To solve the question "The sum/difference of a rational and an irrational number is _______," we will analyze the properties of rational and irrational numbers step by step.
### Step-by-Step Solution:
1. **Define Rational and Irrational Numbers**:
- A **rational number** is any number that can be expressed as the quotient or fraction \( \frac{p}{q} \) of two integers, where \( p \) is an integer and \( q \) is a non-zero integer. Examples include \( 1, -2, \frac{1}{2}, 0.75 \), etc.
- An **irrational number** is a number that cannot be expressed as a simple fraction. It cannot be written as \( \frac{p}{q} \). Examples include \( \sqrt{2}, \pi, e \), etc.
2. **Choose Examples**:
- Let's take a rational number, for example, \( 2 \) (which is rational because it can be expressed as \( \frac{2}{1} \)).
- Let's take an irrational number, for example, \( \sqrt{3} \).
3. **Calculate the Sum**:
- Now, calculate the sum of the rational and irrational number:
\[
2 + \sqrt{3}
\]
- We will assume that this sum is a rational number.
4. **Assumption and Contradiction**:
- If \( 2 + \sqrt{3} \) is rational, then we can express it as \( r \) (where \( r \) is a rational number).
- Rearranging gives us:
\[
\sqrt{3} = r - 2
\]
- Since \( r - 2 \) is rational (as \( r \) and \( 2 \) are both rational), this implies that \( \sqrt{3} \) must also be rational.
5. **Conclusion from Contradiction**:
- However, \( \sqrt{3} \) is known to be irrational. This leads to a contradiction, meaning our assumption that \( 2 + \sqrt{3} \) is rational must be false.
- Therefore, the sum \( 2 + \sqrt{3} \) is irrational.
6. **Calculate the Difference**:
- Now, let's calculate the difference:
\[
2 - \sqrt{3}
\]
- We can apply the same reasoning as above. Assume \( 2 - \sqrt{3} \) is rational.
- Rearranging gives us:
\[
\sqrt{3} = 2 - r
\]
- Again, since \( 2 - r \) is rational, this implies that \( \sqrt{3} \) must be rational, leading to the same contradiction.
7. **Final Conclusion**:
- Since both the sum and difference of a rational number and an irrational number lead to contradictions when assumed to be rational, we conclude that:
- The sum and difference of a rational and an irrational number is always **irrational**.
### Final Answer:
The sum/difference of a rational and an irrational number is **irrational**.
