Demonstration 1 of 4
One allowed case is enough
Does multiplication always make a number bigger?
The top bar is the starting 7. The bottom bar is 7 times the chosen multiplier. When the bottom bar fails to reach past the dashed line, the claim has met a case it allowed and could not keep.
\[7 × 0 = 0\]
The starting number is 7. The multiplier is the second factor. "Always" means every allowed multiplier, and the claim as written puts no limit on it.
Predict first. Before changing the multiplier to 1, predict: is 7 x 1 bigger than 7?
Choose an example
Constructed example: the chapter's opening claim and its case 7 x 0 = 0, with other multipliers.
Calculated values
- Calculation
- 7 x 0 = 0
- Arithmetic correct?
- yes
- Result compared with 7
- smaller
- Against "always bigger"
- counterexample
7 x 0 = 0, which is smaller than 7. The claim allows a multiplier of 0, and the result is smaller, so this allowed case is a counterexample.
Use the idea
When a sentence says always, every or all, try the smallest allowed cases first. Zero and one end many overconfident sentences.
Where the conclusion applies
The claim as written, with no condition on either factor. The arithmetic must be correct for a case to count. Many cases that agree, like 7 x 2 = 14, still do not prove the claim.
Check your understanding: Is 6 x 1 = 6 a counterexample to "multiplication always makes a number bigger"?
Chapter 3 source: section "A claim has a domain". Demonstration C03-D01.
Demonstration 2 of 4
Same calculation, different domain
Why does 1 x 1 = 1 disprove one squaring claim but not the other?
The shaded squares are the claim's domain. A case outside the shading cannot refute the claim, however correct its arithmetic. Inside the shading, a result that is not strictly larger is a counterexample.
\[1\times1=1\]
\[2\times2=4>2\]
Squaring a whole number means multiplying it by itself. The domain is the set of cases the claim agreed to handle: whole numbers greater than one, or every whole number.
Predict first. Keep 1 x 1 = 1 on screen and switch the claim to "every whole number". Does the verdict change, even though the arithmetic does not?
Choose an example
Constructed example: the chapter's worked squaring claim, tested at 0, 1, 2 and 3.
Calculated values
- Calculation
- 1 x 1 = 1
- Strictly larger than the start?
- no (1 equals 1)
- Case allowed by this claim?
- no
- Verdict
- not a counterexample
1 x 1 = 1. The arithmetic is correct, but one is not greater than one, so the case is outside the domain: not a counterexample.
Use the idea
A good counterexample report has three parts: quote the claim, show the case is allowed, and show the conflicting result.
Where the conclusion applies
Whole numbers only, and "larger" means strictly larger. If the claim said "at least as large", 0 and 1 would agree with it, because equality counts.
Check your understanding: Is 0 x 0 = 0 a counterexample to "for whole numbers greater than one, squaring gives a larger number"?
Chapter 3 source: section "Worked example: a true calculation outside the domain". Demonstration C03-D02.
Demonstration 3 of 4
Odd numbers that build squares
What do 1 + 3, 1 + 3 + 5 and 1 + 3 + 5 + 7 have in common?
Each new odd number is an L shape that wraps around the corner of the square already built, making a square one tile wider on each side.
\[1 + 3 + 5 + 7 = 16\]
The odd numbers 1, 3, 5, 7, 9 are added in order, starting from 1. Each colour is one odd number of tiles. A square number is a whole number times itself.
Predict first. Before choosing five odd numbers, predict the total of 1 + 3 + 5 + 7 + 9.
Choose an example
Constructed example: the chapter's sums 4, 9 and 16, extended by one more odd number.
Calculated values
- Odd numbers added
- 1 + 3 + 5 + 7
- Total
- 16
- Square
- 4 x 4 = 16
- Status
- a conjecture supported by examples
1 + 3 + 5 + 7 = 16, and 4 x 4 = 16. The new odd number 7 = 3 + 3 + 1 wraps the old 3 by 3 square into a 4 by 4 square. Each example agrees, but checked examples alone do not prove every case.
Use the idea
When examples line up, write the pattern as a conjecture, then look for the structure that would make it true in every case.
Where the conclusion applies
Consecutive odd whole numbers, starting from 1. Leaving out a number or starting from 3 breaks the pattern: 3 + 5 = 8 is not a square. Agreeing examples are evidence for the conjecture, not a proof.
Check your understanding: What is 1 + 3 + 5 + 7 + 9 + 11, and which square is it?
Chapter 3 source: section "A pattern invites a conjecture". Demonstration C03-D03.
Demonstration 4 of 4
Two leftovers make a pair
Why is the sum of two odd whole numbers always even?
Each odd row brings exactly one leftover. Two leftovers join into one more pair, whatever the rows' lengths, so the argument covers every pair of odd numbers.
\[5 + 3 = 8\]
An even number splits completely into pairs. An odd number is pairs plus one leftover tile. A boxed group is a pair and red marks leftovers. A red pair in the combined row is two leftovers joined.
Predict first. Change the second row from 3 to 4 tiles. Will the total have a leftover tile?
Choose an example
Constructed example: the chapter's rows of five and three tiles, with other row lengths.
Calculated values
- First row
- 2 pairs + 1 leftover
- Second row
- 1 pair + 1 leftover
- Together
- 4 pairs + 0 leftover
- Total
- 5 + 3 = 8, even
5 + 3 = 8. Pairs: 2 + 1 = 3, leftovers: 1 + 1 = 2. The two leftovers join to make one more pair, so no tile is left over. 8 = 4 x 2 + 0, so the total is even.
Use the idea
To explain why a parity claim holds, count leftovers instead of trying more examples.
Where the conclusion applies
Whole numbers of tiles. The reason works because every odd number has the same structure, pairs plus one leftover. An odd row plus an even row has only one leftover, so that total is odd.
Check your understanding: Using leftovers, is 7 + 9 even or odd?
Chapter 3 source: section "Two leftovers make a pair". Demonstration C03-D04.