Let $x$ be a real number such that $x + \frac{1}{x}$ is an integer. Which of the following statements must be true? A) $x$ is a rational number. B) It is possible for $x$ to be eit... Let $x$ be a real number such that $x + \frac{1}{x}$ is an integer. Which of the following statements must be true? A) $x$ is a rational number. B) It is possible for $x$ to be either rational or irrational. C) $x$ is an irrational number. D) $x$ is an integer. Read more

Steps to Solve

1. Define the given condition

Let $x + \frac{1}{x} = n$, where $n$ is an integer.

2. Rearrange the equation

Multiply both sides of the equation by $x$ to get rid of the fraction: $$x^2 + 1 = nx$$ Rearrange the terms to form a quadratic equation: $$x^2 - nx + 1 = 0$$

3. Solve the quadratic equation for $x$

Use the quadratic formula to solve for $x$: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ In our case, $a = 1$, $b = -n$, and $c = 1$. Plugging these values into the quadratic formula, we get: $$x = \frac{n \pm \sqrt{n^2 - 4}}{2}$$

4. Analyze the discriminant

The discriminant of the quadratic equation is $n^2 - 4$. The nature of $x$ depends on the value of this discriminant.

5. Case 1: $n^2 - 4$ is a perfect square

If $n^2 - 4$ is a perfect square, then $\sqrt{n^2 - 4}$ is an integer. In this case, $x$ is a rational number since it is the ratio of integers. For example, if $n = 2$, then $x = \frac{2 \pm \sqrt{2^2 - 4}}{2} = \frac{2 \pm 0}{2} = 1$, which is an integer. If $n = 3$, then $x = \frac{3 \pm \sqrt{3^2 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2}$. In this case, $x$ is irrational. If $n = -2$, then $x = \frac{-2 \pm \sqrt{(-2)^2 - 4}}{2} = \frac{-2 \pm 0}{2} = -1$, which is an integer. If $n = -3$, then $x = \frac{-3 \pm \sqrt{(-3)^2 - 4}}{2} = \frac{-3 \pm \sqrt{5}}{2}$. In this case, $x$ is irrational.

If $n^2 - 4$ results in a perfect square, then $\sqrt{n^2 - 4}$ is an integer, say $k$. Then $x = \frac{n \pm k}{2}$. If $n$ and $k$ are both even, or both odd, then $\frac{n \pm k}{2}$ will be an integer or a rational number.

6. Case 2: $n^2 - 4$ is not a perfect square

If $n^2 - 4$ is not a perfect square, then $\sqrt{n^2 - 4}$ is irrational. In this case, $x$ is irrational.

7. Test values of n to observe the results If $n = 0$, then $x = \frac{0 \pm \sqrt{-4}}{2}$, but x must be real. If $n = 1$, then $x = \frac{1 \pm \sqrt{-3}}{2}$, but x must be real.

8. Conclusion

From the quadratic formula $x = \frac{n \pm \sqrt{n^2 - 4}}{2}$, we can see that $x$ can be rational (including integers) or irrational. However, if $x$ is rational, then $x$ could be an integer.

If $x$ is an integer, then $x = 1$ or $x = -1$. If $x = 1$, then $x + \frac{1}{x} = 1 + 1 = 2$, which is an integer. If $x = -1$, then $x + \frac{1}{x} = -1 + (-1) = -2$, which is an integer.

We consider when $x$ is of the form $a + b\sqrt{c}$, where $a$ and $b$ are rational and $c$ is an integer, to see if it can be rewritten in such a way that $n$ is an integer. If $x = \frac{n + \sqrt{n^2 - 4}}{2}$, then $\frac{1}{x} = \frac{2}{n + \sqrt{n^2 - 4}} = \frac{2(n - \sqrt{n^2 - 4})}{(n + \sqrt{n^2 - 4})(n - \sqrt{n^2 - 4})} = \frac{2(n - \sqrt{n^2 - 4})}{n^2 - (n^2 - 4)} = \frac{2(n - \sqrt{n^2 - 4})}{4} = \frac{n - \sqrt{n^2 - 4}}{2}$.

Then $x + \frac{1}{x} = \frac{n + \sqrt{n^2 - 4}}{2} + \frac{n - \sqrt{n^2 - 4}}{2} = \frac{2n}{2} = n$.

If $x = \frac{n - \sqrt{n^2 - 4}}{2}$, then $\frac{1}{x} = \frac{2}{n - \sqrt{n^2 - 4}} = \frac{2(n + \sqrt{n^2 - 4})}{(n - \sqrt{n^2 - 4})(n + \sqrt{n^2 - 4})} = \frac{2(n + \sqrt{n^2 - 4})}{n^2 - (n^2 - 4)} = \frac{2(n + \sqrt{n^2 - 4})}{4} = \frac{n + \sqrt{n^2 - 4}}{2}$.

Then $x + \frac{1}{x} = \frac{n - \sqrt{n^2 - 4}}{2} + \frac{n + \sqrt{n^2 - 4}}{2} = \frac{2n}{2} = n$.

From the above cases, we can also consider the case when $x$ is irrational. For example, if $x = \frac{3 + \sqrt{5}}{2}$, then $\frac{1}{x} = \frac{2}{3+\sqrt{5}} = \frac{2(3 - \sqrt{5})}{(3 + \sqrt{5})(3-\sqrt{5})} = \frac{2(3 - \sqrt{5})}{9 - 5} = \frac{2(3 - \sqrt{5})}{4} = \frac{3- \sqrt{5}}{2}$. So $x + \frac{1}{x} = \frac{3 + \sqrt{5}}{2} + \frac{3 - \sqrt{5}}{2} = \frac{6}{2} = 3$, which is an integer.

$x$ is rational or irrational depending on $n$.