Integrated Math 1 · Chapter 1 · Functions

Working Problems as a Team

Four students count the same tile border four different ways and write four different-looking expressions. All four are right, and showing that is the whole section.

bookSHelf  ·  Integrated Math 1  ·  §1.1  ·  a self-paced section

bookSHelf · Integrated Math 11 / 41
The tile problem is the vehicle; the real subject is comparing methods.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

Outline — by the end of this section you will be able to

Objectives

  1. Count the tiles in a bordered pattern by breaking the picture into pieces §1.1.2
  2. Write an expression for a center square of any size, not only the size shown Def 1.1.3
  3. Explain why one picture produces different-looking expressions Table 1.1.3
  4. Decide whether two expressions are equivalent, and why expanding beats testing Def 1.1.4
  5. Trace an expression back to the picture it came from, term by term §1.1.4
  6. Present your method, and follow someone else's well enough to locate their pieces §1.1.6
bookSHelf · Integrated Math 12 / 41
The six section SLOs, condensed to one line each.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

§1.1.1 — The cafeteria tile problem

Two questions about one floor

Question A — this floor

How many colored tiles are in the border? Find a method quick enough to trust without counting one tile at a time, and be ready to justify that it counts every tile exactly once.

Question B — every floor

Center squares come in all sizes. Find an expression for the colored border tiles around an NN by NN center, where NN is however many tiles run along one side.

A good method for the 5 by 5 case is one that still makes sense when you cannot see the picture.

123456789
1■□■□■□■□■
2□■□■□■□■□
3■□·····□■
4□■·····■□
5■□·····□■
6□■·····■□
7■□·····□■
8□■□■□■□■□
9■□■□■□■□■

Table 1.1.1: The sample design — a checkerboard border two rows wide around a 5 by 5 center.

bookSHelf · Integrated Math 13 / 41
A is arithmetic you could brute-force. B is not, and that is why A comes first.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

§1.1.1 — Definition

Checkerboard Border

Definition 1.1.1 — Checkerboard Border

A checkerboard border of width two is a ring of tiles around a square block of center tiles. The ring is exactly two tiles thick on every side. It is colored so that no colored tile shares an edge with another colored tile.

Note what that rules out: not two tiles total, and not two tiles on one side only.

A checkerboard border is a ring exactly two tiles thick on every side, colored so that no colored tile touches another colored tile edge to edge A square grid is built up in three stages and everything stays on screen. First a square 5 by 5 block of 25 center tiles appears, tinted blue. Then a first row of plain tiles wraps all the way around it, and then a second row wraps around that, making a 9 by 9 floor; a label reads that the ring is exactly two tiles thick on every side. A bracket under the grid marks 2 border tiles, then 5 center tiles, then 2 border tiles, and reads 2 plus 5 plus 2 equals 9 tiles. Finally the border tiles are colored in a checkerboard, 28 of the 56 in the accent color, and one colored tile near the top left is outlined in blue on all four of its edges to show that each of the four tiles it shares an edge with is white. The closing label reads: no colored tile shares an edge with another colored tile. a ring of tiles around a square block of center tiles a square 5 by 5 block of center tiles row one of the ring row two of the ring, so the ring is exactly two tiles thick on every side 2 5 2 2 + 5 + 2 = 9 tiles no colored tile shares an edge with another colored tile

Definition 1.1.1: a ring exactly two tiles thick on every side, colored so that no colored tile touches another edge to edge.

bookSHelf · Integrated Math 14 / 41
Two tiles thick on EVERY side — the width-two condition does the work later.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

Context Pause — the side grows by four, not by two

The border sits on both sides

Walk across one row: 2 border tiles, then the 5 center tiles, then 2 more border tiles.

2+5+2=9 tiles on a side.2 + 5 + 2 = 9 \text{ tiles on a side.}

The side length grows by 2+2=42 + 2 = 4 in total. Getting this wrong is the single most common way to miss this problem, and the error stays hidden all the way through the generalization.

bookSHelf · Integrated Math 15 / 41
Every method downstream depends on the outer dimension being right.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

§1.1.1 — Worked example

The outer size of the floor

Example 1.1.1 — the center block is 5 by 5 and the border is two rows wide; find the finished floor's dimensions and its total tiles

Step 1 — walk across one row. Cross 2 border tiles, the 5 center tiles, then 2 more. Top to bottom works out identically, so the floor is a 9 by 9 square.

