Understanding Inequalities for Numbers Between 0 and 1

The question asks us to determine the correct inequality relation between \(x\), \(\sqrt{x}\), and \(x^2\) when \(x\) is a number strictly between 0 and 1, i.e., \(0 < x < 1\). Understanding how powers and roots behave for numbers in this range is crucial.

Analyzing Powers and Roots for \(0 < x < 1\)

When a number \(x\) is between 0 and 1 (\(0 < x < 1\)):

• Multiplying \(x\) by itself (squaring it) results in a smaller number. For example, if \(x = 0.5\), then \(x^2 = (0.5)^2 = 0.25\). Here, \(0.25 < 0.5\). This holds true for any \(x\) in the range \(0 < x < 1\). We can prove this: since \(0 < x < 1\), multiplying the inequality \(x < 1\) by the positive number \(x\) gives \(x \cdot x < 1 \cdot x\), which simplifies to \(x^2 < x\).
• Taking the square root of \(x\) results in a larger number. For example, if \(x = 0.25\), then \(\sqrt{x} = \sqrt{0.25} = 0.5\). Here, \(0.5 > 0.25\). This holds true for any \(x\) in the range \(0 < x < 1\). We can think of \(\sqrt{x}\) as \(x^{1/2}\). For \(0 < x < 1\), if the exponent is less than 1 (like 1/2), the resulting value is larger than the original number. If the exponent is greater than 1 (like 2), the resulting value is smaller than the original number. Thus, for \(0 < x < 1\), \(x < \sqrt{x}\).

Combining the Inequalities

From the analysis above, for \(0 < x < 1\), we have found two key inequalities:

1. \(x^2 < x\)
2. \(x < \sqrt{x}\)

Combining these two inequalities, we get the relationship \(x^2 < x < \sqrt{x}\).

Checking the Options

Let's check this combined inequality against the given options.

• Option 1: \(x < \sqrt{x} < x^2\). This is incorrect because \(x^2 < x\) and \(\sqrt{x} > x\).
• Option 2: \(\sqrt{x} < x < x^2\). This is incorrect because \(\sqrt{x} > x\) and \(x > x^2\).
• Option 3: \(x^2 < x < \sqrt{x}\). This matches our derived inequality.
• Option 4: \(\sqrt{x} < x^2 < x\). This is incorrect because \(\sqrt{x} > x\) and \(x^2 < x\).

Using an example like \(x = 0.1\) can also help verify: \(x^2 = (0.1)^2 = 0.01\), \(x = 0.1\), \(\sqrt{x} = \sqrt{0.1} \approx 0.316\). Comparing these values, we get \(0.01 < 0.1 < 0.316\), which confirms \(x^2 < x < \sqrt{x}\).

Conclusion

For any number \(x\) such that \(0 < x < 1\), the correct inequality relating \(x\), \(\sqrt{x}\), and \(x^2\) is \(x^2 < x < \sqrt{x}\).

Revision Table: Comparing Powers and Roots

Additional Information: Exponents and Bases

The behavior of powers and roots depends heavily on the base number and the exponent. Here's a summary focusing on positive bases:

• Base \(> 1\): If \(b > 1\), then for exponents \(p\) and \(q\):
  • If \(p > q\), then \(b^p > b^q\). (e.g., \(2^3 > 2^2\))
  • Taking a power greater than 1 increases the number (\(b^p > b\) for \(p>1\)).
  • Taking a power less than 1 (but greater than 0) decreases the number (\(b^p < b\) for \(0 < p < 1\)). Square root is \(x^{1/2}\).
• Base \(0 < \text{Base} < 1\): If \(0 < b < 1\), then for exponents \(p\) and \(q\):
  • If \(p > q\), then \(b^p < b^q\). (e.g., \((0.5)^3 < (0.5)^2\), \(0.125 < 0.25\))
  • Taking a power greater than 1 decreases the number (\(b^p < b\) for \(p>1\)). Squaring is \(x^2 = x\) raised to power 2.
  • Taking a power less than 1 (but greater than 0) increases the number (\(b^p > b\) for \(0 < p < 1\)). Square root is \(x^{1/2}\).

In our problem, the base is \(x\) and \(0 < x < 1\). We are comparing \(x^1\), \(x^2\), and \(x^{1/2}\). The exponents are \(1, 2, 1/2\). Ordering the exponents: \(1/2 < 1 < 2\). Since the base \(x\) is between 0 and 1, the inequality relation for the powers is reversed compared to the exponents. Thus, \(x^{1/2} > x^1 > x^2\), which is \(\sqrt{x} > x > x^2\), or \(x^2 < x < \sqrt{x}\).