1.1 Working Problems as a Team
SLO F-IF
Interpreting Functions. Understand what a function is and use function notation;
Learning Objectives
By the end of this section, you will be able to:
- count the tiles in a bordered pattern by breaking the picture into pieces that are quick to count;
- write an algebraic expression for a center square of any size, not only the size you were shown;
- explain why two people who see the same picture differently write different-looking expressions for the same quantity;
- decide whether two expressions are equivalent, and explain why expanding both beats testing one number;
- trace an expression back to the picture it came from, term by term;
- present your own method to a partner, and follow someone else's well enough to say which piece of the picture each term counts.
Most of the arithmetic you have done up to now had one method and one answer. Somebody showed you how to divide fractions, you divided fractions, and the only question afterward was whether you got the right number. Algebra does not work that way. This section exists to show you that on the first day rather than the fortieth.
Here is the shift. In algebra a problem usually has one answer and many correct routes to it. Two people can look at exactly the same picture, count it in completely different ways, write down two expressions that do not resemble each other at all, and both be right. That is not a loophole or a technicality. It is the central fact of the subject, and it is the reason algebra is worth learning. An expression records how you saw something, and there is more than one honest way to see almost anything.
That fact has a consequence for how this class runs. If a classmate's answer looks nothing like yours, the useful first question is not "which of us is wrong?" It is "what were you looking at?"
Every later chapter asks you to rewrite an expression, and every rewriting raises the same question: is this still the same quantity? Learning to compare two people's work is the first version of that skill. You will use it on equations in Chapter 3 and on exponents in §1.5.
This section works one problem, the checkerboard border, all the way through. You will count a specific border, count it four different ways, turn each of those four ways into an algebraic expression, and then prove that all four expressions are equal even though no two of them look alike. Section 1.2 takes that generalizing move and makes a routine out of it. Section 1.3 gives a name — function — to the relationship between the size of the square and the number of tiles it needs.
1.1.1 The Cafeteria Tile Problem
Before school starts, the school administration plans to replace the tile in the cafeteria. They want a checkerboard border two rows wide running around the outside of a square block of plain center tiles, as a surround for the tables and serving carts.
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.
Definition 1.1.1 — Checkerboard Border: a ring exactly two tiles thick on every side, colored so that no colored tile touches another edge to edge.
Note what that rules out: not two tiles total, and not two tiles on one side only.
The administration's sample design surrounds a square 5 by 5 block of center tiles. Here is that design drawn as a grid, with ■ for a colored border tile, □ for a white border tile, and · for one of the 25 plain center tiles.
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | |
|---|---|---|---|---|---|---|---|---|---|
| 1 | ■ | □ | ■ | □ | ■ | □ | ■ | □ | ■ |
| 2 | □ | ■ | □ | ■ | □ | ■ | □ | ■ | □ |
| 3 | ■ | □ | · | · | · | · | · | □ | ■ |
| 4 | □ | ■ | · | · | · | · | · | ■ | □ |
| 5 | ■ | □ | · | · | · | · | · | □ | ■ |
| 6 | □ | ■ | · | · | · | · | · | ■ | □ |
| 7 | ■ | □ | · | · | · | · | · | □ | ■ |
| 8 | □ | ■ | □ | ■ | □ | ■ | □ | ■ | □ |
| 9 | ■ | □ | ■ | □ | ■ | □ | ■ | □ | ■ |
Two questions come out of this design, and they are the two halves of the section.
Question A. How many colored tiles are in the border of this design? Find a way of calculating it that is quick and efficient — quick enough that you would trust it without counting one tile at a time — and be ready to justify why your method counts every border tile exactly once.
Question B. The contractor hired to lay the tile does not want to re-count for every cafeteria. Center squares come in all sizes. Find an expression for the number of colored border tiles needed around a square center of any dimensions — an \(N\) by \(N\) center, where \(N\) stands for however many tiles run along one side.
Question A is arithmetic you could brute-force. Question B is not, and that is the whole point of asking A first. The method you invent for A is what gets generalized in B, so a method that is only "count them all one by one" generalizes into nothing. A good method for the 5 by 5 case is one that still makes sense when you cannot see the picture.
The size of the whole floor
Before counting anything, get the outer dimensions straight, because every method depends on them.
The center is 5 tiles across. The border adds 2 tiles on the left and 2 tiles on the right, so the full width is
$$2 + 5 + 2 = 9 \text{ tiles.}$$The same is true top to bottom, so the finished floor is a 9 by 9 square. That is why the grid in Table 1.1.1 has 9 rows and 9 columns.
The border is two rows wide, and there is a border on both sides, so the side length grows by \(2 + 2 = 4\) 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.
The center block is 5 tiles by 5 tiles and the border is two rows wide. Find the dimensions of the finished floor, and find how many tiles the whole floor holds.
Solution
Step 1 — Walk across one row. Start at the left edge and move right. You cross 2 border tiles, then the 5 center tiles, then 2 more border tiles on the far side.
$$2 + 5 + 2 = 9$$The floor is 9 tiles across.
Step 2 — Do the same top to bottom. The center block is square and the border is the same thickness on every side, so the height works out identically.
$$2 + 5 + 2 = 9$$The floor is 9 tiles tall, so the finished floor is a 9 by 9 square.
Step 3 — Count the whole floor. A square that is 9 tiles on a side holds
$$9^2 = 81 \text{ tiles.}$$Check it against the grid in Table 1.1.1: 9 rows of 9 symbols each.
Answer: the floor is 9 tiles by 9 tiles, and it holds 81 tiles in all.
1.1.2 Counting the 5 by 5 Border Four Different Ways
Below are four methods. They were produced by four students looking at the same grid, and no two of them saw the same shapes in it. Each method counts the total number of border tiles — colored and white together. Once we have that total, one short argument in §1.1.3 turns it into the number of colored tiles.
Read all four. The goal is not to pick a favorite; it is to be able to look at the grid and see each of the four decompositions in it.
Marisol: subtract the hole
Marisol did not look at the border at all. She looked at the whole floor and removed what was not border.
The whole floor is 9 by 9, so it holds \(9^2 = 81\) tiles. The center block is 5 by 5, so it holds \(5^2 = 25\) tiles. Every tile is either a center tile or a border tile and never both, so
$$81 - 25 = 56 \text{ border tiles.}$$Her method has a nice feature: it is impossible to double-count, because subtraction cannot count anything twice. Its cost is that it never touches the border directly, so it gives you no picture of how the border is shaped.
Jamal: four strips, pinwheel style
Jamal cut the border into four rectangles of equal size by rotating around the square, the way a pinwheel's blades rotate. Each strip is 2 tiles thick, runs the full length of one side plus one corner block, and stops so that the next strip can start.
Each strip is 2 tiles by 7 tiles, because the center's 5 tiles plus one 2-tile corner block gives \(5 + 2 = 7\). So each strip holds \(2 \cdot 7 = 14\) tiles, and
$$4 \cdot 14 = 56 \text{ border tiles.}$$The pinwheel is worth learning because every corner gets assigned to exactly one strip automatically. Nothing is left out and nothing is counted twice — the arrangement takes care of it.