2+5+2=92 + 5 + 2 = 9

Step 2 — count the whole floor. A square 9 tiles on a side holds

92=81 tiles.9^2 = 81 \text{ tiles.}

Answer: the floor is 9 tiles by 9 tiles and holds 81 tiles in all — 9 rows of 9 symbols in Table 1.1.1.

bookSHelf · Integrated Math 16 / 41
Check the arithmetic against the printed grid, not against itself.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

§1.1.2 — Counting the 5 by 5 border four different ways

Four students, four ways to cut one ring

Marisol — subtract the hole

Whole floor minus the center block. Subtraction cannot double-count, but it never touches the border directly. 92−52=569^2 - 5^2 = 56

Jamal — four pinwheel strips

Four equal strips, each 2 by 7, rotating like a pinwheel so every corner lands in exactly one strip. 4(2⋅7)=564(2 \cdot 7) = 56

Priya — sides plus corners

Four 2-by-5 side strips and four 2-by-2 corner blocks, counted as two different kinds of piece. 40+16=5640 + 16 = 56

Wen — long strips and short

Two full-width 2-by-9 strips take the corners; two 2-by-5 strips fill the sides. 18+18+10+10=5618 + 18 + 10 + 10 = 56

The goal is not to pick a favorite. It is to look at the grid and see each of the four decompositions in it.

bookSHelf · Integrated Math 17 / 41
Each method counts the TOTAL border — colored and white together.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

§1.1.2 — The scoreboard

Four sets of arithmetic sharing almost no numbers

StudentHow they saw itArithmeticBorder tiles
Marisolwhole floor minus the center hole92−529^2 - 5^256
Jamalfour equal 2-by-7 pinwheel strips4(2⋅7)4(2 \cdot 7)56
Priyafour 2-by-5 sides plus four 2-by-2 corners4(2⋅5)+4(2⋅2)4(2 \cdot 5) + 4(2 \cdot 2)56
Wentwo 2-by-9 strips plus two 2-by-5 strips2(2⋅9)+2(2⋅5)2(2 \cdot 9) + 2(2 \cdot 5)56

Table 1.1.2: Four students, four decompositions of the same border.

That agreement is not luck. Every method chops up the same 56 tiles, so the answers were forced to match — and nobody had to trust anybody else's arithmetic to believe the total.

bookSHelf · Integrated Math 18 / 41
Four independent routes landing on one number is the evidence.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

§1.1.2 — Worked example

Finding the double count

Example 1.1.2 — a fifth student says the border is four sides, each a 2-by-9 rectangle, giving 4(2⋅9)=724(2 \cdot 9) = 72; find the error and say exactly how many tiles it added

Step 1 — measure the gap. The claimed total is 72 and the correct total is 56, so 72−56=1672 - 56 = 16 tiles were counted that should not have been.

Step 2 — look at a corner. The "top side" rectangle runs the full 9-tile width and contains the top-left 2-by-2 block. The "left side" rectangle runs the full 9-tile height and contains it too. Both claim the same 4 tiles.

Step 3 — count how often. Four corners, 4 tiles each, each counted twice.

4⋅4=16 extra tiles,72−16=56.4 \cdot 4 = 16 \text{ extra tiles,} \qquad 72 - 16 = 56.

Different-looking work is often the same answer seen differently. It is not automatically the same answer — sometimes it is a mistake, and you tell the difference by tracing every piece back to the picture.

bookSHelf · Integrated Math 19 / 41
16 is not a random number: four corners of four tiles each.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

§1.1.2 — Definition

Valid Decomposition

Definition 1.1.2 — Valid Decomposition

A decomposition of a figure is a way of cutting it into pieces. The decomposition is valid when every tile belongs to exactly one piece — none is left out, and none is counted twice.

Adding up the pieces of a valid decomposition always gives the total.

A decomposition is valid when every tile belongs to exactly one piece; four full-length sides fail that test because each corner block belongs to two The question, does every tile belong to exactly one piece, stands over two copies of the same 56-tile border. On the left, Priya cuts it into four 2-by-5 side strips and four 2-by-2 corner blocks; the pieces meet without overlapping, the arithmetic reads 4 times 2 times 5 plus 4 times 2 times 2 equals 40 plus 16 equals 56, and the verdict reads that every tile belongs to exactly one piece. On the right, one more student counting quickly cuts it into four sides, each one a 2-by-9 rectangle, giving 4 times 2 times 9 equals 72. The four 2-by-2 corner blocks then light up in the accent color because each of them lies inside two rectangles at once; the note reads that each 2 by 2 corner block is counted twice, adding 4 times 4 equals 16 extra tiles, and the repair reads 72 minus 16 equals 56. A closing line reads: checking for overlap and checking for gaps are two separate checks. Does every tile belong to exactly one piece? Priya four 2-by-5 sides plus four 2-by-2 corners 4(2 · 5) + 4(2 · 2) = 40 + 16 = 56 every tile belongs to exactly one piece one more student, counting quickly four sides, each one a 2-by-9 rectangle 4(2 · 9) = 72 each 2 by 2 corner block is counted twice 4 · 4 = 16 extra tiles 72 − 16 = 56 Checking for overlap and checking for gaps are two separate checks.

