Math Problem Statement

  1. 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.]

Solution

To prove by contradiction that there is no rational number rr such that r3+r+1=0r^3 + r + 1 = 0, we assume the contrary: that there exists a rational number r=abr = \frac{a}{b} in its lowest terms (where aa and bb are integers with no common factors other than 1, and b0b \neq 0) that satisfies the equation.

Starting with the equation: r3+r+1=0r^3 + r + 1 = 0

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

Multiply through by b3b^3 to clear the denominators: a3+ab2+b3=0a^3 + ab^2 + b^3 = 0

Rearrange this to: a3+ab2=b3a^3 + ab^2 = -b^3

This equation shows that b3b^3 must divide a3+ab2a^3 + ab^2. Let's analyze the parity (odd or even nature) of aa and bb:

  1. Case 1: Both aa and bb are even.

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

  2. Case 2: Both aa and bb are odd.

    If aa and bb are both odd, then a3a^3, ab2ab^2, and b3b^3 are all odd (since the product of odd numbers is odd). The sum of three odd numbers is odd. Thus, a3+ab2+b3a^3 + ab^2 + b^3 would be odd, which contradicts the equation a3+ab2=b3a^3 + ab^2 = -b^3, since the right-hand side b3-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 aa is odd and bb is even. If bb is even, then b3b^3 is even, and ab2ab^2 is even (since it includes b2b^2). However, a3a^3 is odd. Therefore, a3+ab2a^3 + ab^2 is the sum of an odd number and an even number, which is odd. This again contradicts the fact that b3-b^3 (right-hand side) is even.

Since all possible cases lead to a contradiction, our initial assumption that r=abr = \frac{a}{b} is rational must be false. Therefore, there is no rational number rr such that r3+r+1=0r^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.

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Math Problem Analysis

Mathematical Concepts

Proof by Contradiction
Rational Numbers
Polynomial Equations

Formulas

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Theorems

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Suitable Grade Level

Grades 10-12