If 0 . Which of the following must be less than 1. (r/s rs s - r

1. Translate the problem requirements: We have two numbers where \(\mathrm{r}\) is between 0 and 1 (like 0.5), and \(\mathrm{s}\) is between 1 and 2 (like 1.5). We need to determine which of three expressions (\(\mathrm{r/s}\), \(\mathrm{rs}\), \(\mathrm{s-r}\)) must always be less than 1, regardless of the specific values we choose for \(\mathrm{r}\) and \(\mathrm{s}\) within their given ranges.
2. Test boundary behavior with strategic values: Choose values that push each expression toward its maximum possible value to see if it can reach or exceed 1.
3. Analyze each expression systematically: For each of the three expressions, determine whether it's possible for the expression to equal or exceed 1, or whether it must always stay below 1.
4. Eliminate options based on counterexamples: Use specific examples to show which expressions can actually reach or exceed 1, thus eliminating them from consideration.

Execution of Strategic Approach

1. Translate the problem requirements

Let's understand what we have here. We're told that \(\mathrm{r}\) is a number between 0 and 1, so \(\mathrm{r}\) could be something like 0.3, 0.7, or 0.9. We're also told that \(\mathrm{s}\) is a number between 1 and 2, so \(\mathrm{s}\) could be something like 1.2, 1.5, or 1.8.

We need to figure out which of these three expressions will always be less than 1, no matter what specific values we pick for \(\mathrm{r}\) and \(\mathrm{s}\):

• I. \(\mathrm{r/s}\) (\(\mathrm{r}\) divided by \(\mathrm{s}\))
• II. \(\mathrm{rs}\) (\(\mathrm{r}\) times \(\mathrm{s}\))
• III. \(\mathrm{s - r}\) (\(\mathrm{s}\) minus \(\mathrm{r}\))

The key word here is "must" - we're looking for expressions that are guaranteed to be less than 1.

Process Skill: TRANSLATE - Converting the constraint language into concrete understanding

2. Test boundary behavior with strategic values

To really test these expressions, let's be smart about our choices. We want to see if any of these expressions can get close to 1 or even exceed 1. So let's pick values that might push each expression to its limits.

For trying to make expressions as large as possible:

• Choose \(\mathrm{r}\) close to 1 (like \(\mathrm{r = 0.9}\))
• Choose \(\mathrm{s}\) close to 1 (like \(\mathrm{s = 1.1}\)) or close to 2 (like \(\mathrm{s = 1.9}\)) depending on the expression

Let's test with \(\mathrm{r = 0.9}\) and \(\mathrm{s = 1.1}\) first:

• I. \(\mathrm{r/s = 0.9/1.1 ≈ 0.82}\)
• II. \(\mathrm{rs = 0.9 × 1.1 = 0.99}\)
• III. \(\mathrm{s - r = 1.1 - 0.9 = 0.2}\)

Now let's try \(\mathrm{r = 0.9}\) and \(\mathrm{s = 1.9}\):

• I. \(\mathrm{r/s = 0.9/1.9 ≈ 0.47}\)
• II. \(\mathrm{rs = 0.9 × 1.9 = 1.71}\)
• III. \(\mathrm{s - r = 1.9 - 0.9 = 1.0}\)

Interesting! We're already seeing that expression II can exceed 1, and expression III can equal 1.

3. Analyze each expression systematically

Now let's think about each expression more carefully:

Expression I: \(\mathrm{r/s}\)

Since \(\mathrm{r}\) is less than 1 and \(\mathrm{s}\) is greater than 1, we're dividing a number smaller than 1 by a number larger than 1. This will always give us something smaller than the original \(\mathrm{r}\), which was already less than 1. So \(\mathrm{r/s}\) must always be less than 1.

Expression II: \(\mathrm{rs}\)

Here we're multiplying \(\mathrm{r}\) (less than 1) by \(\mathrm{s}\) (greater than 1). As we saw in our example, when \(\mathrm{r = 0.9}\) and \(\mathrm{s = 1.9}\), we get \(\mathrm{rs = 1.71}\), which is definitely greater than 1. So expression II doesn't always stay below 1.

Expression III: \(\mathrm{s - r}\)

We're subtracting \(\mathrm{r}\) (less than 1) from \(\mathrm{s}\) (greater than 1). In our example with \(\mathrm{r = 0.9}\) and \(\mathrm{s = 1.9}\), we got exactly 1.0. Since \(\mathrm{s}\) can be as large as 1.99... and \(\mathrm{r}\) can be as small as 0.01..., this difference can definitely be 1 or larger. So expression III doesn't always stay below 1.

Process Skill: CONSIDER ALL CASES - Testing boundary conditions to find counterexamples

4. Eliminate options based on counterexamples

Let's confirm our counterexamples:

For Expression II: Choose \(\mathrm{r = 0.8}\) and \(\mathrm{s = 1.5}\)

\(\mathrm{rs = 0.8 × 1.5 = 1.2 > 1}\) ✗

For Expression III: Choose \(\mathrm{r = 0.1}\) and \(\mathrm{s = 1.9}\)

\(\mathrm{s - r = 1.9 - 0.1 = 1.8 > 1}\) ✗

For Expression I: Let's verify it's always less than 1

Since \(0 < \mathrm{r} < 1\) and \(1 < \mathrm{s} < 2\), we have \(\mathrm{r < 1 < s}\), so \(\mathrm{r/s < 1/s < 1/1 = 1}\) ✓

Therefore, only Expression I must always be less than 1.

