Question

Prove or disprove the following statement: "For all integers \(b, c\), and \(d\), if \(x\) is a rational number such that \(x^{2}+b x+c=d\), then \(x\) is an integer." (Hints: Are all the quantifiers given explicitly? It is okay, but not necessary, to use the quadratic formula.)

Short answer

The statement is false; a rational solution of the equation is not always an integer.

Step-by-step solution

Understand the Problem Statement

We need to examine whether, for all integers \(b, c,\) and \(d\), if \(x\) is a rational number satisfying the equation \(x^2 + bx + c = d\), then \(x\) must also be an integer. To disprove this statement, we have to find a counterexample where \(x\) is rational but not integer.

Set Up the Quadratic Equation

The equation given is \(x^2 + bx + c = d\), which can be rearranged into a standard quadratic form: \(x^2 + bx + (c - d) = 0\). This is a standard quadratic equation form \(ax^2 + bx + c = 0\).

Apply the Quadratic Formula

We use the quadratic formula \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\) to solve this equation, where \(a = 1\), \(b = b\), and \(c = c - d\). Thus, \(x = \frac{{-b \pm \sqrt{{b^2 - 4(c-d)}}}}{2}\).

Analyze the Discriminant

The discriminant of the quadratic equation is \(b^2 - 4(c-d)\). For \(x\) to always be an integer, \(\sqrt{{b^2 - 4(c-d)}}\) must be an integer. Therefore, \(b^2 - 4(c-d)\) must be a perfect square.

Find a Counterexample

Let's take specific values: let \(b = 3\), \(c = 3\), and \(d = 2\). Then the equation becomes \(x^2 + 3x + 1 = 0\). Applying the quadratic formula gives \(x = \frac{{-3 \pm \sqrt{{9 - 4}}}}{2} = \frac{{-3 \pm \sqrt{5}}}{2}\). Since \(\sqrt{5}\) is irrational, \(x\) is not an integer. Thus, \(x = \frac{{-3 + \sqrt{5}}}{2}\) or \(x = \frac{{-3 - \sqrt{5}}}{2}\) are rational but non-integer solutions.

Conclusion

Since we found a specific case where the solution is rational but not an integer, the statement is proven false by counterexample. For arbitrary integer values of \(b, c,\) and \(d\), \(x\) does not always result in an integer when it is a rational solution of \(x^2 + bx + c = d\).

Key concepts

Rational Numbers

Rational numbers are numbers that can be expressed as the quotient of two integers, where the numerator is an integer and the denominator is a non-zero integer. Examples of rational numbers include fractions like \( \frac{1}{2} \), \(-\frac{3}{4} \), and whole numbers like \(3\), which can be expressed as \(\frac{3}{1}\). A key aspect of rational numbers is that they have a repeating or terminating decimal representation.

In the context of quadratic equations, a solution is considered rational if it can be simplified into a form where both the numerator and denominator are integer values. This is crucial because not all quadratic equations will have solutions that are clear integres; some might end up in fraction forms, thus making them rational but not integer.

Integer Solutions

An integer solution is a solution to an equation where the answer is a whole number. Integers include negative numbers, zero, and positive numbers, without fractions or decimals. For quadratic equations, solving them can sometimes yield whole numbers. This happens when the square root in the quadratic formula is a perfect square.

For our original exercise's equation \(x^{2} + bx + c = d\), if solutions \(x = \frac{{-b \pm \sqrt{{b^2 - 4(c-d)}}}}{2}\) yield integer results, both the discriminant (inside the square root) and the entire expression, after properly substituting values, must result in integers. These conditions might not always be satisfied, hence not every quadratic equation with rational roots will have integer solutions.

Discriminant Analysis

The discriminant of a quadratic equation, expressed as \(b^2 - 4ac\), is a vital component in determining the nature and number of the roots. If the discriminant is:

• Positive and a perfect square, the roots are rational and distinct integers.
• Positive but not a perfect square, the roots are rational but not integer (typically irrational numbers result).
• Zero, the roots are rational and equal.
• Negative, the roots are complex (having a real and an imaginary part).

In our given exercise, the condition \(b^2 - 4(c-d)\) must be an integer perfect square to ensure integer solutions. If not a perfect square, the resulting root from the quadratic formula will be a non-integer even when rational, disproving that solution's integer nature.

Counterexample Method

The counterexample method is a logical approach used to prove a statement false. Given a universal statement (something true for all cases), we can disprove it by providing at least one specific example where the statement does not hold. This single example is called a "counterexample."

In solving the original problem, finding a counterexample was key to showing that the assertion "all rational solutions are integers" is incorrect. By choosing specific values for \(b, c,\) and \(d\) which result in \(x = \frac{{-3 \pm \sqrt{5}}}{2}\) confirmed that \(x\) is rational but not an integer, effectively disproving the statement.

This method is powerful, as you only need one instance to prove a universal statement false, making it a useful tool in mathematical reasoning.