Problem 111111 worked answer

Can There Be Two Ones?

Designed for Grade 9 · Grade 10 · Grade 11 · Grade 12 · Uses HS DISC math · About 8–20 minutes

Complete to collectJade CabochonLevel 3

At a glance

Part 1: Only $𝑀 1$ is the identity machine. The four values of $𝑀 𝑠 (1)$ are $1, 4, 11, 14$ .
Part 2: For example, $1 \to 4 \to 1$ through $𝑀 4$ . One pass changes the input, while two passes return it.
Part 3: If $𝐴$ and $𝐵$ are both identities, then $𝐴 ⋆ 𝐵 = 𝐵$ because $𝐴$ is an identity and $𝐴 ⋆ 𝐵 = 𝐴$ because $𝐵$ is an identity. Therefore $𝐴 = 𝐵$ .

Key idea: returning is not standing still

A self-undoing action may move an object and then reverse that move. An identity action never moves the object at all. The distinction explains how several elements can square to the identity even though the identity itself is unique.

1. Find the identity machine

For input $1$ , multiplying by $𝑠$ gives

𝑀 𝑠 (1) \equiv 𝑠 (m o d 15) .

The completed table is:

$𝑠$	$𝑀 𝑠 (1)$	Identity machine?
1	1	yes
4	4	no
11	11	no
14	14	no

The machine $𝑀 1$ works for every input because

𝑀 1 (𝑥) \equiv 1 𝑥 \equiv 𝑥 (m o d 15) .

Each of the other three machines changes the particular input $1$ . Therefore none of them can leave every input unchanged. This both constructs a working identity and excludes every other candidate in the table.

2. Show the difference between the promises

One valid example uses $𝑀 4$ and begins at $1$ :

1 \to 4 \to 16 \equiv 1 (m o d 15) .

The first pass changes $1$ to $4$ , so $𝑀 4$ is not the identity. The second pass returns $4$ to $1$ , so $𝑀 4$ is self-undoing.

Other valid examples include $1 \to 11 \to 1$ through $𝑀 11$ and $1 \to 14 \to 1$ through $𝑀 14$ .

3. Prove that an identity is unique

Suppose $𝐴$ and $𝐵$ both satisfy

𝐼 ⋆ 𝑋 = 𝑋 ⋆ 𝐼 = 𝑋

whenever $𝐼$ is replaced by either $𝐴$ or $𝐵$ .

Because $𝐴$ is an identity, it leaves $𝐵$ unchanged when placed on the left:

𝐴 ⋆ 𝐵 = 𝐵 .

Because $𝐵$ is an identity, it leaves $𝐴$ unchanged when placed on the right:

𝐴 ⋆ 𝐵 = 𝐴 .

The same object $𝐴 ⋆ 𝐵$ equals both $𝐴$ and $𝐵$ . Therefore

𝐴 = 𝐵 .

This proves that an operation can have at most one two-sided identity. The proof did not assume associativity, commutativity, numbers, multiplication, or inverses.

Check

The concrete dial agrees with the abstract theorem: four multiplier machines are self-undoing, but the one-pass identity test leaves exactly one candidate, $𝑀 1$ .

Technical fit and rating

MJ HS:DISC.5 · C3 · W2 Stretch challenge · Moderate workload

Algebraic structure and equations

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