Problem 9 - Algebra and Trigonometry

Question

Let \(S=\left\\{-2,-1,0, \frac{1}{2}, 1, \sqrt{2}, 2,4\right\\} .\) Determine which elements of \(S\) satisfy the inequality. $$ \frac{1}{x} \leq \frac{1}{2} $$

Step-by-Step Solution

Answer

The elements of \( S \) that satisfy the inequality are \{-2, -1, 2, 4\}.

Step 1: Understand the inequality

We need to find which elements of the set \( S = \{-2,-1,0, \frac{1}{2}, 1, \sqrt{2}, 2, 4\} \) satisfy the inequality \( \frac{1}{x} \leq \frac{1}{2} \). This means that we are looking for elements \( x \) such that their reciprocal is less than or equal to \( \frac{1}{2} \).

Step 2: Ignore zero

Zero cannot be used as a divisor, so \( x = 0 \) needs to be excluded from our consideration for valid solutions.

Step 3: Rearrange the inequality

Since \( \frac{1}{x} \leq \frac{1}{2} \), this inequality can be rearranged, based on the sign of \( x \). If \( x \) is positive, the inequality is \( x \geq 2 \). If \( x \) is negative \( x \) is left in its current form without rearranging.

Step 4: Analyze positive numbers

For positive elements \( \frac{1}{2}, 1, \sqrt{2}, 2, 4 \), consider \( x \geq 2 \). This means the solution includes \{2, 4\}, as these are the elements that are greater than or equal to 2.

Step 5: Analyze negative numbers

For negative elements \(-2, -1\), we note that any negative number has a denominator that is a more negative value. So, \( \frac{1}{x} \) is less than zero, meaning its reciprocal is automatically less than \( \frac{1}{2} \). Therefore, include \{-2, -1\}.

Step 6: Reconcile the results

After considering both positive and negative solutions, the elements of \( S \) that satisfy the inequality \( \frac{1}{x} \leq \frac{1}{2} \) are \{-2, -1, 2, 4\}.

Key Concepts

Reciprocal

The concept of a reciprocal plays a crucial role in this exercise. A reciprocal of a number is 1 divided by that number. In simpler terms, if you have a number \( x \), its reciprocal is \( \frac{1}{x} \). Reciprocals are particularly useful when working with fractions and inequalities. They flip the number around, but do not change the inequality direction for positive numbers unless the sign of \( x \) changes.
This exercise requires finding which elements in the set have reciprocals less than or equal to \( \frac{1}{2} \). Understanding reciprocals helps us rearrange and solve inequalities by allowing us to manipulate the terms while keeping relationships intact.

Positive and Negative Numbers

Positive and negative numbers behave differently, especially in inequalities.

• Multiplying or dividing by a positive number keeps the direction of the inequality unchanged.
• Multiplying or dividing by a negative number, however, reverses the inequality direction.

In this problem, we note that reciprocals of negative numbers are also negative, and therefore always less than a positive \( \frac{1}{2} \). Hence, all negative numbers in the set automatically satisfy the inequality. For positive numbers, careful consideration of their size relative to 2 is necessary, as smaller values would not satisfy the inequality when reciprocals are considered.

Solution Set

The solution set is the collection of all values that satisfy a given inequality. In this exercise, we identify which elements of the set \( S \) satisfy the inequality \( \frac{1}{x} \leq \frac{1}{2} \).
We start by excluding zero since its reciprocal is undefined. Then, separately evaluate positive and negative numbers:

• Positive elements must be checked if they are greater than or equal to 2 since reciprocals of values less than 2 would not meet the inequality requirement.
• All negative elements satisfy the condition, as their reciprocals are less than any positive number.

After this evaluation, the solution set becomes \(-2, -1, 2, 4\) as they meet the inequality criteria established.

Inequality Rearrangement

Rearranging inequalities is an art in mathematics that allows us to explore the relationships between different numbers or expressions. The inequality \( \frac{1}{x} \leq \frac{1}{2} \) can be altered based on whether \( x \) is positive or negative.

• For positive numbers, we convert the inequality to \( x \geq 2 \), allowing us to focus on elements that are equal to or greater than 2.
• Negative values, on the other hand, remain without rearrangement due to their nature of producing negative reciprocals always less than a positive \( \frac{1}{2} \).

By strategically rearranging these, it becomes possible to solve inequalities in a more structured way, providing clarity to which elements of a set meet the condition. This makes handling similar problems straightforward by applying these principles.