Definition 1.1.2: every tile in exactly one piece, which is the test the four full-length sides fail at the corners.

bookSHelf · Integrated Math 110 / 41
This is the property the 72-tile method lacks.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

§1.1.2 — Your turn

Try It Now 1.1.1

Try It Now 1.1.1 — use Priya's decomposition on the design in Table 1.1.1; say which piece the tile at row 1, column 1 belongs to, then do the same for row 1, column 5

Row 1, column 1 is the outermost corner of the top-left corner block, so it belongs to a corner piece.

Row 1, column 5 is in the middle of the top edge, directly above the center block, so it belongs to the top side strip.

Answer: row 1, column 1 is a corner tile; row 1, column 5 is a side-strip tile.

bookSHelf · Integrated Math 111 / 41
Locating a named tile is the check that a decomposition is understood, not memorized.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

§1.1.3 — Exactly half the border tiles are colored

Every pair splits one and one

Claim: in a checkerboard border two rows wide, exactly half the tiles are colored

Every strip is exactly 2 thick, so it slices into pairs across that thickness. The two tiles in a pair share an edge, and in a checkerboard tiles that share an edge always have different colors — so every pair holds one colored tile and one white one, with no exceptions anywhere in the border.

562=28 colored tiles.\frac{56}{2} = 28 \text{ colored tiles.}

Slicing across the thickness is what makes this work at every size: a side strip 5 tiles tall could not be cut into standing pairs at all, but cut across its 2-tile width it splits perfectly.

bookSHelf · Integrated Math 112 / 41
Answer to Question A: 28 colored tiles.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

Insight — socks come in pairs

You can know the ratio without emptying the drawer

If a drawer holds nothing but pairs, and every pair has one black sock and one white sock, half the drawer is black. You do not need to know how many pairs there are.

The two-tile columns in this border are those pairs — which is why the halving survives a change of size, and a row-by-row count does not.

bookSHelf · Integrated Math 113 / 41
The analogy is doing real work: it isolates what the argument does not depend on.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

Context Pause — one argument replaces four recounts

A proof scales; a count does not

Four students produced four totals for the border. Proving the half-and-half fact once means none of them has to count colored tiles separately — each divides their own total by 2.

One good argument saves four pieces of work, and it keeps saving them at every new size. A count tells you what happened this time; the argument tells you why it had to happen.

bookSHelf · Integrated Math 114 / 41
This is the first appearance of "prove once, apply everywhere" in the course.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

§1.1.3 — Worked example

Confirming 28 by direct count

Example 1.1.3 — count the colored tiles in Table 1.1.1 row by row, and check the result against the halving argument

Step 1 — read the ■ symbols across each row.

5+4+2+2+2+2+2+4+55 + 4 + 2 + 2 + 2 + 2 + 2 + 4 + 5

Step 2 — add them up. Two full border rows top and bottom, five middle rows showing only the left and right edges.

9+10+9=289 + 10 + 9 = 28

The rows are lopsided — the top row has 5 colored tiles and the row below it has 4 — and nothing in the row count explains why they still add to exactly half. The pairing argument does, because it never looks at rows at all.

bookSHelf · Integrated Math 115 / 41
Agreement with 56÷2=2856 \div 2 = 28 is the point; the lopsided rows are the lesson.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

§1.1.3 — Your turn

Try It Now 1.1.2

Try It Now 1.1.2 — a cafeteria has a 7 by 7 center block and the same two-row checkerboard border; (a) use Marisol's method for the total border tiles, (b) find the colored border tiles

(a) Subtract the hole. The border adds 2 at each end of a row, so the floor is 7+4=117 + 4 = 11 on a side.

112−72=121−49=72 border tiles.11^2 - 7^2 = 121 - 49 = 72 \text{ border tiles.}

(b) Halve it. The border is two rows wide and checkerboarded, so it is made entirely of stacked pairs.

