of integers. Similarly for [a, b].

(DM7): Let d = (a, b) and a = a ^′ d and b = b ^′ d. Then (a ^′, b ^′) = 1.

In fact, d = (a, b) = (a ^′ d, b ^′ d) = (a ^′, b ^′) · d, by (DM6), and so (a ^′, b ^′) = 1.

(DM8): If (a, b) = 1 then [a, b] =  | a · b|

Proof.

Write 1 = ra + sb, r, s ∈ ℤ. Let m ^′ be a common multiple of a and b: m ^′ = ua = vb, with u, v ∈ ℤ. Then m ^′ = m ^′ ra + m ^′ sb = vbra + uasb = (vr + us)ab, so ab | m ^′. Since ab is visibly a common multiple of a and b, it follows that |ab |  = [a, b].

(DM9): |a · b |  = (a, b) · [a, b].

Proof.

m = [a ^′ d, b ^′ d] = [a ^′, b ^′] · d by (DM7)

$= \mid {a}^\prime \cdot {b}^\prime \mid \cdot \, d \, by \, (\textrm{DM8 \, and \, 9})$

so

$$ d\cdot m=\mid {a}^{\prime}\cdot {b}^{\prime}\cdot {d}^2\mid =\mid {a}^{\prime }d\cdot {b}^{\prime }d\mid =\mid a\cdot b\mid . $$

(DM10): If (a, b) = 1 = (a, c), then (a, bc) = 1.

Proof.

Note that a, b, c ∈ ℤ. Write 1 = ra + sb = ua + vc, with r, s, u, v ∈ ℤ. Then

$$ 1=\left( ra+ sb\right)\left( ua+ vc\right)=\left( rua+ rvc+ sb u\right)a+(sv) bc. $$

(DM11): Given a₁ a₂, …, a_n and b₁ b₂, …, b_m with

    (a_i b_j) = 1 for 1 ≤ i ≤ n and 1 ≤ j ≤ m,

   it follows that (a₁ a₂…a_n, b₁ b₂…b_m) = 1.

This follows from (DM10) by induction on max(n, m), as follows. In the case n = m = 1 there is nothing to prove. Suppose that m ≥ 2. Then, by induction, we have that

(a₁ a₂…a_n, b₁ b₂…b_m − 1) = 1 = (a₁ a₂…a_n, b_m)

and so the result follows from (DM11).

(DM12): Suppose that (a, b) = 1 = (c, d). Then (ad, bc) = (a, c) ⋅ (b, d).

Proof.

Let =ℤad + ℤbc. We want to show that:

$$ A=\left(\mathrm{\mathbb{Z}}a+\mathrm{\mathbb{Z}}c\right)\cdot \left(\mathrm{\mathbb{Z}}b+\mathrm{\mathbb{Z}}d\right)=\mathrm{\mathbb{Z}} ab+\mathrm{\mathbb{Z}} ad+\mathrm{\mathbb{Z}} bc+\mathrm{\mathbb{Z}} bd $$

Clearly the right side contains the left side. For the reverse inclusion, we must show that ab, bd ∈ A. Write 1 = ra + sb and 1 = uc + vd with r, s, u, v ∈ ℤ. Then

$$ ab= abuc+ abvd=(au)(bc)+(bv)(ad)\in A. $$

Similarly, bd ∈ A.

(DM13): Suppose that a/b and c/d are reduced fractions.

    ${\displaystyle \begin{array}{l}\text{Then}\ \left(\boldsymbol{a}/\boldsymbol{b},\boldsymbol{c}/\boldsymbol{d}\right)=\left(\boldsymbol{a},\boldsymbol{c}\right)/\left[\boldsymbol{b},\boldsymbol{d}\right]\\ {}\qquad\ \ \ \qquad \qquad =\mathbf{\gcd}\left(\mathbf{numerators}\right)/\mathbf{\operatorname{lcm}}\left(\mathbf{denominators}\right)\end{array}}$.

