is this true: If x is an irrational number, then x^2 + 1 is an irrational number

Yes, this statement is **true**.

Here's why:

### Proof:
Assume \( x \) is an irrational number. We want to show that \( x^2 + 1 \) is also irrational.

### Contradiction approach:
Suppose \( x^2 + 1 \) is rational, i.e., \( x^2 + 1 = r \), where \( r \) is some rational number. Then:
\[
x^2 = r - 1
\]
Since \( r \) is rational, \( r - 1 \) is also rational (because the difference of two rational numbers is rational). This implies that \( x^2 \) is rational.

However, if \( x^2 \) is rational, then \( x \) must be rational, because the square root of a rational number (if real) is either rational or irrational. But this contradicts our assumption that \( x \) is irrational.

Therefore, the assumption that \( x^2 + 1 \) is rational must be false. Hence, \( x^2 + 1 \) is irrational.

### Conclusion:
If \( x \) is irrational, then \( x^2 + 1 \) is irrational.

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Would you like more details or have any questions on this proof? Here are some related questions for further exploration:

1. Can an irrational number ever become rational by adding a constant?
2. Is \( x^2 \) always irrational if \( x \) is irrational?
3. What are examples of irrational numbers where \( x^2 + 1 \) stays irrational?
4. Can a sum of two irrational numbers ever be rational?
5. How does the property of irrational numbers affect algebraic operations?

**Tip:** When working with irrational numbers, always remember that the result of most arithmetic operations (besides specific exceptions) will generally also be irrational.

Is this true: If x is an irrational number, then x^2 + 1 is an irrational number

Explore the mathematical proof showing that if x is an irrational number, then x^2 + 1 is also irrational. This detailed explanation uses the contradiction method and properties of rational and irrational numbers, making it suitable for students in Grades 10-12.

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