Demonstration 1 of 4
Same arithmetic, different workspace
If 47 + 38 stays the same, what changes when more of its lines are written down?
The five lines of the place-value route are revealed one stage at a time. Every middle quantity that is not yet on the page sits in the held in mind box.
\[47 + 38\]
\[47 = 40 + 7\]
47 and 38 are the starting quantities and + is the operation. Writing 47 = 40 + 7 splits 47 into four tens and seven ones. Each written line is an external representation: part of the work placed where it can be seen.
Predict first. Once the two expansions are written, how many middle quantities does memory still have to carry, and which ones are they?
Choose an example
Constructed example: the chapter's 47 + 38 and its five place-value lines.
Calculated values
- Lines on the page
- 5 of 5
- Middle quantities held in mind
- 0 (none)
- Answer visible
- yes, 85
- Answer
- 85
With 5 of the 5 lines written, the page shows 7 quantities. Every middle quantity now lives on the page, so memory only has to follow the lines. The arithmetic is the same either way: 40 + 30 = 70, 7 + 8 = 15, and 70 + 15 = 85.
Use the idea
Before a calculation with several pieces, decide which middle quantity you would most like to stop carrying, and write that one down first.
Where the conclusion applies
The count of quantities held in mind follows only the place-value route shown here; another route would hold different quantities. It describes the task, not the person doing it, and says nothing about why a thread goes missing.
Check your understanding: If only the line 7 + 8 = 15 were missing, which quantity would memory have to carry?
Chapter 2 source: section "Same arithmetic, different workspace". Demonstration C02-D01.
Demonstration 2 of 4
A tens-and-ones record
When the ones add up past nine, where does the extra ten go?
Tens join tens and ones join ones. When the ones reach ten or more, ten of them regroup into one new rod, which keeps its full value of 10 in the record.
\[70+10+4=84\]
Each rod is one ten and each small square is one. 70 is the tens total, 10 is the new ten made from the ones, and 4 is the ones left over.
Predict first. For 41 + 37, will a hatched new ten appear? Check by adding the ones before you switch.
Choose an example
Constructed example: the chapter's 56 + 28 and 47 + 38, its practice sum 57 + 26, and one added case, 41 + 37.
Calculated values
- Tens
- 50 + 20 = 70
- Ones
- 6 + 8 = 14
- New ten from the ones
- yes, 1
- Total
- 84
Tens: 50 + 20 = 70. Ones: 6 + 8 = 14. 14 is one ten and 4 ones, so 70 + 10 + 4 = 84. The hatched rod is that new ten, written with its full value.
Use the idea
When you carry a 1 in a stacked sum, say it as "ten" to yourself. Writing its full value keeps it from disappearing.
Where the conclusion applies
Two-digit whole numbers, added by place value. 41 + 37 is a constructed extra case with no new ten. With three-digit numbers, a hundreds column joins the record.
Check your understanding: In 68 + 17, what are the tens total, the ones total, and the final sum?
Chapter 2 source: section "What a tens-and-ones record means". Demonstration C02-D02.
Demonstration 3 of 4
Two routes to the same sum
Can a route that overshoots on purpose still land on the right answer?
Compensation jumps to a convenient round number, then steps back by the amount it overshot. The tens then ones route makes two forward jumps. Both stop at the same point, so each checks the other.
\[47 + 38 = 47 + 40 - 2\]
\[70 + 15 = 85\]
47 + 40 - 2 replaces 38 with the nearby 40, then subtracts the extra 2. 70 + 15 is the place-value record: tens total plus ones total.
Predict first. For 36 + 49 by compensation, which round number will you add, and what will the correction be?
Choose an example
Constructed example: the chapter's 47 + 38 and 56 + 28, with its practice sums 36 + 49 and 57 + 26.
Calculated values
- Problem
- 47 + 38
- Route
- compensation
- Middle stop
- 87
- Result
- 85
- Correction
- subtract 2
Compensation: 47 + 38 = 47 + 40 - 2. Adding 40 added 2 more than 38, so the correction subtracts 2: 47 + 40 = 87, then 87 - 2 = 85. Both routes land on 85, so each one checks the other.
Use the idea
When a number sits just below a round number, add the round number and correct. Name whether you added too much; that decides the correction's sign.
Where the conclusion applies
The round number here is the next ten above the second number. If the correction's direction is flipped, the answer is off by twice the correction.
Check your understanding: Using compensation, what is 46 + 37?
Chapter 2 source: section "Put the middle on paper". Demonstration C02-D03.
Demonstration 4 of 4
Leave a route back
After a subtraction, how can you return to where you started?
Both routes remove a total of the second number. The tens then ones route moves back twice. Compensation moves back too far, then forward by the extra. The middle stop is a restart line you can name.
\[64 - 27 = 64 - 30 + 3\]
\[37 + 27 = 64\]
64 - 30 + 3 removes 30, then restores the extra 3 that was removed. 37 + 27 = 64 is the inverse check: adding back what was taken away rebuilds the start.
Predict first. Keep 64 - 27 and switch the route to compensation. Will the correction after subtracting 30 add or subtract, and by how much?
Choose an example
Constructed example: the chapter's 64 - 27 and 72 - 36, with its practice subtractions 91 - 47 and 120 - 68.
Calculated values
- Problem
- 64 - 27
- Route
- tens then ones
- Restart line
- 44 remains; 7 still to subtract
- Correction
- none needed
- Result
- 37
- Inverse check
- 37 + 27 = 64
Remove the tens, then the ones: 64 - 20 = 44, then 44 - 7 = 37. Route back: 37 + 27 = 64, which returns to the start.
Use the idea
Write a restart note that keeps the next operation and its quantity, such as "44 remains; subtract 7." Then check by adding back.
Where the conclusion applies
Whole numbers with the first number larger, so the result is not negative. A check that repeats the same route can repeat the same slip; inverse addition comes from a different direction.
Check your understanding: Using compensation, what is 83 - 29, and how do you check it?
Chapter 2 source: section "Leave a route back". Demonstration C02-D04.