Ex. 5 Let x be a real number, show that
i. oneofx<0,x=0,x>0istrue,

To prove that one of the statements \( x < 0 \), \( x = 0 \), or \( x > 0 \) is always true for any real number \( x \), we need to rely on the **trichotomy property** of real numbers.

### Proof:

The **trichotomy property** states that for any real number \( x \), exactly one of the following three conditions holds:
1. \( x \) is **negative**: \( x < 0 \)
2. \( x \) is **zero**: \( x = 0 \)
3. \( x \) is **positive**: \( x > 0 \)

These three cases cover all possible values for a real number \( x \). This means that for any given real number, \( x \), it must fall into exactly one of these categories.

- If \( x = -2 \), for example, then \( x < 0 \).
- If \( x = 3 \), then \( x > 0 \).
- If \( x = 0 \), then \( x = 0 \).

Thus, one of these statements will always be true for any real number \( x \).

### Conclusion:
For every real number \( x \), one (and only one) of the conditions \( x < 0 \), \( x = 0 \), or \( x > 0 \) is always true, by the trichotomy property of real numbers.

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Would you like a deeper explanation or have any questions on this?

Here are 5 related questions to explore:
1. What is the **law of trichotomy** and how does it differ from other properties of real numbers?
2. How is the **trichotomy property** used in proofs of inequalities?
3. Can you prove similar properties for complex numbers?
4. How does the trichotomy property apply in practical real-world problems?
5. Why can't a real number \( x \) satisfy more than one of the conditions at the same time?

**Tip**: The trichotomy property is foundational in real analysis and helps ensure that we can define order relations consistently on real numbers.

Let x be a real number, show that one of x < 0, x = 0, x > 0 is true.

Explore the proof that for any real number x, one of the statements x < 0, x = 0, or x > 0 is always true. This problem demonstrates the trichotomy property, a fundamental concept in real analysis. Suitable for high school students in Grades 9-12.

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