Demonstration 1 of 4
More than one number can be the odd one out
In 4, 6, 8, 9, which number does not belong, and what rule makes that choice checkable?
Each rule asks all four numbers the same question. Exactly one says yes, so each rule picks a different number and each choice comes with a reason anyone can test.
\[4, 6, 8, 9\]
\[8 = 2 × 2 × 2\]
A cube number is a whole number multiplied by itself three times, so 8 = 2 x 2 x 2. A square number is a whole number times itself, like 3 x 3 = 9. Even means it splits into pairs with nothing left; odd means one is left.
Predict first. Switch the rule to "is odd". Which tile lights up, and how many pairs does it make first?
Choose an example
Constructed example: the chapter's opening numbers 4, 6, 8, 9 and the four rules it states.
Calculated values
- Rule
- the number is a cube number
- Numbers that pass
- {8}
- Numbers that fail
- {4, 6, 9}
- Chosen as the one that does not belong
- 8
Ask every number the same question: the number is a cube number, yes or no? Only 8 says yes: 8 = 2 x 2 x 2, while 4 = 2 x 2 and 9 = 3 x 3 are squares and 6 is neither. The rule is public, so a stranger testing 4, 6, 8 and 9 one at a time makes the same choice.
Use the idea
When a choice feels obvious, write the yes-or-no question behind it. Then a second person can apply it without guessing what you meant.
Where the conclusion applies
Only the four numbers on the page are tested. A rule picks out one number only if exactly one passes: "is even" fails as an odd-one-out rule because three numbers pass.
Check your understanding: Under the rule "is even", how many of 4, 6, 8, 9 pass, and does that rule single one out?
Chapter 1 source: section "The number words in the opening". Demonstration C01-D01.
Demonstration 2 of 4
A list records, a rule decides
If the boundary moves, which keeps up on its own: the list or the rule?
The shaded squares are what the rule decides. The old list is fixed, so whenever the boundary moves past a new multiple of 3, or back before one, the list falls out of date.
\[\{3, 6, 9, 12\}\]
The braces hold the members of a set. The rule is "whole numbers from 1 through the boundary divisible by 3": a number belongs when it splits into groups of 3 with none left.
Predict first. Move the boundary from 12 to 15. Does the old list {3, 6, 9, 12} still match the rule?
Choose an example
Constructed example: the chapter's multiples of 3 through 12 and through 15, plus two other boundaries.
Calculated values
- Boundary
- 1 through 12
- Members by the rule
- {3, 6, 9, 12}
- Number of members
- 4
- The list {3, 6, 9, 12}
- matches
Through 12, the rule keeps every number that splits into groups of 3: 3 = 1 x 3, 6 = 2 x 3, 9 = 3 x 3, 12 = 4 x 3. The rule and the list agree, so either one can be checked against the other.
Use the idea
When a category might grow, keep the rule written next to the list, so the list can always be rechecked.
Where the conclusion applies
Whole numbers only, starting at 1, with "through" meaning the boundary itself counts. Change "through 15" to "less than 15" and 15 drops out.
Check your understanding: With the boundary through 21, how many members does the rule give, and what is the largest?
Chapter 1 source: section "A rule has to do some work". Demonstration C01-D02.
Demonstration 3 of 4
Two rules on six cards
Which cards meet both rules, at least one, or neither?
The circles only display decisions already made in the table. Each question shades a different region, and the bold cards are the ones whose two answers fit it.
\[K \cap C = \{C\}\]
\[K = \{A, C, D\}\]
K is the set of cards sent by a listed contact. C is the set of cards asking for a one-time code. The symbol ∩ means "in both". U is the universe: the six cards on the table.
Predict first. Switch to "outside K". Which three cards are shaded, and is Card B one of them?
Choose an example
Constructed example: the chapter's six message cards A to F and its two rules K and C.
Calculated values
- Question
- in K and in C (intersection)
- Cards that qualify
- {C}
- Cards left out
- {A, B, D, E, F}
- Count
- 1 of 6
Read each card's two answers (A: K yes, C no; B: K no, C yes; C: K yes, C yes; D: K yes, C no; E: K no, C no; F: K no, C no). Being in K and in C (intersection) needs yes for K and yes for C, which gives {C}: 1 cards, and 1 + 5 = 6 accounts for every card in the universe.
Use the idea
When two conditions are combined, test each one separately first, then decide whether you need both, either, or the outside.
Where the conclusion applies
The universe is exactly six invented cards; it is a sorting exercise, not advice about real messages. Without the stated universe, "outside K" would have no edge.
Check your understanding: How many cards are outside both K and C, and which are they?
Chapter 1 source: section "Two rules can overlap". Demonstration C01-D03.
Demonstration 4 of 4
Every number gets one location
In U = {2, 3, 4, 5, 6, 7}, where does each number land when two rules ask their questions?
The grid shows each number's two answers. Each pair of answers points to exactly one location, so the four locations always add back to six.
\[E = \{2, 4, 6\}\]
\[G = \{5, 6, 7\}\]
E is the even numbers in U. G is the numbers in U greater than the chosen boundary. Two yes-or-no answers give four locations: both, E only, G only, neither.
Predict first. Raise the boundary from 4 to 5. Where does 5 move, and does the overlap {6} change?
Choose an example
Constructed example: the chapter's universe 2 through 7 with G at the book's boundary 4, plus three other boundaries.
Calculated values
- E (even)
- {2, 4, 6}
- G (greater than 4)
- {5, 6, 7}
- Overlap: in both
- {6}
- Outside both
- {3}
- Check: locations add to 6
- 1 + 2 + 2 + 1 = 6
Even numbers: {2, 4, 6}. Greater than 4: {5, 6, 7}. The overlap is {6}: 6 is even and 6 > 4. Outside both: {3}. Count the four locations: 1 + 2 + 2 + 1 = 6, so every number has exactly one place.
Use the idea
When a sorting rule leaves something out, check for the "neither" row; that is where edge cases like 3 are waiting.
Where the conclusion applies
The universe is fixed at 2 through 7 and "greater than" excludes the boundary itself. If the boundary passes 7, G is empty, and so is the overlap.
Check your understanding: If G meant "at least 4" instead of "greater than 4", what would the overlap be?
Chapter 1 source: section "A collection with no familiar story attached". Demonstration C01-D04.