Which Statement About Rational And Irrational Numbers Is True

In the realm of mathematics, understanding the nature of numbers is fundamental. Numbers can be classified into various categories, such as rational, irrational, integers, and more. To accurately answer the question, "Which of the following is true?", we need to dissect each option, applying our knowledge of number classifications to determine the correct statement. This article provides a comprehensive analysis of each option, ensuring a clear understanding of the concepts involved.

Delving into Rational and Irrational Numbers

Before we dissect the options, it's crucial to establish a solid understanding of rational and irrational numbers. Rational numbers are numbers that can be expressed as a fraction p/q, where p and q are integers and q is not zero. This category includes integers, fractions, terminating decimals, and repeating decimals. On the other hand, irrational numbers cannot be expressed as a simple fraction. They are characterized by their non-repeating, non-terminating decimal representations. Classic examples include the square root of 2 ($\sqrt{2}$) and pi (π). The distinction between these two categories is paramount when evaluating mathematical statements.

Analyzing Option A: 0 is neither a rational number nor an irrational number

Option A posits that 0 is neither rational nor irrational. To evaluate this statement, we must consider the definition of rational numbers. Can 0 be expressed as a fraction p/q? The answer is yes. Zero can be written as 0/1, 0/2, 0/n, where n is any non-zero integer. Since 0 fits the definition of a rational number, being expressible as a fraction of two integers, the first part of the statement is false. To further solidify understanding, we can delve into the classification of integers. Integers include all whole numbers and their negatives (..., -2, -1, 0, 1, 2, ...). Zero is undeniably an integer. Since every integer can be expressed as a fraction with a denominator of 1 (e.g., 0 = 0/1), 0 is inherently a rational number. The misconception might arise from the unique properties of zero in arithmetic operations, such as its role as the additive identity and its behavior in multiplication. However, these operational characteristics do not negate its classification as a rational number. Therefore, the claim that 0 is not a rational number is demonstrably false. Furthermore, since a number cannot be both rational and irrational, and 0 is rational, it cannot be irrational. Thus, the complete statement of Option A is incorrect, making it an unsuitable answer to our question.

Dissecting Option B: $\sqrt{2}$ is a rational number

Now, let's turn our attention to Option B, which asserts that the square root of 2 ($\sqrt{2}$) is a rational number. To tackle this, we need to investigate whether $\sqrt{2}$ can be expressed in the form p/q, where p and q are integers. The square root of 2 is one of the most classic examples of an irrational number, a fact that has been known since ancient times. There are several ways to prove this, one of the most common being a proof by contradiction. The proof starts by assuming that $\sqrt{2}$ is rational, meaning it can be written as a/b, where a and b are integers with no common factors other than 1 (i.e., the fraction is in simplest form). Squaring both sides gives 2 = a^2/*b*2, which can be rearranged to 2b^2 = a^2. This equation implies that a^2 is an even number, since it is a multiple of 2. If a^2 is even, then a must also be even (because the square of an odd number is odd). Therefore, we can write a = 2k for some integer k. Substituting this back into the equation 2b^2 = a^2 gives 2b^2 = (2k)^2 = 4k^2. Dividing both sides by 2, we get b^2 = 2k^2. Now, this equation implies that b^2 is also even, and consequently, b is even. We have now shown that both a and b are even, which contradicts our initial assumption that a and b have no common factors other than 1. This contradiction demonstrates that our initial assumption that $\sqrt{2}$ is rational must be false. Therefore, $\sqrt{2}$ is not a rational number; it is an irrational number. This makes Option B incorrect.

Evaluating Option C: $1.\overline{3}$ is a rational number but not an integer

Option C presents $1.\overline3}$ and claims it is a rational number but not an integer. Here, $1.\overline{3}$ represents a repeating decimal, specifically 1.3333..., where the 3s continue infinitely. To determine if this is a rational number, we need to see if it can be expressed as a fraction. Let x = 1.3333.... Multiplying both sides by 10, we get 10x = 13.3333.... Now, subtracting the original equation from this new equation, we have 10x - x = 13.3333... - 1.3333..., which simplifies to 9x = 12. Dividing both sides by 9, we get x = 12/9. This fraction can be simplified to 4/3. Since $1.\overline{3}$ can be expressed as the fraction 4/3, it is indeed a rational number. The second part of the statement claims that $1.\overline{3}$ is not an integer. An integer is a whole number (without any fractional part). The number 4/3 is equivalent to 1 and 1/3, or 1.3333..., which clearly has a fractional component. Therefore, $1.\overline{3}$ is not an integer. Both parts of Option C's statement are true $1.\overline{3$ is a rational number, and it is not an integer. This makes Option C a strong candidate for the correct answer.

Examining Option D: $-\sqrt{16}$ is an irrational number

Finally, we analyze Option D, which states that $-\sqrt{16}$ is an irrational number. To evaluate this claim, we first need to simplify $-\sqrt{16}$. The square root of 16 is 4, because 4 multiplied by itself equals 16. Therefore, $-\sqrt{16}$ is equal to -4. Now, we must determine if -4 is an irrational number. Recall that irrational numbers cannot be expressed as a fraction of two integers. The number -4 can be expressed as -4/1, which fits the definition of a rational number. It is an integer and therefore also a rational number. Thus, $-\sqrt{16}$ is not an irrational number; it is a rational number. This means that Option D is incorrect. The misconception might stem from confusing square roots of non-perfect squares (like $\sqrt{2}$, $\sqrt{3}$, $\sqrt{5}$) with square roots of perfect squares (like $\sqrt{16}$, $\sqrt{25}$, $\sqrt{36}$). The former are irrational, while the latter yield integers, which are rational. Therefore, the statement in Option D is false.

Conclusion: Identifying the Correct Answer

Having meticulously analyzed each option, we can now confidently identify the correct answer. Option A is incorrect because 0 is a rational number. Option B is incorrect because $\sqrt{2}$ is an irrational number. Option D is incorrect because $-\sqrt{16}$ is a rational number. Option C, stating that $1.\overline{3}$ is a rational number but not an integer, is the only statement that holds true. Therefore, the correct answer is C. Understanding the properties and classifications of numbers is essential in mathematics. This detailed explanation not only provides the answer but also reinforces the foundational concepts of rational and irrational numbers, integers, and decimal representations. This comprehensive approach ensures a deeper understanding of the subject matter, promoting mathematical literacy and problem-solving skills.

Keywords

rational number, irrational number, integers, repeating decimals, non-repeating decimals, square root of 2, fraction, mathematics