722=36 colored tiles.\frac{72}{2} = 36 \text{ colored tiles.}
bookSHelf · Integrated Math 116 / 41
Answers: 72 border tiles, 36 of them colored.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

§1.1.4 — Generalizing to an NN by NN center

Replace the 5 with a letter and change nothing else

Marisol — floor minus hole

The floor is (N+4)(N+4) by (N+4)(N+4), the hole is NN by NN: (N+4)2−N2(N + 4)^2 - N^2

Jamal — four pinwheel strips

Each 2 thick and N+2N + 2 long — the center's NN plus one 2-tile corner: 4⋅2(N+2)4 \cdot 2(N + 2)

Priya — sides plus corners

Four 2-by-NN sides and four 2-by-2 corners: 4(2N)+4(4)4(2N) + 4(4)

Wen — long strips and short

Two 2-by-(N+4)(N+4) strips, two 2-by-NN strips: 2⋅2(N+4)+2⋅2N2 \cdot 2(N + 4) + 2 \cdot 2N

Each student's reasoning carries over word for word — only the numbers turn into letters, and the letters sit exactly where the 5 used to sit. The floor is now N+4N + 4 on a side, which gives 9 at N=5N = 5.

bookSHelf · Integrated Math 117 / 41
These are still TOTAL border tiles; the halving comes on the next table.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

Insight — a recipe, not a meal

A number feeds one household; a rule feeds any

"Two eggs and 300 grams of flour" feeds one household. "Two eggs per 300 grams of flour, scaled to the number of guests" feeds any household.

The number 28 is a meal. The expression you are about to build is the recipe.

bookSHelf · Integrated Math 118 / 41
Keep this distinction available all year — it is what a variable buys.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

§1.1.4 — Definition

Variable

Definition 1.1.3 — Variable

A variable is a letter used to stand for a number whose value is not fixed. Writing an expression with a variable in it describes every case at once, so the expression can be evaluated for any particular value you are handed.

Here NN stands for the side length of the center square in tiles. Writing NN instead of 5 is a promise that nothing in the work depends on the number being 5, and every step from here has to keep that promise.

The first thing to redo is the outer dimension: the center is NN across and the border adds 2 on each side, so the floor is N+4N + 4 tiles on a side. Everything else follows from that.

bookSHelf · Integrated Math 119 / 41
Choosing a letter is a promise, not a decoration.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

Context Pause — check the general answer against the case you already know

Keep one known case in your pocket

You counted 28 colored tiles for the 5 by 5 design before any letters appeared. That number is now a test: any expression you write must produce 28 when N=5N = 5.

Keeping one known case in your pocket catches most generalization errors on the spot — including the "grows by two" mistake, which is invisible in the algebra and obvious at N=5N = 5.

bookSHelf · Integrated Math 120 / 41
Cheap, fast, and catches the common errors — but it cannot confirm equivalence.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

§1.1.4 — Halving for the colored tiles

Four answers to the contractor's question

StudentTotal border tilesColored
Marisol(N+4)2−N2(N+4)^2 - N^2(N+4)2−N22\dfrac{(N+4)^2 - N^2}{2}
Jamal4⋅2(N+2)4 \cdot 2(N+2)4(N+2)4(N + 2)
Priya4(2N)+4(4)4(2N) + 4(4)4N+84N + 8
Wen2⋅2(N+4)+2⋅2N2 \cdot 2(N+4) + 2 \cdot 2N2(N+4)+2N2(N+4) + 2N

Table 1.1.3: Each method's total, and the colored count that is half of it.

Read each term back into the picture:

In 4(N+2)4(N+2), the 44 is four strips and the N+2N+2 is the colored tiles in one strip — one per column of two.

In 4N+84N + 8, the 4N4N is four straight sides at NN colored tiles each, and the 88 is four corner blocks at 2 each.

In 2(N+4)+2N2(N+4) + 2N, the first term is the top and bottom strips and the second is the left and right.

In (N+4)2−N22\dfrac{(N+4)^2 - N^2}{2}, the numerator is the whole floor with the center removed, and the 22 is the halving.

bookSHelf · Integrated Math 121 / 41
Answer to Question B: any row of the right-hand column. None of them is THE answer.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

§1.1.4 — Worked example

Carrying Wen's method into the general case

Example 1.1.4 — rebuild Wen's two-long-strips-and-two-short method for an NN by NN center, write the expression for total border tiles, and check it at N=5N = 5

Step 1 — the long strips. The top strip runs the full width of the floor, now N+4N + 4 tiles, and is still 2 tall: 2(N+4)2(N + 4) tiles. The bottom strip is identical.