Priya: four sides plus four corners
Priya separated the straight parts from the corners and counted them as two different kinds of piece.
Along each edge of the center square there is a 2-by-5 strip, holding \(2 \cdot 5 = 10\) tiles. There are four of those, one per side: \(4 \cdot 10 = 40\) tiles.
At each of the four corners there is a 2-by-2 block, holding 4 tiles. There are four of those: \(4 \cdot 4 = 16\) tiles.
$$40 + 16 = 56 \text{ border tiles.}$$This is the method that most closely matches how the border actually looks, and it is also the one that most clearly explains where the corners go. They are their own separate pieces, counted once each.
Wen: two long strips and two short strips
Wen cut horizontally. The top of the border is a rectangle running the full width of the floor: 2 tiles tall and 9 tiles wide, so \(2 \cdot 9 = 18\) tiles. The bottom is the same, another 18.
What is left on the left and right sides is only as tall as the center block, because the top and bottom strips already took the corners. Each of those is 2 tiles wide and 5 tiles tall: \(2 \cdot 5 = 10\) tiles, and there are two of them, so 20.
$$18 + 18 + 10 + 10 = 56 \text{ border tiles.}$$Wen's pieces are not all the same size, which looks less tidy than Jamal's. It is exactly as correct.
The scoreboard
Four descriptions, four sets of arithmetic that share almost no numbers, one answer.
| Student | How they saw it | Arithmetic | Border tiles |
|---|---|---|---|
| Marisol | whole floor minus the center hole | \(9^2 - 5^2\) | \(56\) |
| Jamal | four equal 2-by-7 pinwheel strips | \(4(2 \cdot 7)\) | \(56\) |
| Priya | four 2-by-5 sides plus four 2-by-2 corners | \(4(2 \cdot 5) + 4(2 \cdot 2)\) | \(56\) |
| Wen | two 2-by-9 strips plus two 2-by-5 strips | \(2(2 \cdot 9) + 2(2 \cdot 5)\) | \(56\) |
That agreement is not luck. Every method is a way of chopping up the same 56 tiles, so the answers were forced to match. Nobody in the group had to trust anybody else's arithmetic to believe the total, because four independent routes landed on the same number.
The method that does not work, and why
One more student, counting quickly, said: the border has four sides, each side is a 2-by-9 rectangle, so the total is \(4(2 \cdot 9) = 72\) tiles.
That is 16 too many, and 16 is not a random number.
A student says the border is four sides, each one a 2-by-9 rectangle, giving \(4(2 \cdot 9) = 72\) tiles. The other four methods all gave 56. Find the error, and explain exactly how many tiles it added.
Solution
Step 1 — Measure the gap. The claimed total is 72 and the correct total is 56.
$$72 - 56 = 16$$Sixteen tiles were counted that should not have been.
Step 2 — Look at a corner. Take the top-left corner 2-by-2 block. The "top side" rectangle runs the full 9-tile width, so it contains that corner block. The "left side" rectangle runs the full 9-tile height, so it contains that corner block too. Both rectangles claim the same 4 tiles.
Step 3 — Count how often that happens. The floor has four corners, each a 2-by-2 block holding 4 tiles, and each corner sits in two full-length strips at once. So each corner block gets counted twice instead of once, and each one contributes 4 extra tiles.
$$4 \cdot 4 = 16 \text{ extra tiles.}$$Step 4 — Repair the count. Subtract the overlap from the inflated total.
$$72 - 16 = 56$$That matches all four working methods.
Answer: the four full-length sides overlap at the corners. Each 2-by-2 corner block is counted twice, adding \(4 \cdot 4 = 16\) tiles, and \(72 - 16 = 56\).
Keep that case nearby, because it is the counterexample to a lazy version of this section's message. Different-looking work is often the same answer seen differently. But it is not automatically the same answer. Sometimes it is a mistake, and the way you tell the difference is by tracing every piece back to the picture and asking whether any tile got counted twice or skipped.
A decomposition of a figure is a way of cutting it into pieces. The decomposition is valid when every tile in the figure 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.
Definition 1.1.2 — Valid Decomposition: every tile in exactly one piece, which is the test the four full-length sides fail at the corners.
Use Priya's decomposition on the 5 by 5 design in Table 1.1.1. Point at the tile in row 1, column 1 of the grid and say which piece it belongs to. Then do the same for the tile in row 1, column 5.
Solution
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.
1.1.3 Exactly Half the Border Tiles Are Colored
All four methods counted 56 border tiles, colored and white together. Question A asked only for the colored ones. Rather than redo four counts, we prove one fact and apply it once.
in a checkerboard border two rows wide, exactly half the tiles are colored.
If a drawer holds nothing but pairs, and every pair has one black sock and one white sock, you know half the drawer is black without emptying it. You do not need to know how many pairs there are. The two-tile columns in this border are those pairs.
Look at the border as Jamal did — four strips, each 2 tiles thick. Because every strip is exactly 2 thick, it can be sliced into pairs across that thickness: the top and bottom strips into standing pairs, the left and right strips into lying-down pairs. Either way 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 contains exactly one colored tile and exactly one white tile. There are no exceptions anywhere in the border.
The whole border is nothing but those pairs, laid end to end. Half the tiles in each pair are colored, so half the tiles overall are colored. Notice that slicing across the thickness is what makes this work for every center block: 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.
$$\frac{56}{2} = 28 \text{ colored tiles.}$$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.
The pairing argument is doing something a row-by-row count cannot do. A count tells you what happened this time. The argument tells you why it had to happen, and that is what makes it survive a change of size.
Count the colored tiles in Table 1.1.1 row by row, and check the result against the halving argument.
Solution
Step 1 — Count each row of the grid. Read the ■ symbols across each row in turn.
$$5 + 4 + 2 + 2 + 2 + 2 + 2 + 4 + 5$$Step 2 — Add them up. The first two rows are full border rows, the middle five rows show only the left and right edges, and the last two rows are full again.
$$5 + 4 = 9, \qquad 2 + 2 + 2 + 2 + 2 = 10, \qquad 4 + 5 = 9$$ $$9 + 10 + 9 = 28$$Step 3 — Compare with the halving argument. The total border was 56 tiles, and half of 56 is 28. The two results agree.
Step 4 — Notice what the row count could not tell you. The rows are lopsided: the top row has 5 colored tiles and the row below it has 4. Nothing in the row-by-row count explains why those uneven rows still add to exactly half. The pairing argument does explain it, because it never looks at rows at all — it looks at the stacked pairs, and every pair splits one and one.
Answer: 28 colored tiles, matching \(56 \div 2 = 28\).
Answer to Question A: 28 colored tiles.
A cafeteria has a 7 by 7 block of center tiles and the same two-row checkerboard border.
a) Use Marisol's method to find the total number of border tiles.
b) Find the number of colored border tiles.
