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If \(\mathrm{n}\) denotes a number to the left of \(0\) on the number line such that \(\mathrm{n}^2 < \frac{1}{100}\), then \(\frac{1}{\mathrm{n}}\) must be
Let's break down what we're told in plain English:
So we're looking for the range of \(\frac{1}{\mathrm{n}}\) when n is negative and \(\mathrm{n}^2 < \frac{1}{100}\).
Process Skill: TRANSLATE - Converting the problem statement into clear mathematical understanding
Since n is negative and \(\mathrm{n}^2 < \frac{1}{100}\), let's think about what this means:
First, what does \(\mathrm{n}^2 < \frac{1}{100}\) tell us? Since \(\frac{1}{100} = 0.01\), we need \(\mathrm{n}^2 < 0.01\).
Now, what numbers when squared give us something less than 0.01?
So for the square to be less than \(\frac{1}{100}\), the number n must be closer to zero than \(-\frac{1}{10}\).
In other words: n must be between \(-\frac{1}{10}\) and 0 (since n is negative).
Mathematically: \(-\frac{1}{10} < \mathrm{n} < 0\)
Process Skill: APPLY CONSTRAINTS - Determining the valid range for n
Now let's think about what happens when we take the reciprocal of a small negative number:
Notice a pattern: As n gets closer to zero (but stays negative), the reciprocal \(\frac{1}{\mathrm{n}}\) becomes a larger and larger negative number.
Key insight: When you take the reciprocal of a small negative number, you get a large negative number.
Since we established that \(-\frac{1}{10} < \mathrm{n} < 0\), let's see what this means for \(\frac{1}{\mathrm{n}}\):
Therefore: \(\frac{1}{\mathrm{n}} < -10\)
Let's verify with a specific example:
If \(\mathrm{n} = -\frac{1}{50} = -0.02\), then \(\mathrm{n}^2 = 0.0004 < 0.01\) ✓
And \(\frac{1}{\mathrm{n}} = \frac{1}{-0.02} = -50\), which is indeed less than -10 ✓
Process Skill: INFER - Drawing the non-obvious conclusion about the reciprocal's range
The reciprocal of n must be less than -10.
This corresponds to answer choice A. Less than -10.
Our reasoning confirms this: since n is a small negative number (between \(-\frac{1}{10}\) and 0), its reciprocal \(\frac{1}{\mathrm{n}}\) must be a large negative number (less than -10).
1. Misunderstanding the constraint \(\mathrm{n}^2 < \frac{1}{100}\)
Students often struggle to translate "the square of n is less than \(\frac{1}{100}\)" into the correct range for n. They may incorrectly think that since \(\mathrm{n}^2 < \frac{1}{100}\), then \(\mathrm{n} < \frac{1}{100}\), forgetting that when we take the square root of both sides of an inequality involving a squared term, we need to consider both positive and negative roots. Since n is negative, they need to recognize that \(|\mathrm{n}| < \frac{1}{10}\), which means \(-\frac{1}{10} < \mathrm{n} < 0\).
2. Forgetting that n is explicitly stated to be negative
The problem clearly states that n is "to the left of 0 on the number line," but students may lose track of this crucial constraint when working through the algebra. This leads them to consider positive values of n as well, which completely changes the reciprocal analysis and leads to wrong answer choices.
3. Not recognizing the reciprocal behavior for small negative numbers
Students may not immediately grasp that when you take the reciprocal of a small negative number (close to zero), you get a large negative number. This conceptual gap in understanding reciprocal properties makes it difficult to set up the correct approach for finding the range of \(\frac{1}{\mathrm{n}}\).
1. Incorrect boundary analysis for the reciprocal
When students find that \(-\frac{1}{10} < \mathrm{n} < 0\), they may incorrectly conclude that the reciprocal \(\frac{1}{\mathrm{n}}\) falls between -10 and negative infinity, but fail to recognize that this means \(\frac{1}{\mathrm{n}} < -10\). They might mistakenly think the reciprocal is bounded above by -10 rather than bounded above by negative infinity.
2. Sign errors when computing reciprocals
Students may make computational errors with negative reciprocals, such as calculating \(\frac{1}{-0.05} = 20\) instead of -20, or lose track of the negative sign when working with fractional examples. These arithmetic mistakes lead to selecting answer choices involving positive ranges.
3. Confusion about inequality direction with reciprocals
When taking reciprocals of negative numbers, students often forget that the inequality direction can be tricky to track. Since we're dealing with negative numbers throughout, some students may incorrectly flip inequality signs when they shouldn't, or fail to flip them when they should.
1. Selecting a range that includes the boundary value
Students might calculate that when n approaches \(-\frac{1}{10}\), the reciprocal \(\frac{1}{\mathrm{n}}\) approaches -10, and then incorrectly select answer choice B ("Between -1 and \(-\frac{1}{10}\)") or think that -10 itself could be included in the answer. They fail to recognize that since n cannot equal \(-\frac{1}{10}\) (due to the strict inequality \(\mathrm{n}^2 < \frac{1}{100}\)), the reciprocal cannot equal -10 and must be strictly less than -10.
2. Picking the reciprocal range of n instead of the reciprocal itself
Some students correctly find that \(-\frac{1}{10} < \mathrm{n} < 0\), but then mistakenly select answer choice C ("Between \(-\frac{1}{10}\) and 0") because they confuse the range of n with the range of \(\frac{1}{\mathrm{n}}\). This is a classic error where students answer the wrong question - they find the range of the original variable instead of the range of its reciprocal.