Step 2 — the short strips. The long strips already took the corners, so each side strip is only as tall as the center: 2N2N tiles.

Step 3 — assemble and check at N=5N = 5.

2⋅2(N+4)+2⋅2N⟶4(9)+4(5)=36+20=562 \cdot 2(N + 4) + 2 \cdot 2N \qquad\longrightarrow\qquad 4(9) + 4(5) = 36 + 20 = 56

It matches the 56 tiles counted in §1.1.2, so the generalization kept its promise: nothing in the reasoning depended on the number being 5.

bookSHelf · Integrated Math 122 / 41
The check at a known value is the habit, not the answer.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

§1.1.4 — Your turn

Try It Now 1.1.3

Try It Now 1.1.3 — a cafeteria has a 12 by 12 center block and the same border; (a) use Priya's expression for the colored tiles, (b) use Jamal's on the same cafeteria

(a) Priya's 4N+84N + 8, with N=12N = 12.

4(12)+8=48+8=56 colored tiles.4(12) + 8 = 48 + 8 = 56 \text{ colored tiles.}

(b) Jamal's 4(N+2)4(N + 2), same NN.

4(12+2)=4(14)=56 colored tiles.4(12 + 2) = 4(14) = 56 \text{ colored tiles.}

The two agree, which is what we expect from two valid decompositions of one border.

bookSHelf · Integrated Math 123 / 41
Answer: 56 colored tiles by either method.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

§1.1.5 — Showing the expressions are equivalent

Four expressions, no two alike. Are they really equal?

Testing is evidence

Substitute a value into all four and compare. Agreement at two values does not settle it — the contractor needs a formula that works for every NN, including sizes nobody tested.

Expanding is proof

Rewrite each expression in the same form using the properties from §0.5, and see whether the forms coincide. They do — every one of them is 4N+84N + 8.

(N+4)2−N22  =  4(N+2)  =  4N+8  =  2(N+4)+2N\frac{(N+4)^2 - N^2}{2} \;=\; 4(N+2) \;=\; 4N + 8 \;=\; 2(N+4) + 2N

Now the claim is settled for every NN at once, not for the two values we happened to try. That is the difference between checking and proving.

bookSHelf · Integrated Math 124 / 41
The next four slides are the evidence and then the proof, in that order.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

§1.1.5 — Testing a value is evidence, not proof

All four at N=5N = 5

ExpressionSubstitutedValue
(N+4)2−N22\dfrac{(N+4)^2 - N^2}{2}92−522=81−252\dfrac{9^2 - 5^2}{2} = \dfrac{81 - 25}{2}28
4(N+2)4(N+2)4(7)4(7)28
4N+84N + 820+820 + 828
2(N+4)+2N2(N+4) + 2N2(9)+10=18+102(9) + 10 = 18 + 1028

Table 1.1.4: All four expressions evaluated at N=5N = 5.

All 28 — and matching the tile count we made by hand before any letters appeared. That is worth having, and it is not yet a proof.

bookSHelf · Integrated Math 125 / 41
One value agreeing rules nothing out; one value disagreeing would rule everything out.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

§1.1.5 — A second value

The same four at N=10N = 10

ExpressionSubstitutedValue
(N+4)2−N22\dfrac{(N+4)^2 - N^2}{2}196−1002=962\dfrac{196 - 100}{2} = \dfrac{96}{2}48
4(N+2)4(N+2)4(12)4(12)48
4N+84N + 840+840 + 848
2(N+4)+2N2(N+4) + 2N2(14)+20=28+202(14) + 20 = 28 + 2048

Table 1.1.5: The same four expressions evaluated at N=10N = 10.

Agreement again — and a disagreement here would have proved immediately that something was wrong. But agreement at two values still says nothing about the third.

bookSHelf · Integrated Math 126 / 41
Two data points, an unbounded claim.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

Context Pause — testing can refute, but it cannot confirm

The asymmetry runs through the whole course

One value where two expressions disagree proves they are not the same, and that is worth a lot.