Solution
Part (a) — Subtract the hole. The border adds 2 tiles at each end of a row, so the finished floor is \(7 + 4 = 11\) tiles on a side, and the center block is 7 by 7.
$$11^2 - 7^2 = 121 - 49 = 72 \text{ border tiles.}$$Part (b) — Halve it. The border is two rows wide and checkerboarded, so it is made entirely of stacked pairs, one colored and one white.
$$\frac{72}{2} = 36 \text{ colored tiles.}$$Answer: 72 border tiles, 36 of them colored.
1.1.4 Generalizing to an \(N\) by \(N\) Center
Now the contractor's question. Replace the 5 with a letter and carry each student's method through unchanged.
"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.
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 the variable is \(N\), and it stands for the side length of the center square in tiles. Choosing a letter is not a decoration. Writing \(N\) instead of 5 is a promise that nothing in the work that follows depends on the number being 5, and every step from here on has to keep that promise.
The first thing to redo is the outer dimension. The center is \(N\) tiles across, the border adds 2 on each side, so the whole floor is
$$N + 4 \text{ tiles on a side.}$$For \(N = 5\) that gives \(9\), matching the sample design. Everything else follows from that.
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 \(N = 5\). Keeping one known case in your pocket catches most generalization errors on the spot.
The four methods in general
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.
Marisol — whole floor minus the center hole. The floor is \((N+4)\) by \((N+4)\) and the hole is \(N\) by \(N\):
$$(N + 4)^2 - N^2$$Jamal — four pinwheel strips, each 2 thick and \(N + 2\) long (the center's \(N\) tiles plus one 2-tile corner):
$$4 \cdot 2(N + 2)$$Priya — four 2-by-\(N\) sides plus four 2-by-2 corners:
$$4(2N) + 4(4)$$Wen — two strips 2 by \((N+4)\) across the top and bottom, plus two strips 2 by \(N\) on the left and right:
$$2 \cdot 2(N + 4) + 2 \cdot 2N$$Halving for the colored tiles
The half argument in §1.1.3 never used the size of the square, so it applies here word for word. Every column of two in the border holds one colored tile and one white one, and half of each expression is the count of colored tiles.
| Student | Total border tiles | Colored border tiles |
|---|---|---|
| Marisol | \((N+4)^2 - N^2\) | \(\dfrac{(N+4)^2 - N^2}{2}\) |
| Jamal | \(4 \cdot 2(N+2)\) | \(4(N + 2)\) |
| Priya | \(4(2N) + 4(4)\) | \(4N + 8\) |
| Wen | \(2 \cdot 2(N+4) + 2 \cdot 2N\) | \(2(N+4) + 2N\) |
Answer to Question B: any row of that right-hand column. All four are correct answers to the contractor's question, and there is no sense in which one of them is the answer.
Each one is also readable back into the picture, which is the test of whether an expression means anything to you:
- In \(4(N+2)\), the \(4\) is four strips and the \(N+2\) is the colored tiles in one strip — one per column of two, and there are \(N + 2\) columns.
- In \(4N + 8\), the \(4N\) is the four straight sides at \(N\) colored tiles each, and the \(8\) is the four corner blocks at 2 colored tiles each.
- In \(2(N+4) + 2N\), the \(2(N+4)\) is the top and bottom strips and the \(2N\) is the left and right strips.
- In \(\dfrac{(N+4)^2 - N^2}{2}\), the numerator is the whole floor with the center removed and the \(2\) is the halving.
If you can do that — point at a term and name the tiles it counts — you understand the expression. If you cannot, you have memorized a string of symbols.
Wen cut the 5 by 5 border into two 2-by-9 strips across the top and bottom and two 2-by-5 strips on the sides. Rebuild that method for an \(N\) by \(N\) center, write the expression for the total border tiles, and check it at \(N = 5\).
Solution
Step 1 — Redo the long strips. Wen's top strip runs the full width of the floor. The floor is now \(N + 4\) tiles wide, and the strip is still 2 tiles tall.
$$2(N + 4) \text{ tiles in the top strip.}$$The bottom strip is identical, so there are two of them.
Step 2 — Redo the short strips. The left and right strips are only as tall as the center block, because the long strips already took the corners. The center block is \(N\) tiles tall and each strip is 2 tiles wide.
$$2N \text{ tiles in the left strip,}$$and the right strip matches it.
Step 3 — Assemble the expression. Two long strips plus two short strips:
$$2 \cdot 2(N + 4) + 2 \cdot 2N$$Step 4 — Check it at \(N = 5\). Substitute and compare against the 56 border tiles counted in §1.1.2.
$$2 \cdot 2(5 + 4) + 2 \cdot 2(5) = 4(9) + 4(5) = 36 + 20 = 56$$It matches, so the generalization kept its promise: nothing in the reasoning depended on the number being 5.
Answer: \(2 \cdot 2(N + 4) + 2 \cdot 2N\), which gives 56 when \(N = 5\).
A cafeteria has a 12 by 12 block of center tiles and the same two-row checkerboard border.
a) Use Priya's expression to find the number of colored border tiles.
b) Use Jamal's expression on the same cafeteria.
Solution
Part (a) — Priya's expression. Her colored-tile count is \(4N + 8\), and here \(N = 12\).
$$4(12) + 8 = 48 + 8 = 56 \text{ colored tiles.}$$Part (b) — Jamal's expression. His colored-tile count is \(4(N + 2)\), with the same \(N = 12\).
$$4(12 + 2) = 4(14) = 56 \text{ colored tiles.}$$The two agree, which is what we expect from two valid decompositions of one border.
Answer: 56 colored tiles by either method.
1.1.5 Showing the Expressions Are Equivalent
Four expressions, no two alike. Are they really equal?
Testing a value is evidence, not proof
Substitute \(N = 5\) into all four and compare against the 28 we already counted.
| Expression | Substituted | Value |
|---|---|---|
| \(\dfrac{(N+4)^2 - N^2}{2}\) | \(\dfrac{9^2 - 5^2}{2} = \dfrac{81 - 25}{2}\) | \(28\) |
| \(4(N+2)\) | \(4(7)\) | \(28\) |
| \(4N + 8\) | \(20 + 8\) | \(28\) |
| \(2(N+4) + 2N\) | \(2(9) + 10 = 18 + 10\) | \(28\) |
All 28. Try a second value, \(N = 10\):
| Expression | Substituted | Value |
|---|---|---|
| \(\dfrac{(N+4)^2 - N^2}{2}\) | \(\dfrac{196 - 100}{2} = \dfrac{96}{2}\) | \(48\) |
| \(4(N+2)\) | \(4(12)\) | \(48\) |
| \(4N + 8\) | \(40 + 8\) | \(48\) |
| \(2(N+4) + 2N\) | \(2(14) + 20 = 28 + 20\) | \(48\) |
Agreement again — and a disagreement here would have proved immediately that something was wrong. But agreement at two values does not settle the matter: the contractor needs a formula that works for every \(N\), including sizes nobody in the room tested.
Expanding settles it
The proof is to rewrite each expression in the same form using the properties from Section 0.5, and see whether the forms coincide.
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. This asymmetry runs through the whole course.
