Problem 37 worked answer

The Safe Zone

Designed for Grade 9 · Grade 10 · Uses HS DISC math · About 11–28 minutes

Complete to collectEmerald CrystalLevel 2

At a glance

The first values are $𝑦 1 = 1$ , $𝑦 2 = 54$ , and $𝑦 3 = 8964$ . More importantly, any current value in $[0, 2]$ produces a next value in $[1, 2]$ , which is contained in $[0, 2]$ . Since $𝑦 0 = 0$ starts inside, every later value stays inside. The smaller interval $[0, 32]$ fails because the allowed input $32$ produces $2516 > 32$ .

Key idea

Calculating several examples can suggest a safe zone, but it cannot check infinitely many steps. The proof has two parts: verify that the starting value is inside, and prove the if-then statement that every value inside produces another value inside. Those two finite checks control the whole recurrence.

Worked solution, part by part

1. Calculate three values

𝑦 1 = 1 + 024 = 1, 𝑦 2 = 1 + 124 = 54, 𝑦 3 = 1 + (5 / 4) 2 4 = 1 + 2564 = 8964 .

All three values lie in $[0, 2]$ , so the examples support the conjecture but do not yet prove it.

2. Test every current value in the interval

Assume $0 \leq 𝑦 𝑛 \leq 2$ . Squaring preserves these nonnegative bounds:

0 \leq 𝑦 2 𝑛 \leq 4 .

Dividing by 4 and then adding 1 gives

0 \leq 𝑦 2 𝑛 4 \leq 1 .

Therefore,

1 \leq 𝑦 𝑛 + 1 \leq 2 .

Every number in $[1, 2]$ is also in $[0, 2]$ , so the rule cannot send a current value out of the proposed safe zone.

3. Control all later values

The starting value $𝑦 0 = 0$ is in $[0, 2]$ . Part 2 says that whenever one value is in $[0, 2]$ , the next value is also in $[0, 2]$ . Apply that statement repeatedly: $𝑦 1$ is inside, then $𝑦 2$ is inside, and so on. Therefore every $𝑦 𝑛$ is in $[0, 2]$ .

4. Test the smaller interval

The input $32$ is allowed by $[0, 32]$ , but

1 + (3 / 2) 2 4 = 1 + 916 = 2516 .

Since $2516 > 2416 = 32$ , this one allowed input is sent outside. Thus $[0, 32]$ is not invariant under the rule.

Proof and completeness

The proof for $[0, 2]$ covers every possible current value by starting with an arbitrary $𝑦 𝑛$ satisfying the interval bounds. The counterexample for $[0, 32]$ needs only one allowed input, because a single failure disproves the claim that every input stays inside.

Check

The general bound predicts $1 \leq 𝑦 𝑛 + 1 \leq 2$ , and all three computed non-starting values satisfy it. The endpoint test for the smaller interval gives $1.5625$ , consistent with the exact fraction $2516$ .

Technical fit and rating

MJ HS:DISC.4 · C2 · W2 Standard challenge · Moderate workload

Discrete mathematics and logic

How the rating works →

Worksheet arc · Guided Practice · 3 of 5 on Complete arc

Edge of Escape

Open the proof branch by developing invariant intervals and distinguishing examples from a universal argument.