Problem 111 worked answer

The Missing Rung

Designed for Grade 8 · Grade 9 · Uses Late Grade 8 math · About 12–28 minutes

Complete to collectPolished Smoky QuartzLevel 3

At a glance

Part 1: The missing rung is $𝑃 0 = 1$ . Dividing by $𝑎$ is legal only because $𝑎 \neq 0$ .
Part 2: For every nonzero real $𝑎$ , $𝑎 0 = 1$ , and both evaluations of $𝑎 111 / 𝑎 111$ give $1$ .
Part 3: The product of an empty block is $1$ , the unique value that leaves every other block product unchanged.

Key idea: multiplication needs a starting value

Repeated addition begins at $0$ because adding $0$ changes nothing. Repeated multiplication begins at $1$ because multiplying by $1$ changes nothing.

The zero in $𝑎 0$ counts copies of the factor $𝑎$ . It does not say that the value of the product is zero. With no copies present, the product must begin at the multiplicative identity.

1. Complete the ladder

The neighboring-rung rule is

𝑃 𝑛 + 1 = 𝑎 𝑃 𝑛 .

At $𝑛 = 0$ , this requires

𝑃 1 = 𝑎 𝑃 0 .

The worksheet defines $𝑃 1 = 𝑎$ , so

𝑎 = 𝑎 𝑃 0 .

Because $𝑎 \neq 0$ , divide both sides by $𝑎$ :

𝑃 0 = 1 .

This is the only possible value. If $𝑎 = 0$ , the equation would become

0 = 0 \cdot 𝑃 0,

which is true for every value of $𝑃 0$ and therefore does not determine a unique missing rung. That is exactly where the nonzero condition is used.

2. Interpret the zero exponent

Since $𝑃 𝑛 = 𝑎 𝑛$ , the missing rung gives

𝑎 0 = 𝑃 0 = 1

for nonzero $𝑎$ .

Now evaluate the displayed quotient as an ordinary division. Since $𝑎 \neq 0$ , its power $𝑎 111$ is also nonzero, so

𝑎 111 𝑎 111 = 1 .

The exponent-division rule gives the same result:

𝑎 111 𝑎 111 = 𝑎 111 - 111 = 𝑎 0 = 1 .

The ladder and quotient are two views of the same required identity value.

3. Find the empty product

Let $𝑃 \emptyset$ mean the product of the empty block. Suppose the other block has product $𝑞$ . The splitting rule requires

𝑞 = 𝑃 \emptyset 𝑞 .

To force the value, choose a one-factor block containing $1$ , whose product is $𝑞 = 1$ . Then

1 = 𝑃 \emptyset \cdot 1,

𝑃 \emptyset = 1 .

Conversely, this value always works because

1 \cdot 𝑞 = 𝑞

for every block product $𝑞$ . Thus $1$ is both necessary and sufficient, and no other value can serve as the empty product.

Check

Insert $𝑃 0 = 1$ into the bottom neighboring-rung rule:

𝑎 𝑃 0 = 𝑎 \cdot 1 = 𝑎 = 𝑃 1 .

It fits the ladder. It also preserves exponent multiplication at the boundary,

𝑎 𝑚 𝑎 0 = 𝑎 𝑚 \cdot 1 = 𝑎 𝑚 = 𝑎 𝑚 + 0,

and preserves every split product with an empty block.

Boundary note

This worksheet does not assign a universal value to $00$ . The ladder cancellation and quotient check deliberately assume $𝑎 \neq 0$ . Some discrete contexts define $00$ as $1$ , while other contexts leave it undefined or treat a related limiting form as indeterminate. None of those conventions is needed for the exact claims proved here.

Technical fit and rating

MJ 8.9 · C3 · W2 Stretch challenge · Moderate workload

Number and quantity

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