A hundred values where they agree proves nothing about the hundred-and-first. "But I plugged in a number and it worked" is a common way to stay wrong.

bookSHelf · Integrated Math 127 / 41
Counterexamples are cheap; confirmations are not available at all by testing.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

§1.1.5 — Expanding settles it

Marisol's, step by step

StepExpressionReason
1(N+4)2−N22\dfrac{(N+4)^2 - N^2}{2}starting expression
2N2+8N+16−N22\dfrac{N^2 + 8N + 16 - N^2}{2}expand (N+4)2(N+4)^2
38N+162\dfrac{8N + 16}{2}N2−N2=0N^2 - N^2 = 0
44N+84N + 8divide each term by 2

Table 1.1.6: Rewriting Marisol's expression step by step.

Jamal's takes one line — 4(N+2)=4N+84(N + 2) = 4N + 8 by the distributive property — and Priya's is already there.

bookSHelf · Integrated Math 128 / 41
Every step names a property from §0.5.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

§1.1.5 — Expanding settles it

Wen's, step by step

StepExpressionReason
12(N+4)+2N2(N + 4) + 2Nstarting expression
22N+8+2N2N + 8 + 2Ndistributive property
32N+2N+82N + 2N + 8commutative property of addition
44N+84N + 8combine like terms

Table 1.1.7: Rewriting Wen's expression step by step.

Four routes in, one form out. The four expressions were never in competition — they were four descriptions of one ring of tiles.

bookSHelf · Integrated Math 129 / 41
Landing on the same form is what "equivalent" means operationally.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

Insight — same word, four languages

Translate all four and they turn out to be one word

Four people say a word in four languages and it sounds like four different words. Translate all four into one shared language and they turn out to be the same word.

Expanding is that translation, and 4N+84N + 8 is the shared language.

bookSHelf · Integrated Math 130 / 41
Simplified form is where comparisons happen because it is the common tongue.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

§1.1.5 — Definition

Equivalent Expressions

Definition 1.1.4 — Equivalent Expressions

Two expressions are equivalent when they produce the same value for every allowed value of the variable.

You can show two expressions are not equivalent with a single value where they disagree. Showing that they are equivalent requires rewriting one into the other using properties that preserve value.

Four expressions built from four different cuts of the same border all rewrite to 4N plus 8, so they agree at every N, while a fifth that only matched at N equals 5 breaks at N equals 10 Under the question, four expressions, no two alike, are they really equal, four rows appear. Each row starts with a small diagram of the same square ring: Marisol's has no cut lines at all and its centre hole is outlined, Jamal's is cut into four pinwheel strips, Priya's has eight short cuts separating four corners from four sides, and Wen's has two straight horizontal cuts. Beside each diagram is that student's expression for the colored tiles: the quantity N plus 4 squared minus N squared, all over 2; then 4 times the quantity N plus 2; then 4N plus 8; then 2 times the quantity N plus 4, plus 2N. Four arrows then converge from the four expressions onto a single large 4N plus 8 in the accent color. Below a rule, a counterexample: one classmate writes 4N plus 8 and another writes N squared plus 3, and both give 28 when N is 5; taking N as 10, a cafeteria size nobody in the room checked, gives 48 and 103. The closing line reads: equivalence means matching at every allowed value, so a single mismatch rules it out permanently. Four expressions, no two alike. Are they really equal? Marisol Jamal Priya Wen (N + 4)2 − N2 2 4(N + 2) 4N + 8 2(N + 4) + 2N 4N + 8 One classmate writes 4N + 8 and another writes N2 + 3. Both give 28 when N = 5. Take N = 10, a cafeteria size nobody in the room checked. 4(10) + 8 = 48 and 102 + 3 = 103 Equivalence means matching at every allowed value, so a single mismatch rules it out permanently.

Definition 1.1.4: four different cuts, four different-looking expressions, one form they all rewrite to.

bookSHelf · Integrated Math 131 / 41
The two halves of the definition are not symmetric, and that asymmetry is the point.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

§1.1.5 — Your turn

Try It Now 1.1.4

Try It Now 1.1.4 — one classmate writes 4N+84N + 8 and another writes N2+3N^2 + 3; both give 28 at N=5N = 5, so show the two are still not equivalent

Confirm the agreement first.

4(5)+8=28and52+3=284(5) + 8 = 28 \qquad \text{and} \qquad 5^2 + 3 = 28

Try a second value — N=10N = 10, a cafeteria size nobody checked.

4(10)+8=48and102+3=1034(10) + 8 = 48 \qquad \text{and} \qquad 10^2 + 3 = 103

One disagreement is enough, permanently. The agreement at N=5N = 5 was a coincidence — which is exactly why testing can never confirm equivalence on its own.

bookSHelf · Integrated Math 132 / 41
Answer: they agree at 5 but give 48 and 103 at 10, so they are not equivalent.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

§1.1.5 — Definition

Simplified Form

Definition 1.1.5 — Simplified Form

The simplified form of an expression is the version with the fewest terms and no parentheses left. It is the form two equivalent expressions land on when each is rewritten as far as it will go, which is why it is the natural place to compare them.

