Question

Let \(c\) be any real number and \(1 < c\), then:

• A. \(c^2 < c\)
• B. \(c^2 < 1\)
• C. \(c < c^2\)
• D. \(c^{-1} > 1\)

Hint:

For any real number \(c\):
- If \(c > 1\), then powers of \(c\) grow: \(c < c^2 < c^3 < \dots\)
- If \(0 < c < 1\), then powers of \(c\) shrink: \(c > c^2 > c^3 > \dots\)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem deals with the multiplicative properties of inequalities for real numbers greater than 1.

Step 3: Detailed Explanation:
We are given that \(c\) is a real number such that:
\[ c > 1 \]
This implies that \(c\) is a positive number.
Let us evaluate each of the given options:
- Option (A): \(c^2 < c\)
If we divide both sides of this inequality by the positive number \(c\), we get \(c < 1\), which contradicts our given condition \(c > 1\). Thus, Option (A) is incorrect.
- Option (B): \(c^2 < 1\)
Since \(c > 1\), multiplying both sides by itself must yield a value greater than 1. Thus, Option (B) is incorrect.
- Option (C): \(c < c^2\)
We start with our given inequality:
\[ 1 < c \]
Since \(c > 1\), \(c\) is a positive real number.
Multiplying both sides of the inequality by the positive value \(c\) preserves the direction of the inequality:
\[ 1 \cdot c < c \cdot c \]
\[ c < c^2 \]
This statement is mathematically correct.
- Option (D): \(c^{-1} > 1\)
Since \(c > 1\), taking the reciprocal of both sides reverses the inequality:
\[ c^{-1} = \frac{1}{c} < 1 \]
Thus, Option (D) is incorrect.

Step 4: Final Answer:
Therefore, the correct option is (C).