Marisol's. Expand the square, then subtract, then divide.
| Step | Expression | Reason |
|---|---|---|
| 1 | \(\dfrac{(N+4)^2 - N^2}{2}\) | starting expression |
| 2 | \(\dfrac{N^2 + 8N + 16 - N^2}{2}\) | expand \((N+4)^2\) |
| 3 | \(\dfrac{8N + 16}{2}\) | \(N^2 - N^2 = 0\) |
| 4 | \(4N + 8\) | divide each term by 2 |
Jamal's.
$$4(N + 2) = 4N + 8 \qquad \text{distributive property}$$Priya's. Already there: \(4N + 8\).
Wen's.
| Step | Expression | Reason |
|---|---|---|
| 1 | \(2(N + 4) + 2N\) | starting expression |
| 2 | \(2N + 8 + 2N\) | distributive property |
| 3 | \(2N + 2N + 8\) | commutative property of addition |
| 4 | \(4N + 8\) | combine like terms |
Every one of them is \(4N + 8\).
$$\frac{(N+4)^2 - N^2}{2} \;=\; 4(N+2) \;=\; 4N + 8 \;=\; 2(N+4) + 2N$$Now the claim is settled for every \(N\) at once, not for the two values we happened to try. That is the difference between checking and proving, and it is a distinction this course will keep returning to.
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 \(4N + 8\) is the shared language.
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.
Definition 1.1.4 — Equivalent Expressions: four different cuts, four different-looking expressions, one form they all rewrite to.
One classmate writes \(4N + 8\) for the colored tiles and another writes \(N^2 + 3\). Both give 28 when \(N = 5\). Show that the two expressions are still not equivalent.
Solution
Confirm the agreement first. Substitute \(N = 5\) into each expression.
$$4(5) + 8 = 20 + 8 = 28 \qquad \text{and} \qquad 5^2 + 3 = 25 + 3 = 28$$They do agree there, so that one test rules nothing out.
Try a second value. Take \(N = 10\), a cafeteria size nobody in the room checked.
$$4(10) + 8 = 48 \qquad \text{and} \qquad 10^2 + 3 = 103$$Draw the conclusion. The two expressions disagree at \(N = 10\), and one disagreement is enough. Equivalence means matching at every allowed value, so a single mismatch rules it out permanently. The agreement at \(N = 5\) was a coincidence, which is exactly why testing can never confirm equivalence on its own.
Answer: they agree at \(N = 5\) but give 48 and 103 at \(N = 10\), so they are not equivalent.
The simplified form, and why it is not always the best form
The form \(4N + 8\) has a name worth knowing early.
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.
Definition 1.1.5 — Simplified Form: the fewest terms and no parentheses left, which is not the same thing as the best form for the job.
But do not mistake "simplified" for "best." Priya's \(4N + 8\) is the tidiest to compute with. Jamal's \(4(N+2)\) is the one that most obviously says four strips of the same size, which is more useful if you are the person cutting the tile. Marisol's \(\dfrac{(N+4)^2 - N^2}{2}\) is the ugliest and the easiest to adapt if the border width changes from 2 to something else. Different forms of the same expression are good at different jobs. Choosing a form on purpose is a skill, and simplifying on reflex is not.
A contractor needs to order tile for a cafeteria whose center is 30 by 30, and also wants to know how many equal strips to cut.
a) Which of the four forms would you use to compute the number of colored tiles, and why?
b) Which form tells the cutter how many strips of what length, and why?
c) Compute the number of colored tiles.
Solution
Part (a) — For computing. Use Priya's \(4N + 8\). It is the simplified form, so it has the fewest operations: one multiplication and one addition, with no parentheses to keep track of.
Part (b) — For cutting. Use Jamal's \(4(N + 2)\). The \(4\) says four strips, and the \(N + 2\) says each strip carries \(N + 2\) colored tiles, one for every stacked pair along its length. The structure of the expression is the instruction.
Part (c) — The count. With \(N = 30\):
$$4(30) + 8 = 120 + 8 = 128 \text{ colored tiles.}$$Check with Jamal's form:
$$4(30 + 2) = 4(32) = 128$$Answer: \(4N + 8\) for computing, \(4(N+2)\) for cutting, and 128 colored tiles either way.
1.1.6 How to Work a Problem as a Team
The mathematics above only happened because four people compared four methods. That comparison does not happen by itself, and this last part is about the habits that make it happen.
Explaining a method
When you present your method, you are not reporting the answer. Everyone will get the answer. You are explaining what you looked at.
A weak explanation is "I got 28." A strong one is "I split the border into four strips the same size, each one 2 tiles thick and 7 tiles long, turning like a pinwheel so each strip takes one corner." The second version can be checked, argued with, and generalized. The first cannot be anything but agreed or disagreed with.
Three things belong in every explanation:
- What you split the picture into — the pieces, and their dimensions.
- Why nothing was counted twice or missed — how you know every tile lands in exactly one piece.
- Where each number in your arithmetic came from — which piece of the picture each factor measures.
Item 2 is the one people skip, and it is the one that caught the 72-tile error in §1.1.2.
Listening to a method
Following someone else's method is harder than presenting your own, because you have to set down the picture in your head and pick up theirs. The way to tell whether you have actually done it: can you point at the grid and show their pieces? If you can restate their method in your own words and locate their shapes on the diagram, you have it. If all you can do is nod, you do not.
A question that is always fair, and never rude: "Where does the 7 come from?" Asking someone to connect a number back to the picture is not a challenge to their competence. It is the main way both of you find out whether the method is sound.
Comparing, not competing
When your expression does not match a partner's, work through this in order:
- Test both at a value. Substitute the same number into each. If they give different results, at least one is wrong, and you have a specific case to examine.
- If they agree, expand both. Rewrite each into simplified form. Landing on the same form means they were the same expression wearing different clothes.
- If the forms differ, find the disagreement in the picture. Somebody's pieces overlap or leave a gap. Trace the tiles.
Step 1 can only rule things out; step 2 is what actually settles a question. Getting these in the wrong order is why "but I plugged in a number and it worked" is such a common way to stay wrong.
Your partner hands you \(2(2N + 4)\) as the number of colored border tiles. You wrote \(4N + 8\). Work through the three steps and report what you find.
Solution
Step 1 — Test both at a value. Use \(N = 5\), where the answer is already known to be 28.
$$2(2 \cdot 5 + 4) = 2(10 + 4) = 2(14) = 28$$ $$4(5) + 8 = 20 + 8 = 28$$They agree, so nothing has been ruled out yet. Testing cannot confirm, so go on to step 2.
Step 2 — Expand both. Distribute the outside 2 in your partner's expression.
$$2(2N + 4) = 4N + 8$$Yours is already in simplified form: \(4N + 8\).
Step 3 — Compare the forms. Both are \(4N + 8\), so the two expressions are equivalent and step 3 of the routine is not needed. Nobody made an error.
Read it back into the picture. Your partner's \(2(2N+4)\) says: take one long side plus one short side of the border, which together hold \(2N + 4\) colored tiles, and double it because the border has two of each. That is a fifth valid way of seeing the same ring.
Answer: equivalent. Both expand to \(4N + 8\), and your partner's grouping counts one long strip plus one short strip, doubled.
