2) Use a proof by contradiction to show that there is no rational number
r for which r
3 + r + 1 = 0.[Hint:Assume that r = a/b is a root, where
a and b are integers and a/b is in lowest terms. Obtain an equation
involving integers by multiplying by b
3
. Then look at whether a and b
are each odd or even.]

To prove by contradiction that there is no rational number \( r \) such that \( r^3 + r + 1 = 0 \), we assume the contrary: that there exists a rational number \( r = \frac{a}{b} \) in its lowest terms (where \( a \) and \( b \) are integers with no common factors other than 1, and \( b \neq 0 \)) that satisfies the equation. 

Starting with the equation:
\[ r^3 + r + 1 = 0 \]

Substitute \( r = \frac{a}{b} \):
\[ \left(\frac{a}{b}\right)^3 + \frac{a}{b} + 1 = 0 \]

Multiply through by \( b^3 \) to clear the denominators:
\[ a^3 + ab^2 + b^3 = 0 \]

Rearrange this to:
\[ a^3 + ab^2 = -b^3 \]

This equation shows that \( b^3 \) must divide \( a^3 + ab^2 \). Let's analyze the parity (odd or even nature) of \( a \) and \( b \):

1. **Case 1: Both \( a \) and \( b \) are even.**

   If \( a \) and \( b \) are both even, then they share a common factor of 2. This contradicts our assumption that \( \frac{a}{b} \) is in lowest terms.

2. **Case 2: Both \( a \) and \( b \) are odd.**

   If \( a \) and \( b \) are both odd, then \( a^3 \), \( ab^2 \), and \( b^3 \) are all odd (since the product of odd numbers is odd). The sum of three odd numbers is odd. Thus, \( a^3 + ab^2 + b^3 \) would be odd, which contradicts the equation \( a^3 + ab^2 = -b^3 \), since the right-hand side \(-b^3\) is odd and the left-hand side should then be even.

3. **Case 3: One is odd and the other is even.**

   Without loss of generality, assume \( a \) is odd and \( b \) is even. If \( b \) is even, then \( b^3 \) is even, and \( ab^2 \) is even (since it includes \( b^2 \)). However, \( a^3 \) is odd. Therefore, \( a^3 + ab^2 \) is the sum of an odd number and an even number, which is odd. This again contradicts the fact that \( -b^3 \) (right-hand side) is even.

Since all possible cases lead to a contradiction, our initial assumption that \( r = \frac{a}{b} \) is rational must be false. Therefore, there is no rational number \( r \) such that \( r^3 + r + 1 = 0 \).

Would you like more details or have any questions?

Here are 5 more questions you might consider:

1. How can we use proof by induction to show a given statement?
2. What is the difference between direct proof and proof by contradiction?
3. Can you provide an example of a rational number that satisfies a polynomial equation?
4. How can we determine the roots of a polynomial equation?
5. What is the significance of the Rational Root Theorem in polynomial equations?

**Tip:** When tackling proof problems, always start by clearly stating your assumptions and method of proof. This helps in systematically reaching the conclusion.

Explore a proof by contradiction showing that there is no rational number r such that r^3 + r + 1 = 0. This problem involves understanding rational numbers, polynomial equations, and proof techniques. Suitable for advanced high school students.

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