Question

it has been demonstrated that \\(sqrt{96}\\) is between 9.79 and 9.80. if we were to continue this process, it would because \\(sqrt{96}\\) is .

Explanation:

To solve this, we analyze the nature of \(\sqrt{96}\):

Step 1: Recall the property of irrational numbers

Irrational numbers are non - repeating and non - terminating decimals. The square root of a non - perfect square number is an irrational number. Since 96 is not a perfect square (\(9^2 = 81\) and \(10^2=100\), and 96 is between 81 and 100 but not a perfect square), \(\sqrt{96}\) is an irrational number.

Step 2: Analyze the process of approximating \(\sqrt{96}\)

When we approximate an irrational number like \(\sqrt{96}\), we can use methods such as the Newton - Raphson method or successive approximation. Since \(\sqrt{96}\) is irrational, its decimal expansion is non - repeating and non - terminating. So, if we continue the process of approximating \(\sqrt{96}\) (for example, finding more decimal places), the process would never end (or be an infinite process) because \(\sqrt{96}\) is an irrational number (or a non - repeating, non - terminating decimal).

For the first blank (the action of the process): The process would "never end" (or similar phrases like "continue infinitely", "not terminate").

For the second blank: \(\sqrt{96}\) is "an irrational number" (or "a non - repeating, non - terminating decimal").

Answer:

To solve this, we analyze the nature of \(\sqrt{96}\):

Step 1: Recall the property of irrational numbers

Irrational numbers are non - repeating and non - terminating decimals. The square root of a non - perfect square number is an irrational number. Since 96 is not a perfect square (\(9^2 = 81\) and \(10^2=100\), and 96 is between 81 and 100 but not a perfect square), \(\sqrt{96}\) is an irrational number.

Step 2: Analyze the process of approximating \(\sqrt{96}\)

When we approximate an irrational number like \(\sqrt{96}\), we can use methods such as the Newton - Raphson method or successive approximation. Since \(\sqrt{96}\) is irrational, its decimal expansion is non - repeating and non - terminating. So, if we continue the process of approximating \(\sqrt{96}\) (for example, finding more decimal places), the process would never end (or be an infinite process) because \(\sqrt{96}\) is an irrational number (or a non - repeating, non - terminating decimal).

For the first blank (the action of the process): The process would "never end" (or similar phrases like "continue infinitely", "not terminate").

For the second blank: \(\sqrt{96}\) is "an irrational number" (or "a non - repeating, non - terminating decimal").