Two more habits
The stance that works is a plain one: 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. That is a joint investigation rather than a contest, and it produces better mathematics than either of you would produce alone.
In groups of three or four, 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.
A classmate says: "I counted the border of the 5 by 5 design and got 28."
a) Explain why that sentence is a weak explanation, using the three parts listed above.
b) Rewrite it as a strong explanation for one specific way of cutting the border, and give the expression that goes with it.
c) Your classmate's partner got 32 for the same design. Describe what the group should do first, and why testing a value cannot settle it on its own.
Solution
a) What is missing. The sentence reports an answer and nothing else. It names none of the three parts: it does not say what the picture was split into, it does not say why nothing was double-counted or missed, and it gives no expression that could be checked against another. Everyone in the group will get 28, so the number is the least informative thing that could have been said.
b) A strong version. For example, using the four-sides-and-four-corners cut:
"I took the four straight sides, each 2 tiles thick and 5 tiles long, so \(4(2 \cdot 5) = 40\) tiles — but that leaves the four corner blocks out, and each corner is 2 by 2, so that is \(4(4) = 16\) more. Every tile is in exactly one piece because the sides stop where the corners begin."
The expression is \(4(2N) + 4(4)\), which for \(N = 5\) gives \(40 + 16 = 56\) border tiles, of which 28 are colored.
Any of the four cuts in §1.1.2 works here; what makes it strong is naming the pieces, their dimensions, and why they do not overlap.
c) What to do about the 32. Do not start by testing another value — 28 and 32 already disagree at \(N = 5\), so a test has told you there is an error without telling you where it is. Testing can only ever rule a method out.
The group should have the person who got 32 explain their pieces and their dimensions, while the others point at the grid and locate each piece. The error will be a tile counted twice or a tile missed, and it lives in a specific piece — which only the explanation can expose.
Answer: a) it names no pieces, no reason nothing was double-counted, and no expression; b) e.g. \(4(2N) + 4(4)\) with the sides-and-corners description; c) have the disagreeing person explain their pieces against the picture, because a test locates no error even when it finds one.
Everything in this section came out of one move: describing a pattern in a way that no longer depends on the specific size you were shown. The 5 by 5 design gave 28 colored tiles, which is a fact about one cafeteria. The expression \(4N + 8\) is a fact about every cafeteria, and it is worth incomparably more.
Section 1.2 makes that move the routine rather than the finale. You will meet patterns that grow step by step — figures, tables, sequences of drawings — and build the rule from a table of values instead of from a single picture, so that generalizing becomes a procedure you can run rather than an insight you have to wait for.
Section 1.3 supplies the name. The relationship between \(N\) and \(4N + 8\) — every center size producing exactly one tile count — is called a function, and \(4N + 8\) is the rule that computes it. You have already built one. Section 1.3 defines what you built, gives it notation, and starts asking which relationships qualify and which do not.
Problem Set 1.1
Problem 1. A square center block is 6 tiles on a side and gets a checkerboard border two rows wide. Find the dimensions of the finished floor.
Solution
Step 1 — Apply the outer-dimension rule: the border is 2 tiles wide and runs on both sides of the center, so the floor's side length is the center's side length plus \(2 + 2 = 4\).
$$6 + 4 = 10$$Answer: the finished floor is 10 tiles by 10 tiles.
Problem 2. A square center block is 20 tiles on a side and gets a checkerboard border two rows wide. Find the dimensions of the finished floor.
Solution
Step 1 — Apply the same rule: add 4 to the center's side length.
$$20 + 4 = 24$$Answer: the finished floor is 24 tiles by 24 tiles.
Problem 3. Explain in one or two sentences why a two-row border makes the side length of the floor grow by 4 rather than by 2.
Solution
Step 1 — Count the sides of the border, not just its width: the border is 2 tiles wide, but it runs along both the left and right (or top and bottom) of the center, not just one side.
Step 2 — Add the two contributions: each side of the center gains 2 tiles of border, and there are two opposite sides per dimension, so the side length grows by \(2 + 2 = 4\), not by the border's width of 2 alone.
Answer: the side length grows by 4 because a border of width 2 is added on both sides of the center in each direction, not on just one side.
Problem 4. A finished floor is 15 tiles on a side and the border is two rows wide. Find the dimensions of the center block.
Solution
Step 1 — Reverse the outer-dimension rule: the floor's side length is the center's side length plus 4, so the center's side length is the floor's side length minus 4.
$$15 - 4 = 11$$Answer: the center block is 11 tiles by 11 tiles.
Problem 5. Use Marisol's method to find the total number of border tiles around a 6 by 6 center.
Solution
Step 1 — Find the floor size: with a 6 by 6 center and a border 2 wide, the floor is \(6 + 4 = 10\) tiles on a side.
Step 2 — Subtract the hole (Marisol's method): the floor holds \(10^2 = 100\) tiles and the center hole holds \(6^2 = 36\) tiles. Every tile is either center or border, never both, so subtracting removes exactly the center.
$$100 - 36 = 64$$Answer: 64 border tiles.
Problem 6. Use Jamal's method to find the total number of border tiles around a 6 by 6 center, and check your answer against Problem 1.1.5.
Solution
Step 1 — Build one pinwheel strip (Jamal's method): each strip is 2 tiles thick and runs the center's side plus one 2-tile corner block, so its length is \(6 + 2 = 8\).
$$2 \cdot 8 = 16 \text{ tiles per strip}$$Step 2 — Multiply by the four strips: the pinwheel has four strips of equal size, and the rotation assigns every corner to exactly one strip.
$$4 \cdot 16 = 64 \text{ border tiles}$$Step 3 — Check against Problem 1.1.5: Marisol's method also gave 64.
$$64 = 64 \checkmark$$Answer: 64 border tiles, matching Problem 1.1.5.
Problem 7. Use Priya's method to find the total number of border tiles around a 9 by 9 center.
Solution
Step 1 — Count the side strips (Priya's method): each of the four sides is a 2-by-9 strip, holding \(2 \cdot 9 = 18\) tiles, so \(4 \cdot 18 = 72\) tiles from the sides.
Step 2 — Count the corners: each of the four corners is a 2-by-2 block, holding \(4\) tiles, so \(4 \cdot 4 = 16\) tiles from the corners.
Step 3 — Add the two kinds of piece: sides and corners together make up the whole border with nothing overlapping and nothing missing.
$$72 + 16 = 88 \text{ border tiles}$$Answer: 88 border tiles.
Problem 8. Use Wen's method to find the total number of border tiles around a 9 by 9 center, and check it against Problem 1.1.7.
Solution
Step 1 — Find the floor size: with a 9 by 9 center, the floor is \(9 + 4 = 13\) tiles on a side.
Step 2 — Build the top and bottom strips (Wen's method): each runs the full floor width, 2 tiles tall by 13 tiles wide, so \(2 \cdot 13 = 26\) tiles each, and there are two of them: \(2 \cdot 26 = 52\).