But do not mistake "simplified" for "best." Different forms of the same expression are good at different jobs.

Three equal forms of one expression, each one good at a different job: 4N plus 8 is the tidiest to compute with, 4 times the quantity N plus 2 says four strips of the same size, and the fraction form is the easiest to adapt if the border width changes Under the line, different forms of the same expression are good at different jobs, three forms are stacked. The first, 4N plus 8, is tagged fewest terms, no parentheses left; a bracket under 4N is labelled the four straight sides and a bracket under the 8 is labelled the four corner blocks, and its job reads the tidiest to compute with. The second, 4 times the quantity N plus 2, has a bracket under the 4 labelled four strips and a bracket under the parentheses labelled the colored tiles in one strip; its job reads the one that most obviously says four strips of the same size. The third is the quantity N plus 4 squared minus N squared, all over 2; its job reads the ugliest and the easiest to adapt if the border width changes from 2 to something else. A closing line reads: choosing a form on purpose is a skill, and simplifying on reflex is not. Different forms of the same expression are good at different jobs. fewest terms, no parentheses left 4N + 8 the four straight sides the four corner blocks the tidiest to compute with 4 (N + 2) four strips the colored tiles in one strip the one that most obviously says four strips of the same size (N + 4)2 − N2 2 the ugliest and the easiest to adapt if the border width changes from 2 to something else Choosing a form on purpose is a skill, and simplifying on reflex is not.

Definition 1.1.5: the fewest terms and no parentheses left, which is not the same thing as the best form for the job.

4N+84N + 8 is tidiest to compute with; 4(N+2)4(N+2) most obviously says four strips of the same size, which matters if you are cutting tile; Marisol's fraction is ugliest and the easiest to adapt if the border width changes. Choosing a form on purpose is a skill; simplifying on reflex is not.

bookSHelf · Integrated Math 133 / 41
Simplified is a destination, not a verdict.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

§1.1.5 — Your turn

Try It Now 1.1.5

Try It Now 1.1.5 — a contractor's cafeteria has a 30 by 30 center; (a) which form would you compute with, (b) which form tells the cutter how many strips of what length, (c) compute the colored tiles

(a) For computing — Priya's 4N+84N + 8: the simplified form, one multiplication and one addition, no parentheses to track.

(b) For cutting — Jamal's 4(N+2)4(N + 2): the 44 says four strips, the N+2N + 2 says each carries that many colored tiles. The structure of the expression is the instruction.

(c) The count, with N=30N = 30.

4(30)+8=128and4(30+2)=1284(30) + 8 = 128 \qquad \text{and} \qquad 4(30 + 2) = 128
bookSHelf · Integrated Math 134 / 41
Answer: 128 colored tiles, either way.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

§1.1.6 — How to work a problem as a team

Explaining, listening, comparing

Explaining

You are not reporting the answer — everyone gets the answer. Say what you split the picture into, why nothing was double-counted or missed, and where each number came from. The middle one is the item people skip, and it is what caught the 72-tile error.

Listening

Set down the picture in your head and pick up theirs. The test: can you point at the grid and show their pieces? "Where does the 7 come from?" is always fair and never rude.

Comparing

Test both at a value. If they agree, expand both. If the forms still differ, find the disagreement in the picture — somebody's pieces overlap or leave a gap.

The stance that works: your method and my method are two accounts of the same picture, so they should agree; let's find out whether they do, and if they don't, let's find out where.

bookSHelf · Integrated Math 135 / 41
Step 1 rules out; step 2 settles. Getting the order backwards is how people stay wrong.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

§1.1.6 — Worked example

Running the comparison routine

Example 1.1.5 — your partner hands you 2(2N+4)2(2N + 4) for the colored border tiles and you wrote 4N+84N + 8; work through the three steps and report what you find

Step 1 — test both at N=5N = 5, where the answer is already known to be 28.

2(2⋅5+4)=2(14)=28and4(5)+8=282(2 \cdot 5 + 4) = 2(14) = 28 \qquad \text{and} \qquad 4(5) + 8 = 28

Step 2 — expand both. Distribute the outside 2; yours is already simplified. 2(2N+4)=4N+82(2N + 4) = 4N + 8

Step 3 — compare the forms. Both are 4N+84N + 8, so they are equivalent and step 3 of the routine is not needed. Nobody made an error.