Proof.

|bd | (a/b, c/d) = (ad, bc) by (DM7)

           = (a, c) ⋅ (b, d) by (DM13)

so

$$ \left(a/b,c/d\right)=\left(a,c\right)\cdot \left(b,d\right)/\mid b\cdot d\mid $$

                = (a, c) ⋅ (b, d)/(b, d) ⋅ [b, d] by (DM9)

$$ \ \ \ =\left(a,c\right)/\left[b,d\right] $$

For the next items, we shall use the following notation: If a and b are real numbers we shall write: ℤ(a, b) = ℤa + ℤb. If a and b are commensurable, then ℤ(a, b) is a discrete additive group, generated by (a, b). If a and b are incommensurable then we have shown that ℤ(a, b) is dense in ℝ.

(DM14): Let a and b be real numbers, and let t be an integer. Then

$$ \mathrm{\mathbb{Z}}\left(a,b+ ta\right)=\mathrm{\mathbb{Z}}\left(a,b\right) $$

In case a and b are commensurable, it follows that:

   (a, b + ta) = (a, b) “Additive translation”

Proof.

Let A = ℤ(a, b), and B = ℤ(a, b + ta). We want to show that A = B. Since a, (b + ta) ∈ A, it follows that B ⊆ A. Writing b = (b + ta) − ta, we see that a, b ∈ B, and so A ⊆ B. Hence, A = B, as claimed.

Example.

The Fibonacci sequence F_n is defined recursively by: F₀ = 0, F₁ = 1, and, for n ≥ 2, F_n = F_n − 1 + F_n − 2 (i.e., 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, …)

It follows from (DM14) that, for all n ≥ 1, gcd(F_n, F_n + 1) = 1.

(DM15): Given real numbers a ≥ b > 0, put c = r_b(a) < b. Then ℤ(b, c) = ℤ(a, b).

   Hence, if a and b are commensurable, then so also are b and c, and

   (b, c) = (a, b).

Proof.

By DwR, a = qb + r with q an integer and 0 ≤ r = r_b(a) < b. Thus, 0 ≤ r = a − qb = c < b, and it follows then from Additive Translation (DM14) that: ℤb + ℤc = (a − qb) = ℤa + ℤb.

(DM16): The Euclidean Algorithm (EA). Let a and b be commensurable real numbers, not both equal to 0. The Euclidean Algorithm (EA) is an algorithm to produce (a, b) =  gcd (a, b). Without loss of generality (DM5), we can assume that a ≥ b ≥ 0, and we shall then write a₀ = a and a₁ = b. Then there is an integer n = n(a₀, a₁) ≥ 0, and a sequence, a₀ ≥ a₁ > a₂ > … > a_n > a_n + 1 = 0 such that (a_j, a_j + 1) = (a₀, a₁) for all j ≤ n. In particular, (a₀, a₁) = (a_n, a_n + 1) = a_n.

Proof.

If a₁ = 0, we set n = 0, and all is clear. So suppose that a₁ > 0. Suppose that j ≥ 1, and that we have constructed a₀ > a₁ > a₂ > … > a_j > 0.

Then, with the notation (of DM16), we set a_j + 1 = r_aj(a_j − 1) and we have (a_j, a_j + 1) = (a_j − 1, a_j).

We continue in this fashion if a_j + 1 > 0. If a_j + 1 = 0, then we set n = j, and stop. All of the properties above follow from (DM16). The process must stop in a finite number of steps since ℤ(a, b) is uniformly discrete, so it cannot contain an infinite decreasing sequence of positive numbers.

(DM17): Suppose that a > b > 0 are incommensurable real numbers. Then we can still apply the Euclidean Algorithm process, but it won’t stop in finitely many steps. Explicitly, set a₀ = a and a₁ = b. Then we can produce an infinite sequence of positive numbers a₀ > a₁ > a₂ > … > a_n > a_n + 1 > … such that ℤ(a_j, a_j + 1) = ℤ(a₀, a₁) for all j ≥ 0.

This follows inductively from (DM16). We can’t have a_j + 1 = 0 since ℤ(a₀, a₁) is not discrete, in fact it is dense in ℝ. It can further be shown that a_n → 0 as n → ∞.

(DM18): Multiple gcds and lcms. Define commensurability for a sequence (a₁, a₂, …, a_n) of real numbers to mean that ℤa₁ + ℤa₂ + … + ℤa_n is discrete.