Step 3 — Build the left and right strips: each is only as tall as the center block (9), because the top and bottom strips already took the corners, so \(2 \cdot 9 = 18\) tiles each, and there are two: \(2 \cdot 18 = 36\).
Step 4 — Add all four pieces and check against Problem 1.1.7:
$$52 + 36 = 88 \text{ border tiles}$$Problem 1.1.7 also gave 88, so the two methods agree.
Answer: 88 border tiles, matching Problem 1.1.7.
Problem 9. State the test that tells you whether a decomposition is valid.
Solution
Step 1 — Recall Definition 1.1.2: a decomposition of a figure is a way of cutting it into pieces.
Step 2 — State the validity test: a decomposition is valid exactly when every tile in the figure belongs to exactly one piece — none left out, and none counted twice. That is the property that lets you trust adding up the pieces to get the total.
Answer: a decomposition is valid if and only if every tile belongs to exactly one piece — nothing is left uncounted, and nothing is counted in more than one piece.
Problem 10. A student cuts the border around a 6 by 6 center into four full-length 2-by-10 sides and gets 80 tiles. Find the correct total and explain exactly where the extra tiles came from.
Solution
Step 1 — Find the true total: for a 6 by 6 center, the floor is \(6+4=10\) tiles on a side, so the border holds \(10^2 - 6^2 = 100 - 36 = 64\) tiles (this matches Problem 1.1.5).
Step 2 — Compare to the student's claimed total: the student got 80 by treating the border as four full-length 2-by-10 sides, \(4(2 \cdot 10) = 80\).
Step 3 — Measure the gap:
$$80 - 64 = 16 \text{ extra tiles}$$Step 4 — Locate the source of the extra tiles: each "full-length" side runs the entire 10-tile floor edge, so it includes a corner 2-by-2 block at each end. Every corner block sits inside two of the four full-length sides at once (e.g., the top-left corner belongs to both the "top" strip and the "left" strip), so it gets counted twice instead of once. There are 4 corners, each a 2-by-2 block of 4 tiles, so the overlap is \(4 \cdot 4 = 16\) tiles — exactly the gap found in Step 3.
Answer: the correct total is 64 border tiles; the extra 16 tiles came from double-counting the four 2-by-2 corner blocks, each of which belongs to two of the "full-length" sides at once.
Problem 11. Explain why Marisol's method can never double-count, and say what she gives up by using it.
Solution
Step 1 — Explain why Marisol's method cannot double-count: her method is a single subtraction, whole floor minus center hole. Subtraction assembles no pieces and never revisits a tile, so there is nothing to "count twice" — each tile is simply classified once as center or as border, and the arithmetic only ever removes the center tiles once.
Step 2 — Explain what the method gives up: because Marisol never looks at the border directly, her method gives no picture of how the border is shaped, how many strips or corners it contains, or how a tile-cutter should actually lay out the pieces. It answers "how many" but not "what does the border look like."
Answer: subtraction can never double-count because it never assembles overlapping pieces — it just removes the center once from the whole. In exchange, Marisol's method gives no insight into the border's actual shape or how to decompose it for cutting.
Problem 12. In Jamal's pinwheel, each strip is 2 tiles by \(N + 2\) tiles. Explain where the \(+\,2\) comes from.
Solution
Step 1 — Identify the two contributions to a pinwheel strip's length: each strip runs along one side of the center block, contributing \(N\) tiles of length, and then continues through one 2-by-2 corner block before the next strip begins.
Step 2 — Identify where the \(+2\) comes from: the corner block is 2 tiles wide in the direction the strip runs, so it adds exactly 2 more tiles of length onto the strip, making the total length \(N + 2\).
Answer: the \(+2\) is the one 2-tile corner block that gets attached onto the end of each pinwheel strip so that, rotating around, every corner is covered by exactly one strip.
Problem 13. Explain why the four 2-by-2 corner blocks in Priya's method are counted once each rather than being part of the side strips.
Solution
Step 1 — Recall how Priya's side strips are sized: each side strip is 2 by \(N\), where \(N\) is the center block's side length — not the full floor side length \(N+4\).
Step 2 — Explain why the corners are excluded from the sides: because each side strip only runs the length of the center block, it stops exactly where the center ends and the corner block begins. It never reaches into the corner.
Step 3 — Explain why the corners need their own piece: if the corners were left out of both the side strips and a separate piece, those tiles would never be counted at all — violating the "nothing left out" half of Definition 1.1.2. Making the corners their own 2-by-2 pieces, one per corner, ensures every corner tile is counted in exactly one piece.
Answer: the corner blocks are their own piece because the side strips are sized to the center's length \(N\) and stop before reaching the corners; folding the corners into the side strips would either miss them or double-count them, so they must be counted separately, once each.
Problem 14. In Wen's decomposition, the left and right strips are only as tall as the center block. Explain why they are not the full height of the floor.
Solution
Step 1 — Recall how Wen's top and bottom strips are sized: the top and bottom strips each run the full width of the floor, \(N+4\) tiles, which already includes the tiles at both far ends — the corners.
Step 2 — Explain why the side strips must stop short: since the top and bottom strips already claimed the corner tiles, giving the left and right strips the full floor height \(N+4\) as well would count those corner tiles a second time.
Step 3 — State the correct height: to avoid that overlap, the left and right strips are only as tall as the center block, \(N\), which is exactly the height left over once the top and bottom strips are removed.
Answer: the left and right strips are only as tall as the center block because the top and bottom strips already cover the full floor width — corners included — so giving the side strips the full height too would double-count those corner tiles.
Problem 15. State the claim proved in §1.1.3 and the two conditions its proof depends on.
Solution
Step 1 — State the claim: in a checkerboard border two tiles wide, exactly half the tiles are colored.
Step 2 — Identify the two conditions the proof depends on: (1) the border is exactly two tiles thick, so it can be sliced into pairs of tiles running across that thickness; and (2) the coloring is a true checkerboard pattern, meaning any two tiles that share an edge always have different colors — which is what forces each pair to contain exactly one colored tile and one white tile.
Answer: the claim is that exactly half the border tiles are colored, and the proof depends on the border being exactly two tiles wide (so it splits into edge-sharing pairs) and the coloring being checkerboard-style (so every such pair has one colored and one white tile).
Problem 16. Explain why the halving argument works even though the top row of the grid has 5 colored tiles and the row below it has only 4.
Solution
Step 1 — Recall how the halving argument is built: it never counts row by row. Instead, it slices the border into pairs of tiles running across the 2-tile thickness — for example, a standing pair in the top strip, or a lying-down pair in a side strip.
Step 2 — Apply the checkerboard rule to each pair: the two tiles in any such pair share an edge, and in a checkerboard, edge-sharing tiles always have different colors. So every pair, without exception, has exactly one colored tile and one white tile.
Step 3 — Explain why the row imbalance is irrelevant: the uneven row counts (5 colored in the top row, 4 in the row below) are a true fact about how those two particular rows split the pairs, but the halving argument never examines rows — it only examines the pairs, and each pair is balanced regardless of which row its tiles land in.
