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If \(\mathrm{m} < \mathrm{p} < \mathrm{r} < \mathrm{t} < \mathrm{z}\) and if the product \(\mathrm{mprtz}\) is positive, which of the following products must be positive?
Let's break down what we know in simple terms:
We have five numbers: m, p, r, t, and z
These numbers are arranged from smallest to largest: \(\mathrm{m} < \mathrm{p} < \mathrm{r} < \mathrm{t} < \mathrm{z}\)
Think of it like five people standing in a line by height - m is the shortest, z is the tallest.
The key constraint is that when we multiply all five numbers together (\(\mathrm{m} \times \mathrm{p} \times \mathrm{r} \times \mathrm{t} \times \mathrm{z}\)), we get a positive result.
We need to figure out which of these three products MUST be positive:
• Product I: \(\mathrm{m} \times \mathrm{p}\) (the two smallest numbers)
• Product II: \(\mathrm{r} \times \mathrm{t}\) (the two middle numbers)
• Product III: \(\mathrm{t} \times \mathrm{z}\) (the two largest numbers)
Process Skill: TRANSLATE - Converting the inequality and product constraints into manageable conditions
Now let's think about what makes a product positive. In everyday terms:
• When you multiply two positive numbers, you get positive
• When you multiply two negative numbers, you also get positive
• When you multiply a positive and negative number, you get negative
Since we're multiplying five numbers and getting a positive result, we need either:
• All five numbers are positive, OR
• An even number of the five numbers are negative
This means we could have 0, 2, or 4 negative numbers among our five variables.
But here's the crucial insight: since the numbers are in order from smallest to largest, if any number is negative, then all numbers to its left must also be negative!
For example, if r is negative, then m and p (which are smaller) must also be negative.
Given the ordering constraint (\(\mathrm{m} < \mathrm{p} < \mathrm{r} < \mathrm{t} < \mathrm{z}\)), let's consider all possible sign patterns that would make \(\mathrm{mprtz}\) positive:
Case 1: All numbers positive
\(\mathrm{m}(+), \mathrm{p}(+), \mathrm{r}(+), \mathrm{t}(+), \mathrm{z}(+)\) → Product is positive ✓
Case 2: Exactly 2 numbers negative (must be the leftmost ones)
\(\mathrm{m}(-), \mathrm{p}(-), \mathrm{r}(+), \mathrm{t}(+), \mathrm{z}(+)\) → Product is positive ✓
Case 3: Exactly 4 numbers negative (must be the leftmost ones)
\(\mathrm{m}(-), \mathrm{p}(-), \mathrm{r}(-), \mathrm{t}(-), \mathrm{z}(+)\) → Product is positive ✓
We cannot have 1, 3, or 5 negative numbers because that would make the overall product negative.
Process Skill: CONSIDER ALL CASES - Systematically examining all valid sign combinations
Now let's check each product in all valid cases:
Testing Product I (\(\mathrm{mp}\)):
• Case 1: \(\mathrm{m}(+) \times \mathrm{p}(+) = \text{positive}\)
• Case 2: \(\mathrm{m}(-) \times \mathrm{p}(-) = \text{positive}\)
• Case 3: \(\mathrm{m}(-) \times \mathrm{p}(-) = \text{positive}\)
Product I is ALWAYS positive ✓
Testing Product II (\(\mathrm{rt}\)):
• Case 1: \(\mathrm{r}(+) \times \mathrm{t}(+) = \text{positive}\)
• Case 2: \(\mathrm{r}(+) \times \mathrm{t}(+) = \text{positive}\)
• Case 3: \(\mathrm{r}(-) \times \mathrm{t}(-) = \text{positive}\)
Product II is ALWAYS positive ✓
Testing Product III (\(\mathrm{tz}\)):
• Case 1: \(\mathrm{t}(+) \times \mathrm{z}(+) = \text{positive}\)
• Case 2: \(\mathrm{t}(+) \times \mathrm{z}(+) = \text{positive}\)
• Case 3: \(\mathrm{t}(-) \times \mathrm{z}(+) = \text{negative}\) ✗
Product III is NOT always positive
Process Skill: APPLY CONSTRAINTS - Using the ordering relationship to determine which sign patterns are possible
Products I (\(\mathrm{mp}\)) and II (\(\mathrm{rt}\)) must always be positive, while Product III (\(\mathrm{tz}\)) can be negative in some valid scenarios.
Therefore, the answer is C. I and II only.
