Question 2

We will assume the opposite of what we want to prove.

Step 1 — Assume the sum is rational

Let's assume $r+x$ is a rational number. A rational number can be written as a fraction. Let $r+x = \frac{a}{b}$. Here, $a$ and $b$ are integers. Also, $b$ cannot be zero.

Step 2 — Express $r$ as a fraction

We know $r$ is a rational number. So, $r$ can be written as $\frac{c}{d}$. Here, $c$ and $d$ are integers. Also, $d$ cannot be zero.

Step 3 — Isolate $x$

Substitute $r = \frac{c}{d}$ into the equation. $\frac{c}{d} + x = \frac{a}{b}$ Now, let's isolate $x$. Subtract $\frac{c}{d}$ from both sides. $x = \frac{a}{b} - \frac{c}{d}$ Find a common denominator. $x = \frac{ad}{bd} - \frac{cb}{bd}$ Combine the fractions. $x = \frac{ad - cb}{bd}$

$\boxed{x = \frac{ad - cb}{bd}}$

Step 4 — Analyze the nature of $x$

Look at the numerator $ad - cb$. Since $a, b, c, d$ are integers, $ad$ is an integer. Also, $cb$ is an integer. The difference of two integers is an integer. So, $ad - cb$ is an integer. Now look at the denominator $bd$. Since $b$ and $d$ are non-zero integers, $bd$ is a non-zero integer. Thus, $x$ is a ratio of two integers. This means $x$ is a rational number.

Step 5 — Conclude the proof

We found that $x$ is a rational number. However, we were given that $x$ is an irrational number. This is a contradiction. Our initial assumption must be false. Therefore, $r+x$ cannot be rational. It must be an irrational number.

Answer

The sum of a rational and an irrational number is an irrational number.