5. Final Answer

Only expression I (\(\mathrm{r/s}\)) must always be less than 1, while expressions II and III can equal or exceed 1 under certain conditions within the given constraints.

The answer is A. I only.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding what "must be less than 1" means: Students often confuse "must be less than 1" with "can be less than 1." They may think they need to find expressions that are sometimes less than 1, rather than expressions that are ALWAYS less than 1 for all valid values of \(\mathrm{r}\) and \(\mathrm{s}\). This leads them to incorrectly include expressions II and III.

2. Not recognizing the need to test boundary cases: Students may only test "nice" middle values like \(\mathrm{r = 0.5}\) and \(\mathrm{s = 1.5}\), missing the critical insight that they need to test extreme cases within the constraints (like \(\mathrm{r}\) close to 1 and \(\mathrm{s}\) close to 2) to see if expressions can approach or exceed 1.

3. Misinterpreting the constraint notation: Students may misread \(0 < \mathrm{r} < 1 < \mathrm{s} < 2\) and think that \(\mathrm{r}\) and \(\mathrm{s}\) can actually equal the boundary values (\(\mathrm{r = 1}\) or \(\mathrm{s = 1}\)), rather than understanding these are strict inequalities where \(\mathrm{r}\) must be less than 1 and \(\mathrm{s}\) must be greater than 1.

Errors while executing the approach

1. Arithmetic errors in testing examples: When calculating expressions like \(\mathrm{rs = 0.9 × 1.9 = 1.71}\) or \(\mathrm{s - r = 1.9 - 0.9 = 1.0}\), students may make basic multiplication or subtraction errors that lead them to incorrect conclusions about whether expressions exceed 1.

2. Not testing sufficient examples: Students may test only one set of values and conclude incorrectly. For instance, testing only \(\mathrm{r = 0.9, s = 1.1}\) gives \(\mathrm{rs = 0.99 < 1}\), leading them to think expression II is always less than 1, without testing larger values of \(\mathrm{s}\).

3. Incorrect algebraic reasoning: When trying to prove that \(\mathrm{r/s < 1}\), students may make logical errors in their inequality manipulations or fail to properly use the given constraints that \(\mathrm{r < 1}\) and \(\mathrm{s > 1}\).

Errors while selecting the answer

1. Confusing which expressions were proven vs. disproven: After finding that expressions II and III can exceed 1 in some cases, students may accidentally select answer choices that include these expressions, forgetting that they need expressions that are ALWAYS less than 1.

2. Misreading the answer choices: Students may correctly identify that only expression I works, but then accidentally select "E. I and III" instead of "A. I only" due to careless reading of the multiple choice options.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose strategic smart numbers within the given constraints

Since \(0 < \mathrm{r} < 1 < \mathrm{s} < 2\), let's choose values that will help us test each expression effectively:

• Let \(\mathrm{r = 0.8}\) (close to 1 to maximize expressions involving \(\mathrm{r}\))
• Let \(\mathrm{s = 1.9}\) (close to 2 to maximize expressions involving \(\mathrm{s}\))

These values are chosen strategically because they push the expressions toward their maximum possible values within the given constraints.

Step 2: Test each expression with our smart numbers

Expression I: \(\mathrm{r/s}\)

\(\mathrm{r/s = 0.8/1.9 ≈ 0.42}\)

Since \(\mathrm{r < 1}\) and \(\mathrm{s > 1}\), we have \(\mathrm{r/s}\) = (something less than 1)/(something greater than 1), which must be less than 1.

Expression II: \(\mathrm{rs}\)

\(\mathrm{rs = 0.8 × 1.9 = 1.52}\)

This is greater than 1! Let's verify this makes sense: when \(\mathrm{r}\) is close to 1 and \(\mathrm{s}\) is close to 2, their product approaches \(1 × 2 = 2\).

Expression III: \(\mathrm{s - r}\)

\(\mathrm{s - r = 1.9 - 0.8 = 1.1}\)

This is also greater than 1! This makes sense because \(\mathrm{s}\) can be close to 2 while \(\mathrm{r}\) can be close to 0, making the difference approach 2.

Step 3: Verify our findings with boundary analysis

To confirm Expression I always stays below 1, note that no matter what values we choose:

• \(\mathrm{r < 1}\) (numerator is always less than 1)
• \(\mathrm{s > 1}\) (denominator is always greater than 1)
• Therefore \(\mathrm{r/s < 1/1 = 1}\)

Step 4: Conclusion

Using our smart numbers, we've shown that:

• Expression I (\(\mathrm{r/s}\)) must always be less than 1
• Expression II (\(\mathrm{rs}\)) can exceed 1
• Expression III (\(\mathrm{s-r}\)) can exceed 1

Therefore, only I must be less than 1.

Answer: A