The Number $(-\sqrt{9})$ Does Not Belong To:A. $Q$ B. $R$ C. $Z$ D. $I$
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Introduction
In mathematics, we often deal with various sets of numbers, each with its own unique properties and characteristics. The sets in question are the set of rational numbers (Q), the set of real numbers (R), the set of integers (Z), and the set of imaginary numbers (I). In this discussion, we will explore which of these sets the number $(-\sqrt{9})$ does not belong to.
Understanding the Sets
Rational Numbers (Q)
Rational numbers are those numbers that can be expressed as the ratio of two integers, i.e., a/b, where a and b are integers and b is non-zero. Examples of rational numbers include 3/4, 22/7, and -5/2.
Real Numbers (R)
Real numbers include all rational numbers and also include irrational numbers, which cannot be expressed as a ratio of two integers. Examples of real numbers include 3.14, -2.5, and π.
Integers (Z)
Integers are whole numbers, either positive, negative, or zero, without a fractional part. Examples of integers include 5, -3, and 0.
Imaginary Numbers (I)
Imaginary numbers are numbers that, when squared, give a negative result. They are typically represented as a multiple of the imaginary unit i, where i = √(-1). Examples of imaginary numbers include 3i, -2i, and i.
Analyzing the Number $(-\sqrt{9})$
The number $(-\sqrt{9})$ can be simplified as follows:
This is because the square root of 9 is 3, and multiplying it by -1 gives -3.
Does $(-\sqrt{9})$ belong to Q?
Since $(-\sqrt{9})$ simplifies to -3, which is an integer, it can be expressed as a ratio of two integers (e.g., -3/1). Therefore, $(-\sqrt{9})$ belongs to the set of rational numbers (Q).
Does $(-\sqrt{9})$ belong to R?
As we have already established that $(-\sqrt{9})$ simplifies to -3, which is a real number, it also belongs to the set of real numbers (R).
Does $(-\sqrt{9})$ belong to Z?
Since $(-\sqrt{9})$ simplifies to -3, which is an integer, it also belongs to the set of integers (Z).
Does $(-\sqrt{9})$ belong to I?
Since $(-\sqrt{9})$ simplifies to -3, which is a real number and not an imaginary number, it does not belong to the set of imaginary numbers (I).
Conclusion
In conclusion, the number $(-\sqrt{9})$ does not belong to the set of imaginary numbers (I). It belongs to the sets of rational numbers (Q), real numbers (R), and integers (Z).
Final Answer
The final answer is D. I.
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Introduction
In our previous discussion, we explored which sets of numbers the number $(-\sqrt{9})$ belongs to. We concluded that it belongs to the sets of rational numbers (Q), real numbers (R), and integers (Z), but not to the set of imaginary numbers (I). In this Q&A article, we will delve deeper into the properties of $(-\sqrt{9})$ and its relationships with these sets.
Q: What is the square root of 9?
A: The square root of 9 is 3, because 3 multiplied by 3 equals 9.
Q: Why is $(-\sqrt{9})$ not an imaginary number?
A: $(-\sqrt{9})$ is not an imaginary number because it can be simplified to a real number, -3. Imaginary numbers are typically represented as a multiple of the imaginary unit i, where i = √(-1). Since $(-\sqrt{9})$ does not involve the imaginary unit i, it is not an imaginary number.
Q: Can $(-\sqrt{9})$ be expressed as a ratio of two integers?
A: Yes, $(-\sqrt{9})$ can be expressed as a ratio of two integers, -3/1. This is because it simplifies to -3, which is an integer.
Q: Is $(-\sqrt{9})$ a rational number?
A: Yes, $(-\sqrt{9})$ is a rational number because it can be expressed as a ratio of two integers, -3/1.
Q: Is $(-\sqrt{9})$ an irrational number?
A: No, $(-\sqrt{9})$ is not an irrational number because it can be expressed as a ratio of two integers, -3/1. Irrational numbers are those that cannot be expressed as a ratio of two integers.
Q: Is $(-\sqrt{9})$ a real number?
A: Yes, $(-\sqrt{9})$ is a real number because it can be expressed as a decimal value, -3.0.
Q: Is $(-\sqrt{9})$ an integer?
A: Yes, $(-\sqrt{9})$ is an integer because it is a whole number, -3.
Q: Can $(-\sqrt{9})$ be used in mathematical operations involving imaginary numbers?
A: No, $(-\sqrt{9})$ cannot be used in mathematical operations involving imaginary numbers because it is a real number and not an imaginary number.
Q: Is $(-\sqrt{9})$ a member of the set of complex numbers?
A: Yes, $(-\sqrt{9})$ is a member of the set of complex numbers because it can be expressed as a combination of real and imaginary parts, -3 + 0i.
Conclusion
In conclusion, the number $(-\sqrt{9})$ has several properties that make it a member of various sets of numbers. It is a rational number, a real number, an integer, and a member of the set of complex numbers. However, it is not imaginary number.
Final Answer
The final answer is D. I.