Read it back into the picture: 2(2N+4)2(2N+4) says take one long side plus one short side, holding 2N+42N + 4 colored tiles together, and double it because the border has two of each. A fifth valid way of seeing the same ring.

bookSHelf · Integrated Math 136 / 41
Agreement at step 1 rules nothing out — go to step 2 anyway.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

§1.1.6 — Your turn

Try It Now 1.1.6

Try It Now 1.1.6 — a classmate says "I counted the border of the 5 by 5 design and got 28." (a) why is that a weak explanation, in terms of the three parts? (b) rewrite it strongly for one specific cut, with its expression; (c) their partner got 32 for the same design — what should the group do first, and why can testing a value not settle it?

(a) It reports an answer and nothing else — no pieces and their dimensions, no reason nothing was double-counted or missed, no expression to check against another. Everyone gets 28, so the number is the least informative thing that could have been said.

(b) For instance, four straight sides 2 tiles thick and 5 tiles long give 4(2⋅5)=404(2 \cdot 5) = 40, and the four 2-by-2 corner blocks add 4(4)=164(4) = 16; every tile is in exactly one piece because the sides stop where the corners begin. Expression: 4(2N)+4(4)4(2N) + 4(4), which at N=5N = 5 gives 56 border tiles, 28 of them colored.

(c) Do not test another value. 28 and 32 already disagree at N=5N = 5, so a test has said an error exists without saying where. Have the person who got 32 explain their pieces and dimensions while the others point at the grid — the error is a tile counted twice or missed, and it lives in a specific piece that only the explanation exposes.

bookSHelf · Integrated Math 137 / 41
Testing rules a method out; only an explanation locates the error.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

§1.1.6 — In groups of three or four

Run the routine on the tile floor

Collaborative Exercise

Take a fresh copy of the design in Table 1.1.1 and have each person count the border a different way, without talking to each other.

Then take turns. Each person explains their pieces using the three parts listed above, while the others point at the grid and locate those pieces. Anyone may ask "where does that number come from?" at any time.

Finish by writing all of your expressions in a row and expanding each one. If any two do not land on the same simplified form, find the tile that was counted twice or the tile that was missed.

bookSHelf · Integrated Math 138 / 41
Open-ended in-class activity — no answer to unveil.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

Key Terminology — the five words this section defines

Key terms

checkerboard border — a ring of tiles two tiles thick on every side, colored so that no colored tile shares an edge with another colored tile.

decomposition — a way of cutting a figure into pieces; it is valid when every tile belongs to exactly one piece.

variable — a letter used to stand for a number whose value is not fixed.

equivalent expressions — two expressions that produce the same value for every allowed value of the variable.

simplified form — the version of an expression with the fewest terms and no parentheses left.

bookSHelf · Integrated Math 139 / 41
Second column reveals on click.
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

The headline result

Every valid decomposition of the border lands on 4N+84N + 8

Four students cut the same ring four ways and wrote four expressions that share almost no numbers. Expanded, all four are one expression — and it answers the contractor's question at every size, not just at 5.

(N+4)2−N22  =  4(N+2)  =  4N+8  =  2(N+4)+2N\frac{(N+4)^2 - N^2}{2} \;=\; 4(N+2) \;=\; 4N + 8 \;=\; 2(N+4) + 2N

† The 5 by 5 design gave 28 colored tiles, which is a fact about one cafeteria. 4N+84N + 8 is a fact about every cafeteria, and it is worth incomparably more.

bookSHelf · Integrated Math 140 / 41
One headline claim; everything else in the section is support for it.
1.1
Working Problems as a Team · bookSHelf Integrated Math 1§1.1

§1.1 — Conclusions

What to carry forward

The one idea

Describe a pattern in a way that no longer depends on the size you were shown. Different-looking expressions built from different cuts of one picture are usually the same expression, and expanding both is what proves it. Every term should trace back to tiles you can point at.

Where it goes wrong

Growing the side by 2 instead of 4; four full-length sides that double-count all four corners; treating "it worked at N=5N = 5" as proof; and simplifying on reflex when the unsimplified form was the one that told the cutter what to do.

Next: §1.2 makes generalizing a procedure you run on a table of values, rather than an insight you wait for — and §1.3 names what you built here a function. Back to start.

bookSHelf · Integrated Math 141 / 41
The relationship between N and 4N + 8 is a function; §1.3 supplies the name.