Then, as above, we can define gcd(a₁, a₂, …, a_n) and lcm(a₁, a₂, …, a_n) to be the nonnegative generators of ℤa₁ + ℤa₂ + … + ℤa_n and of ℤa₁ ∩ ℤa₂ ∩ … ∩ ℤa_n, respectively. These are clearly symmetric functions of their n variables. Moreover, we have recursive descriptions,

gcd(gcd (a₁, a₂, …, a_n − 1), a_n) =  gcd (a₁, a₂, …, a_n)

lcm(lcm(a₁, a₂, …, a_n − 1), a_n) =  lcm (a₁, a₂, …, a_n).

To simplify writing we shall put

$$ \delta \left({a}_1,{a}_2,\dots, {a}_n\right)=\gcd \left({a}_1,{a}_2,\dots, {a}_n\right),\mathrm{and} $$

$$ \mu \left({a}_1,{a}_2,\dots, {a}_n\right)=\operatorname{lcm}\left({a}_1,{a}_2,\dots, {a}_n\right) $$

Note that the a_j are all integers if and only if δ(a₁, a₂, …, a_n) is an integer. Moreover, δ(a₁, a₂, …, a_n) and μ(a₁, a₂, …, a_n) satisfy the analogue of multiplicative scaling (DM7).

(DM19): Writing d = δ(a₁, a₂, …, a_n) and a_j = a_j ^′ ⋅ d for each j, we have

$$ \delta \left({a_1}^{\prime },{a_2}^{\prime },\dots, {a_n}^{\prime}\right)=1. $$

(DM20): Assume that all a_j ≠ 0. If δ(a₁, a₂, …, a_n) = 1, then $\mu \left(\frac{1}{a_1},\frac{1}{a_2},\dots, \frac{1}{a_n}\right)\break =1$. (i.e., if ℤa₁ + ℤa₂ + … + ℤa_n = ℤ, then ℤa₁ ⁻¹ ∩ ℤa₂ ⁻¹ ∩ … ∩ ℤa_n ⁻¹ = ℤ.)

Proof.

Put G = ℤa₁ ⁻¹ ∩ ℤa₂ ⁻¹ ∩ … ∩ ℤa_n ⁻¹. Clearly G contains ℤ. It remains to show that any a in G is an integer. For each j we can write a = s_j/a_j with s_j an integer. By hypothesis we can write 1 = r₁ a₁ + r₂ a₂ + … + r_n a_n, with all r_j integers. Then

$$ a={r}_1{a}_1a+{r}_2{a}_2a+\dots +{r}_n{a}_na $$

$$ =\frac{r_1{a}_1{s}_1}{a_1}+\frac{r_2{a}_2{s}_2}{a_2}+\dots +\frac{r_n{a}_n{s}_n}{a_n} $$

$$ ={r}_1{s}_1+{r}_2{s}_2+\dots +{r}_n{s}_n\in \mathrm{\mathbb{Z}}. $$

(DM21): Put A = a₁ ⋅ a₂ ⋅ … ⋅ a_n. Then $A=\delta \left({a}_1,{a}_2,\dots, {a}_n\right)\cdot \mu \left(\frac{A}{a_1},\frac{A}{a_2},\dots, \frac{A}{a_n}\right)$.

This is a nice, but nonobvious, generalization of the case n = 2, which is just (DM10): a ⋅ b =  gcd (a, b) ⋅  lcm (a, b).

Proof.

Writing a_j = a_j ^′ d, with the notation of (DM20), we can apply (DM21) to obtain:

$\delta \left({a_1}^{\prime },{a_2}^{\prime },\dots, {a_n}^{\prime}\right)=1=\mu \left(\frac{1}{{a_1}^{\prime }},\frac{1}{{a_2}^{\prime }},\dots, \frac{1}{{a_n}^{\prime }}\right).$

Multiplying the right side by $\frac{A}{d}$, and using multiplicative scaling, we get

$\mu \left(\frac{A}{a_1},\frac{A}{a_2},\dots, \frac{A}{a_n}\right)=\frac{A}{d}$, whence the result.