Question
Megan added two numbers and the resulting sum was an irrational number. If the first number was 2/3 , which of the following could be the second number?
A. 6/5
B. 3π
C. 5
D.
A. 6/5
B. 3π
C. 5
D.
Answers
GPT-4o mini
To determine which of the provided options for the second number, when added to \( \frac{2}{3} \), results in an irrational sum, we need to consider the properties of rational and irrational numbers.
1. **Rational Numbers**: A number that can be expressed as the quotient of two integers (like \( \frac{2}{3} \) or \( 5 \)).
2. **Irrational Numbers**: A number that cannot be expressed as a fraction of two integers (like \( \pi \) or \( \sqrt{2} \)).
- The sum of two rational numbers is always rational.
- The sum of a rational number and an irrational number is always irrational.
Given that \( \frac{2}{3} \) is a rational number, in order for the sum to be irrational, the second number must be irrational.
Let's analyze the options:
A. \( \frac{6}{5} \) - This is a rational number.
B. \( 3\pi \) - This is an irrational number (since \( \pi \) is irrational).
C. \( 5 \) - This is a rational number.
Therefore, the second number that Megan could add to \( \frac{2}{3} \) to get an irrational sum is \( \text{B. } 3\pi \).
1. **Rational Numbers**: A number that can be expressed as the quotient of two integers (like \( \frac{2}{3} \) or \( 5 \)).
2. **Irrational Numbers**: A number that cannot be expressed as a fraction of two integers (like \( \pi \) or \( \sqrt{2} \)).
- The sum of two rational numbers is always rational.
- The sum of a rational number and an irrational number is always irrational.
Given that \( \frac{2}{3} \) is a rational number, in order for the sum to be irrational, the second number must be irrational.
Let's analyze the options:
A. \( \frac{6}{5} \) - This is a rational number.
B. \( 3\pi \) - This is an irrational number (since \( \pi \) is irrational).
C. \( 5 \) - This is a rational number.
Therefore, the second number that Megan could add to \( \frac{2}{3} \) to get an irrational sum is \( \text{B. } 3\pi \).