Question
Ms. Arrowood is classifying the number −18−−−−√ . Is Ms. Arrowood correct or incorrect? How do you know?
she say it is real number. there is no i
she say it is real number. there is no i
Answers
GPT-4o mini
To determine whether Ms. Arrowood's classification of the number \(-\sqrt{18}\) as a real number is correct, we need to analyze the expression.
The expression \(-\sqrt{18}\) can be further simplified:
\[
\sqrt{18} = \sqrt{9 \times 2} = \sqrt{9} \cdot \sqrt{2} = 3\sqrt{2}
\]
So, \(-\sqrt{18} = -3\sqrt{2}\).
Since \(\sqrt{2}\) is a real number, \(-3\sqrt{2}\) is also a real number because any real number multiplied by another real number results in a real number.
Thus, \(-\sqrt{18}\) is indeed a real number.
**Conclusion**: Ms. Arrowood is correct in classifying \(-\sqrt{18}\) as a real number, as it is a valid operation that results in a real number without involving imaginary numbers (\(i\)).
The expression \(-\sqrt{18}\) can be further simplified:
\[
\sqrt{18} = \sqrt{9 \times 2} = \sqrt{9} \cdot \sqrt{2} = 3\sqrt{2}
\]
So, \(-\sqrt{18} = -3\sqrt{2}\).
Since \(\sqrt{2}\) is a real number, \(-3\sqrt{2}\) is also a real number because any real number multiplied by another real number results in a real number.
Thus, \(-\sqrt{18}\) is indeed a real number.
**Conclusion**: Ms. Arrowood is correct in classifying \(-\sqrt{18}\) as a real number, as it is a valid operation that results in a real number without involving imaginary numbers (\(i\)).