Answer: the halving argument works because it is built on pairs sliced across the border's 2-tile thickness, not on rows, and every such pair has exactly one colored and one white tile no matter how the rows happen to divide those pairs up.
Problem 17. A two-row checkerboard border holds 96 tiles in total. Find the number of colored tiles.
Solution
Step 1 — Apply the halving argument: in a two-row checkerboard border, exactly half the tiles are colored.
$$\frac{96}{2} = 48$$Answer: 48 colored tiles.
Problem 18. A two-row checkerboard border around a square center holds 40 colored tiles. Find the total number of border tiles.
Solution
Step 1 — Reverse the halving argument: if the colored tiles are exactly half the border, the total is twice the colored count.
$$40 \cdot 2 = 80$$Answer: 80 border tiles in total.
Problem 19. Write an expression for the total number of border tiles around an \(N\) by \(N\) center using Marisol's method.
Solution
Step 1 — Recall Marisol's method: the total border count is the whole floor minus the center hole, using the general floor size \(N+4\) and center size \(N\).
Answer: \((N+4)^2 - N^2\).
Problem 20. Write an expression for the number of colored border tiles around an \(N\) by \(N\) center using Priya's method.
Solution
Step 1 — Recall Priya's colored-tile expression: her total border count is \(4(2N) + 4(4) = 8N + 16\), and halving it (since exactly half the border is colored) gives the colored count.
Answer: \(4N + 8\).
Problem 21. Evaluate \(4N + 8\) at \(N = 3\), \(N = 11\), and \(N = 25\).
Solution
Step 1 — Substitute each value into \(4N+8\) and evaluate:
$$N=3: \quad 4(3) + 8 = 12 + 8 = 20$$ $$N=11: \quad 4(11) + 8 = 44 + 8 = 52$$ $$N=25: \quad 4(25) + 8 = 100 + 8 = 108$$Answer: 20 at \(N=3\), 52 at \(N=11\), and 108 at \(N=25\).
Problem 22. Evaluate \(4(N + 2)\) at \(N = 3\), \(N = 11\), and \(N = 25\), and compare your answers with Problem 1.1.21.
Solution
Step 1 — Substitute each value into \(4(N+2)\) and evaluate:
$$N=3: \quad 4(3+2) = 4(5) = 20$$ $$N=11: \quad 4(11+2) = 4(13) = 52$$ $$N=25: \quad 4(25+2) = 4(27) = 108$$Step 2 — Compare with Problem 1.1.21: the values there were 20, 52, and 108 — identical at every \(N\) tested.
Answer: 20, 52, and 108, exactly matching Problem 1.1.21 at all three values, which is consistent with \(4(N+2)\) and \(4N+8\) being equivalent expressions.
Problem 23. In the expression \(4N + 8\), say which tiles the \(4N\) counts and which tiles the \(8\) counts.
Solution
Step 1 — Trace \(4N\) back to the picture: this term counts the tiles along the four straight sides of the border — \(N\) colored tiles per side (one per stacked pair along the center's length), times 4 sides.
Step 2 — Trace \(8\) back to the picture: this term counts the tiles in the four corner blocks — 2 colored tiles per 2-by-2 corner, times 4 corners, giving \(4 \cdot 2 = 8\).
Answer: \(4N\) counts the colored tiles along the four straight sides; \(8\) counts the colored tiles in the four corner blocks.
Problem 24. In the expression \(2(N + 4) + 2N\), say which tiles each of the two terms counts.
Solution
Step 1 — Trace \(2(N+4)\) back to the picture: this is Wen's top and bottom strips, each running the full floor width \(N+4\), so together they contribute \(2(N+4)\) colored tiles.
Step 2 — Trace \(2N\) back to the picture: this is Wen's left and right strips, each only as tall as the center block \(N\) (since the corners already belong to the top and bottom strips), contributing \(2N\) colored tiles together.
Answer: \(2(N+4)\) counts the colored tiles in the top and bottom strips; \(2N\) counts the colored tiles in the left and right strips.
Problem 25. In the expression \(\dfrac{(N+4)^2 - N^2}{2}\), say what the numerator counts and what the 2 is doing.
Solution
Step 1 — Trace the numerator back to the picture: \((N+4)^2 - N^2\) is Marisol's total-border expression — the whole floor (\(N+4\) by \(N+4\)) with the center hole (\(N\) by \(N\)) removed, so it counts every border tile, colored and white together.
Step 2 — Explain what the 2 is doing: because the border is checkerboard and two tiles wide, exactly half of it is colored (the halving argument of §1.1.3). Dividing the numerator by 2 converts the total border count into just the colored count.
Answer: the numerator counts all the border tiles (colored and white); the 2 halves that total to isolate just the colored tiles, using the fact that exactly half of any two-row checkerboard border is colored.
Problem 26. Expand \(4(N + 2)\) and name the property you used.
Solution
Step 1 — Distribute the 4 across the parentheses:
$$4(N+2) = 4N + 4(2) = 4N + 8$$Answer: \(4N + 8\), using the distributive property.
Problem 27. Expand \(2(N + 4) + 2N\) into simplified form, naming the property behind each step.
Solution
Step 1 — Distribute the outer 2:
$$2(N+4) + 2N = 2N + 8 + 2N \qquad \text{(distributive property)}$$Step 2 — Reorder the terms:
$$2N + 8 + 2N = 2N + 2N + 8 \qquad \text{(commutative property of addition)}$$Step 3 — Combine the like terms:
$$2N + 2N + 8 = 4N + 8 \qquad \text{(combining like terms)}$$Answer: \(4N + 8\).
Problem 28. Expand \(\dfrac{(N+4)^2 - N^2}{2}\) into simplified form, showing every step.
Solution
Step 1 — Expand the square:
$$\frac{(N+4)^2 - N^2}{2} = \frac{N^2 + 8N + 16 - N^2}{2} \qquad \text{(expand } (N+4)^2\text{)}$$Step 2 — Cancel the \(N^2\) terms:
$$\frac{N^2 + 8N + 16 - N^2}{2} = \frac{8N + 16}{2} \qquad \text{(}N^2 - N^2 = 0\text{)}$$Step 3 — Divide each term by 2:
$$\frac{8N+16}{2} = 4N + 8 \qquad \text{(divide each term by 2)}$$Answer: \(4N + 8\).
Problem 29. State what it means for two expressions to be equivalent.
Solution
Step 1 — Recall Definition 1.1.4: equivalent expressions are the mathematical statement of "same quantity, seen differently."
Answer: two expressions are equivalent when they produce the same value for every allowed value of the variable — not just for one or a few values that happen to have been tried.
Problem 30. Explain why testing two expressions at \(N = 5\) and \(N = 10\) is not enough to prove they are equivalent.
Solution
Step 1 — Recall what equivalence requires: two expressions are equivalent only if they agree at every allowed value of \(N\), not merely at a chosen few.
Step 2 — Explain why two matching values fall short: two genuinely different (non-equivalent) expressions can still happen to agree at some particular values — Try It Now 1.1.4 shows exactly this, where \(4N+8\) and \(N^2+3\) both equal 28 at \(N=5\) yet diverge at \(N=10\). Testing at \(N=5\) and \(N=10\) checks only two points out of infinitely many possible values of \(N\); nothing rules out a disagreement at, say, \(N=6\) or \(N=100\).
