If lfloor z rfloor denotes the greatest integer less than or equal to z and lfloor z rfloor = -1...

1. Translate the problem requirements: Understand that \(\lfloor z \rfloor\) represents the floor function (greatest integer less than or equal to z), and we need to find which statement must be true when \(\lfloor z \rfloor = -1\)
2. Identify the range of z values: Determine what values of z would produce \(\lfloor z \rfloor = -1\) by understanding how the floor function works with negative numbers
3. Test boundary conditions: Check specific values at the boundaries to confirm our range is correct
4. Match against answer choices: Compare our derived range with the given options to select the correct statement

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what this problem is asking in plain English. The symbol \(\lfloor z \rfloor\) represents something called the "floor function" or "greatest integer function." Think of it this way: if you have any number z, then \(\lfloor z \rfloor\) gives you the largest whole number that is less than or equal to z.

For example:
• If \(z = 3.7\), then \(\lfloor z \rfloor = 3\) (because 3 is the largest integer ≤ 3.7)
• If \(z = 5\), then \(\lfloor z \rfloor = 5\) (because 5 itself is the largest integer ≤ 5)
• If \(z = -1.3\), then \(\lfloor z \rfloor = -2\) (because -2 is the largest integer ≤ -1.3)

The problem tells us that \(\lfloor z \rfloor = -1\), and we need to figure out what this tells us about the possible values of z.

Process Skill: TRANSLATE - Converting the floor function notation into understandable language

2. Identify the range of z values

Now let's think about this step by step. If \(\lfloor z \rfloor = -1\), that means -1 is the greatest integer that is less than or equal to z.

Let's consider what this means:
• Since -1 is the greatest integer ≤ z, we know that z must be at least -1. If z were less than -1, then the greatest integer ≤ z would be -2, not -1.
• But z also can't be too big. If z were 0 or bigger, then the greatest integer ≤ z would be 0 or bigger, not -1.

Let's test some specific values to see the pattern:
• If \(z = -1\), then \(\lfloor z \rfloor = -1\) ✓ (this works)
• If \(z = -0.5\), then \(\lfloor z \rfloor = -1\) ✓ (this works)
• If \(z = -0.1\), then \(\lfloor z \rfloor = -1\) ✓ (this works)
• If \(z = 0\), then \(\lfloor z \rfloor = 0\) ✗ (this doesn't work)
• If \(z = -1.1\), then \(\lfloor z \rfloor = -2\) ✗ (this doesn't work)

From this pattern, we can see that z must be greater than or equal to -1, but less than 0.

Process Skill: INFER - Drawing conclusions about the range from the floor function definition

3. Test boundary conditions

Let's carefully check the boundaries to make sure we have the right range:

At \(z = -1\):
\(\lfloor z \rfloor = \lfloor -1 \rfloor = -1\) ✓ This works perfectly

As z approaches 0 from below (like \(z = -0.0001\)):
\(\lfloor z \rfloor = \lfloor -0.0001 \rfloor = -1\) ✓ This still works

At \(z = 0\):
\(\lfloor z \rfloor = \lfloor 0 \rfloor = 0\) ✗ This gives us 0, not -1

As z goes below -1 (like \(z = -1.0001\)):
\(\lfloor z \rfloor = \lfloor -1.0001 \rfloor = -2\) ✗ This gives us -2, not -1

This confirms our range: z must be greater than or equal to -1, and less than 0.
In mathematical notation: \(-1 ≤ z < 0\)

Process Skill: CONSIDER ALL CASES - Testing boundary values to ensure our range is complete and accurate

4. Match against answer choices

Now let's look at our answer choices and see which one matches our derived range of \(-1 ≤ z < 0\):

1. \(z = -1\) → This is too restrictive; z can be other values like -0.5
2. \(-2 ≤ z < -1\) → This is wrong; if \(z = -1.5\), then \(\lfloor z \rfloor = -2\), not -1
3. \(-2 < z ≤ -1\) → This is close but wrong; if \(z = -1.0001\), then \(\lfloor z \rfloor = -2\), not -1
4. \(-1 ≤ z < 0\) → This matches exactly what we found!
5. \(-1 < z ≤ 0\) → This excludes \(z = -1\), but we know \(\lfloor -1 \rfloor = -1\) works

Final Answer

The correct answer is D: \(-1 ≤ z < 0\)

This range captures exactly when \(\lfloor z \rfloor = -1\): z can be -1 itself, or any value between -1 and 0 (but not including 0).

