Demonstration 1 of 4
One new guest
Every room is taken. Can one more guest check in without anyone sharing or leaving?
Every guest moves one room up, so room 1 is left over for the new guest. Working backward from any room r of 2 or more finds guest r - 1, so no room is skipped.
\[a + 1 = b + 1\]
n is a current guest's original room number, and the rule moves that guest to room n + 1. a and b are two starting rooms; if they landed in the same room, then a + 1 = b + 1, which forces a = b.
Predict first. Switch the hotel to one that ends at room 6. Which guest is the problem, and which room would that guest need?
Choose an example
Constructed example: the chapter's hotel with rooms 1, 2, 3, ... and its rule n -> n + 1, beside a 6-room hotel for contrast.
Calculated values
- Rule
- old guest n moves to room n + 1
- Rooms in the hotel
- 1, 2, 3, ... with no last room
- Who is in room 1
- the new guest (1 - 1 = 0 is not a label)
- Does everyone fit?
- every original guest has a room
Old guest 5 moves to 5 + 1 = 6. Room 1: no original guest, because n + 1 = 1 needs n = 0, which is not a guest label. The new guest takes it. Every original guest n has room n + 1, and no two collide. Nothing is evicted, because there is no last guest.
Use the idea
When someone says a list is full, ask whether it has a last item. The shift only works when there is no last label to push off the end.
Where the conclusion applies
Rooms are labels 1, 2, 3, ... with no last one, and every guest moves at once by the same rule. A real building, or any device with a fixed number of slots, has a last room, and the rule fails there.
Check your understanding: Under the rule n -> n + 1, which original guest ends up in room 40?
Chapter 26 source: section "One new guest". Demonstration C26-D01.
Demonstration 2 of 4
Any finite number of new guests
If k new guests arrive at once, which rooms open up for them?
Shifting by k leaves rooms 1 through k free. Old guests only use rooms above k, so they never collide with the new guests, and a + k = b + k forces a = b.
\[k + 1, k + 2, k + 3, ...\]
\[a + k = b + k\]
k is the fixed number of new guests. Old guest n moves to room n + k, so old guests land in rooms k + 1, k + 2, k + 3, and onward. New guest j takes room j, for j from 1 to k.
Predict first. With k = 7, which old guest lands in room 12?
Choose an example
Constructed example: the chapter's shift n -> n + k with k = 4 from its practice set and other small values.
Calculated values
- New guests k
- 4
- Rule
- old guest n moves to room n + 4
- Open rooms for new guests
- rooms 1 through 4
- Old guest 1 goes to
- room 1 + 4 = 5
- Room 12 holds
- old guest 12 - 4 = 8
With k = 4, old guest 3 moves to 3 + 4 = 7 and old guest 1 to 1 + 4 = 5. New guest j takes room j, so rooms 1 through 4 hold the 4 new guests. If a + 4 = b + 4 then a = b, so no two old guests collide, and room r of at least 5 came from old guest r - 4.
Use the idea
When records need room at the front of a numbered list, shift every label by the same amount and check one room on each side of the boundary.
Where the conclusion applies
k is stated first and stays fixed. The rule is written after k is known; no single shift handles every possible k at once.
Check your understanding: If k = 4, which rooms are free, and where does old guest 3 go?
Chapter 26 source: section "Any finite number of new guests". Demonstration C26-D02.
Demonstration 3 of 4
A whole second list of guests
A second endless list of guests arrives. How can everyone get exactly one room?
Old guests take the even rooms and new guests take the odd rooms. Working backward, even room r holds old guest r/2 and odd room r holds new guest (r + 1)/2.
\[r/2\]
\[(r + 1)/2\]
Old guest n goes to room 2n and new guest m goes to room 2m - 1. Working backward, even room r holds old guest r/2 and odd room r holds new guest (r + 1)/2. Both lists start at 1, so the tags old and new stay with each number.
Predict first. Switch to the plan that sends both lists to room 2n. What happens in room 8, and what happens in room 17?
Choose an example
Constructed example: the chapter's even and odd room rules, and the broken plan from its find and repair exercise.
Calculated values
- Old guest n goes to
- room 2n
- New guest m goes to
- room 2m - 1
- Room 17 holds
- new guest (17 + 1)/2 = 9
- Problem
- none
Room 17 is odd, so it holds new guest (17 + 1)/2 = 9. Check: 2 x 9 - 1 = 17. Old guests fill every even room and new guests fill every odd room. No number is both even and odd, so nobody collides, and every room is used.
Use the idea
When two numbered lists are merged, give each list its own class of numbers, then test the merge by recovering where any combined number came from.
Where the conclusion applies
Both lists are countably infinite and labeled 1, 2, 3, .... The split works because every whole number is even or odd and never both. A plan that sends both lists to the same class collides.
Check your understanding: In the even and odd plan, who is in room 31?
Chapter 26 source: section "A countably infinite new group". Demonstration C26-D03.
Demonstration 4 of 4
Why 0.999... is 1
Every truncation of 0.999... is below 1. Does that leave a tiny gap at the end?
Each extra nine makes the gap ten times smaller. Any positive gap you name is beaten at some stage and stays beaten, so no positive gap survives the endless string.
\[1-10^{-n}\]
\[0.999\ldots=1\]
n counts the nines written so far. The truncation with n nines equals 1 - 10 to the power -n, so its gap from 1 is one tenth, one hundredth, one thousandth, and so on.
Predict first. Ask for a gap below 0.00001. How many nines do you need, and is 3 nines enough?
Choose an example
Constructed example: the chapter's truncations 0.9, 0.99 and 0.999, extended a few stages, with requested gaps chosen for the reader.
Calculated values
- Truncation
- 0.999
- Gap from 1
- 1 - 0.999 = 0.001
- Below 1?
- yes, every finite truncation is
- Gap below 0.01?
- yes
- First stage with gap below 0.01
- 3 nines (gap 0.001)
- Infinite decimal 0.999...
- names the number 1; there is no final nine
With 3 nines, 1 - 0.999 = 0.001, which is 10 to the power -3. That is already below the requested 0.01. Any positive gap you name is beaten by enough nines, and every later truncation stays below it. No positive gap survives the endless string, so 0.999... = 1.
Use the idea
When two digit strings look different, check the numbers they name before deciding they are different numbers, as with 0.4999... and 0.5.
Where the conclusion applies
An infinite decimal names the limit of its finite truncations. Only finite stages are drawn. Reading 0.999... as a list that stops somewhere would bring the gap back, and that misreads the notation.
Check your understanding: How many nines make the gap from 1 smaller than 0.0001?
Chapter 26 source: section "Why two decimal strings can name one number". Demonstration C26-D04.