Faltering Point 1: Misunderstanding the ordering constraint's impact on sign patterns
Students often fail to recognize that the ordering constraint (\(\mathrm{m} < \mathrm{p} < \mathrm{r} < \mathrm{t} < \mathrm{z}\)) means that if any number is negative, ALL numbers to its left must also be negative. They might think numbers can have random signs regardless of their position in the ordering. For example, they might consider a scenario like \(\mathrm{m}(+), \mathrm{p}(-), \mathrm{r}(+), \mathrm{t}(-), \mathrm{z}(+)\), which violates the ordering principle since p cannot be negative if m is positive when \(\mathrm{p} > \mathrm{m}\).
Faltering Point 2: Not systematically identifying all valid sign combinations
Students may jump into testing products without first determining all possible sign patterns that make \(\mathrm{mprtz}\) positive. They might miss that only 0, 2, or 4 negative numbers are allowed (even count for positive product), or they might not realize these negative numbers must be consecutive starting from the leftmost position due to the ordering constraint.
Faltering Point 3: Incorrectly applying sign rules for multiplication
Students sometimes confuse the basic multiplication rules, especially when dealing with negative numbers. They might incorrectly think that multiplying two negative numbers gives a negative result, or they might make errors when systematically checking each product across different cases. This leads to wrong conclusions about which products are always positive.
Faltering Point 4: Testing products in isolation rather than across all valid cases
Students often test each product (\(\mathrm{mp}\), \(\mathrm{rt}\), \(\mathrm{tz}\)) using only one scenario instead of checking them against ALL valid sign patterns. For instance, they might test \(\mathrm{tz}\) only in Case 1 (all positive) and conclude it's always positive, missing Case 3 where \(\mathrm{t}(-) \times \mathrm{z}(+) = \text{negative}\).
Faltering Point 5: Misreading 'must be positive' vs 'can be positive'
Students sometimes confuse the question requirement. The question asks which products MUST be positive (meaning positive in ALL valid scenarios), but students might select products that are positive in SOME scenarios. This could lead them to incorrectly include Product III (\(\mathrm{tz}\)) in their answer, choosing option E instead of the correct option C.
Step 1: Choose strategic values that satisfy the constraint
We need \(\mathrm{m} < \mathrm{p} < \mathrm{r} < \mathrm{t} < \mathrm{z}\) where \(\mathrm{mprtz} > 0\). Let's test different sign patterns systematically using concrete numbers.
Step 2: Test the critical case with mixed signs
Since the product \(\mathrm{mprtz}\) must be positive, we need an even number of negative values. Let's try the case where the two smallest values are negative:
Let \(\mathrm{m} = -4, \mathrm{p} = -2, \mathrm{r} = 1, \mathrm{t} = 3, \mathrm{z} = 5\)
Check: \(\mathrm{m} < \mathrm{p} < \mathrm{r} < \mathrm{t} < \mathrm{z}\)? Yes: \(-4 < -2 < 1 < 3 < 5\) ✓
Check: \(\mathrm{mprtz} > 0\)? \((-4)(-2)(1)(3)(5) = 8 \times 15 = 120 > 0\) ✓
Step 3: Evaluate each product with our chosen values
I. \(\mathrm{mp} = (-4)(-2) = 8 > 0\) ✓
II. \(\mathrm{rt} = (1)(3) = 3 > 0\) ✓
III. \(\mathrm{tz} = (3)(5) = 15 > 0\) ✓
Step 4: Test another valid case to check consistency
Let's try all positive values: \(\mathrm{m} = 1, \mathrm{p} = 2, \mathrm{r} = 3, \mathrm{t} = 4, \mathrm{z} = 5\)
Check: \(\mathrm{mprtz} = (1)(2)(3)(4)(5) = 120 > 0\) ✓
I. \(\mathrm{mp} = (1)(2) = 2 > 0\) ✓
II. \(\mathrm{rt} = (3)(4) = 12 > 0\) ✓
III. \(\mathrm{tz} = (4)(5) = 20 > 0\) ✓
Step 5: Test edge case with four negatives
Try \(\mathrm{m} = -5, \mathrm{p} = -3, \mathrm{r} = -2, \mathrm{t} = -1, \mathrm{z} = 1\)
Check: \(\mathrm{mprtz} = (-5)(-3)(-2)(-1)(1) = 30 > 0\) ✓
I. \(\mathrm{mp} = (-5)(-3) = 15 > 0\) ✓
II. \(\mathrm{rt} = (-2)(-1) = 2 > 0\) ✓
III. \(\mathrm{tz} = (-1)(1) = -1 < 0\) ✗
Step 6: Conclusion
From our concrete examples, we see that \(\mathrm{mp}\) and \(\mathrm{rt}\) are always positive in all valid scenarios, but \(\mathrm{tz}\) can be negative (as shown in the four-negative case). Therefore, only I and II must be positive.
Answer: C