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the floor function definition

Students often confuse the floor function \(\lfloor z \rfloor\) with rounding or absolute value. They might think \(\lfloor z \rfloor = -1\) means z is close to -1, rather than understanding that \(\lfloor z \rfloor\) gives the greatest integer less than or equal to z. This leads to incorrect reasoning about which values of z are possible.

2. Confusion about "greatest integer less than or equal to"

Students may misinterpret "greatest integer less than or equal to z" as "smallest integer greater than or equal to z" (which is actually the ceiling function). This fundamental misunderstanding of the floor function direction leads to completely wrong ranges.

3. Not recognizing the need to find a range

Some students might think that if \(\lfloor z \rfloor = -1\), then z must exactly equal -1, failing to realize that multiple values of z can produce the same floor function result. They don't understand that they need to find all possible values of z, not just one specific value.

Errors while executing the approach

1. Incorrect boundary analysis

When testing boundary values, students often make errors about whether endpoints should be included or excluded. For example, they might incorrectly conclude that \(z = 0\) should be included because they test \(z = -0.001\) and see it works, without realizing that exactly \(z = 0\) gives \(\lfloor z \rfloor = 0\), not -1.

2. Testing insufficient or wrong test values

Students might only test integer values like \(z = -2, -1, 0\) and miss the crucial insight that non-integer values between -1 and 0 (like -0.5, -0.1) also satisfy \(\lfloor z \rfloor = -1\). This leads to incomplete understanding of the full range.

3. Misapplying the floor function to negative numbers

A common error is incorrectly computing the floor function for negative decimals. Students might think \(\lfloor -1.5 \rfloor = -1\) instead of -2, because they don't understand that for negative numbers, the floor function moves further away from zero, not closer to it.

Errors while selecting the answer

1. Confusing inequality symbols

Even after correctly determining that z ranges from -1 (inclusive) to 0 (exclusive), students often mix up the inequality notation. They might choose answer C (\(-2 < z ≤ -1\)) instead of answer D (\(-1 ≤ z < 0\)), confusing which endpoint should be included or excluded.

2. Selecting the most restrictive correct statement

Students might choose answer A (\(z = -1\)) because they verify that \(z = -1\) works, without recognizing that the question asks for what "must be true" given the condition. They don't understand that while \(z = -1\) is possible, it's not the complete answer since other values like \(z = -0.5\) also work.

Alternate Solutions

Smart Numbers Approach

Strategy: Test specific values systematically to discover the range where \(\lfloor z \rfloor = -1\)

Step 1: Test boundary and key values
Let's test specific values to see when \(\lfloor z \rfloor = -1\):

• \(z = -2\): \(\lfloor z \rfloor = \lfloor -2 \rfloor = -2\) (too small)
• \(z = -1.5\): \(\lfloor z \rfloor = \lfloor -1.5 \rfloor = -2\) (still too small)
• \(z = -1\): \(\lfloor z \rfloor = \lfloor -1 \rfloor = -1\) ✓ (works!)
• \(z = -0.5\): \(\lfloor z \rfloor = \lfloor -0.5 \rfloor = -1\) ✓ (works!)
• \(z = -0.1\): \(\lfloor z \rfloor = \lfloor -0.1 \rfloor = -1\) ✓ (works!)
• \(z = 0\): \(\lfloor z \rfloor = \lfloor 0 \rfloor = 0\) (too large)

Step 2: Identify the pattern
From our testing:

• Values less than -1 give \(\lfloor z \rfloor = -2\)
• Values from -1 up to (but not including) 0 give \(\lfloor z \rfloor = -1\)
• Values of 0 and above give \(\lfloor z \rfloor ≥ 0\)

Step 3: Determine the exact range
Our testing shows \(\lfloor z \rfloor = -1\) when z is in the range [-1, 0), which means:
\(-1 ≤ z < 0\)

Step 4: Verify boundaries

• At \(z = -1\): \(\lfloor z \rfloor = -1\) ✓ (included)
• As z approaches 0 from below: \(\lfloor z \rfloor = -1\) ✓
• At \(z = 0\): \(\lfloor z \rfloor = 0\) ✗ (not included)

Answer: D. \(-1 ≤ z < 0\)