Answer: testing at two values only proves agreement at those two values; because different expressions can coincidentally match at a handful of chosen points, only expanding both expressions to the same simplified form proves they agree for every \(N\).
Problem 31. Explain why a single value at which two expressions disagree is enough to prove they are not equivalent.
Solution
Step 1 — Recall what equivalence requires: it requires matching at every allowed value, with no exceptions.
Step 2 — Apply that to a single disagreement: if two expressions give different values at even one value of \(N\), the "matches at every value" condition is already violated — no amount of agreement elsewhere can restore it.
Answer: a single value where two expressions disagree is enough to disprove equivalence because equivalence demands agreement at every value; one counterexample directly breaks that requirement, regardless of how many other values do agree.
Problem 32. A partner writes \(4(N + 1) + 4\) for the number of colored border tiles. Decide whether it is equivalent to \(4N + 8\), and show your work.
Solution
Step 1 — Expand the partner's expression:
$$4(N+1) + 4 = 4N + 4 + 4 = 4N + 8 \qquad \text{(distributive property, then combine like terms)}$$Step 2 — Compare to the target form: \(4N + 8\) is exactly Priya's simplified expression for the colored border tiles.
Answer: yes, \(4(N+1)+4\) is equivalent to \(4N+8\) — expanding it gives \(4N+8\) exactly.
Problem 33. A partner writes \(2(2N + 3)\) for the number of colored border tiles. Decide whether it is equivalent to \(4N + 8\), and show your work.
Solution
Step 1 — Expand the partner's expression:
$$2(2N+3) = 4N + 6 \qquad \text{(distributive property)}$$Step 2 — Compare to the target form: \(4N+6\) has a different constant term than \(4N+8\) (6 versus 8), so the simplified forms do not match.
Step 3 — Confirm with a test value, \(N=5\):
$$2(2(5)+3) = 2(13) = 26 \qquad \text{versus} \qquad 4(5)+8 = 28$$\(26 \ne 28\), which agrees with the expanded forms disagreeing.
Answer: no, \(2(2N+3)\) is not equivalent to \(4N+8\) — it simplifies to \(4N+6\), which is 2 tiles short at every value of \(N\).
Problem 34. Give the simplified form of \(3(N + 2) + N + 2\).
Solution
Step 1 — Distribute the 3:
$$3(N+2) + N + 2 = 3N + 6 + N + 2 \qquad \text{(distributive property)}$$Step 2 — Combine like terms:
$$3N + 6 + N + 2 = 4N + 8$$Answer: \(4N + 8\).
Problem 35. Explain in two or three sentences why \(4(N + 2)\) might be more useful than \(4N + 8\) to somebody cutting tile, even though \(4N + 8\) is simpler.
Solution
Step 1 — Recall what each form makes visible: \(4(N+2)\) keeps the factor of 4 and the group \(N+2\) intact, while \(4N+8\) has already merged them into two separate terms.
Step 2 — Connect the form to the cutting task: \(4(N+2)\) reads directly as an instruction — "cut 4 strips, each \(N+2\) tiles long" — because the 4 is literally the strip count and the \(N+2\) is literally the strip length. \(4N+8\) is arithmetically simpler to evaluate, but it no longer shows that strip structure; a cutter reading \(4N+8\) would have to reverse-engineer where the two terms came from before knowing how to cut anything.
Answer: \(4(N+2)\) is more useful for cutting tile because its factored form directly names the number of strips (4) and the length of each strip (\(N+2\)), even though \(4N+8\) is the simpler expression to compute with.
Problem 36. Write out the three parts of a strong explanation of a counting method.
Solution
Step 1 — Recall §1.1.6's list of what belongs in a strong explanation.
Answer: (1) what you split the picture into — the pieces, and their dimensions; (2) why nothing was counted twice or missed — how every tile lands in exactly one piece; (3) where each number in your arithmetic came from — which piece of the picture each factor measures.
Problem 37. A classmate says only "I got 56." Write down two questions you could ask that would let you check their method.
Solution
Step 1 — Identify what "I got 56" leaves out: it names an answer but no pieces, no dimensions, and no reasoning about double-counting.
Step 2 — Write two questions that would recover that missing information.
Answer: "What did you split the border into, and what are the dimensions of each piece?" and "How do you know every tile is counted exactly once — what stops a tile from being missed or counted twice?"
Problem 38. List, in order, the three steps for comparing your expression with a partner's, and say which step actually settles the question.
Solution
Step 1 — List the three steps in order: (1) test both expressions at a value; (2) if they agree, expand both into simplified form; (3) if the simplified forms differ, find the disagreement in the picture, tracing which tile was double-counted or missed.
Step 2 — Identify which step actually settles the question: step 1 can only ever rule out equivalence (a disagreement proves they're different) — it can never prove equivalence, since agreement at one value doesn't guarantee agreement everywhere. Step 2, expanding both to simplified form, is what actually settles whether they're equivalent for every \(N\).
Answer: (1) test at a value, (2) expand both to simplified form, (3) trace any remaining disagreement in the picture — and step 2 is the one that actually settles the question.
Problem 39. Explain what is wrong with the reasoning "I plugged in a number and it worked, so my expression must be right."
Solution
Step 1 — Recall what a single successful test actually shows: it shows only that the two expressions agree at that one particular value of \(N\).
Step 2 — Explain why that falls short of proof: two genuinely different, non-equivalent expressions can still agree at some values by coincidence — Try It Now 1.1.4 demonstrates this directly, with \(4N+8\) and \(N^2+3\) both giving 28 at \(N=5\) but 48 and 103 at \(N=10\). Plugging in one number and getting a match rules nothing out about every other value of \(N\).
Answer: the reasoning is flawed because a single matching value is evidence, not proof — two non-equivalent expressions can still agree at that one value by coincidence, so only expanding both to the same simplified form (which holds for every \(N\) at once) actually proves equivalence.
Problem 40. Explain in two or three sentences why two students with different-looking expressions are usually not disagreeing, and give one situation in which they genuinely are.
Solution
Step 1 — Explain why different-looking expressions are usually not in conflict: each expression typically records a different, but equally valid, way of decomposing the very same picture into pieces — as with Marisol's, Jamal's, Priya's, and Wen's four different-looking expressions, all of which expand to the same simplified form \(4N+8\). Different-looking arithmetic does not mean different quantities; it means different pieces were chosen.
Step 2 — Give a situation where they genuinely are disagreeing: they are genuinely disagreeing when expanding both to simplified form produces two different results — as in Problem 1.1.33, where \(2(2N+3)\) simplifies to \(4N+6\), not \(4N+8\). That mismatch signals a real error somewhere in the decomposition: a tile counted twice or a tile left out.
Answer: two different-looking expressions are usually just two valid decompositions of the same picture, and expanding both lands on the same simplified form; they are genuinely disagreeing only when expanding reveals different simplified forms, which points to an actual counting error in one of them.
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.