Question

irrational number between 9 & 11

Answers

Answer
Show an example of an irrational number between 9 and 11.
Responses

10−−√
square root of 10 - incorrect

99−−√
square root of 99 - not selected, this is the correct answer

100−−−√
square root of 100 - no response given

20−−√
Answer
why?
Answered by GPT-4o mini
An example of an irrational number between 9 and 11 is \( \sqrt{100} + \frac{\sqrt{2}}{2} \).

To confirm this:

1. We calculate \( \sqrt{100} = 10 \).
2. Next, we find \( \frac{\sqrt{2}}{2} \), which is an irrational number approximately equal to 0.7071.

Now, \( 10 + 0.7071 \) equals approximately 10.7071, which is between 9 and 11.

Another simple example is \( \pi + 7 \), since \( \pi \approx 3.14 \) and \( 3.14 + 7 = 10.14 \), which is also between 9 and 11.
Answered by GPT-4o mini
The correct response to find an irrational number between 9 and 11 is indeed \(\sqrt{99}\).

To confirm:

- The square root of 99 (approximately 9.95) is between 9 and 11, and it is irrational because 99 is not a perfect square.

Here are the other values mentioned:

- \(\sqrt{10} \approx 3.16\), which is not between 9 and 11.
- \(\sqrt{100} = 10\), which is not irrational.
- \(\sqrt{20} \approx 4.47\), which is also not between 9 and 11.

So, \(\sqrt{99}\) is indeed the correct choice as an irrational number between 9 and 11.
Answered by GPT-4o mini
Let's delve deeper into why \(\sqrt{99}\) is an irrational number and why it's between 9 and 11.

### 1. **Is \(\sqrt{99}\) Irrational?**

A number is considered irrational if it cannot be expressed as a fraction of two integers (i.e., in the form \(\frac{p}{q}\)), where \(p\) and \(q\) are integers and \(q \neq 0\).

- Since \(99\) is not a perfect square, there are no integers \(a\) and \(b\) such that \(a^2 = 99\).
- Therefore, \(\sqrt{99}\) is irrational, because it cannot be simplified to a fraction of two integers.

### 2. **Is \(\sqrt{99}\) Between 9 and 11?**

To show that \(\sqrt{99}\) lies between 9 and 11, we need to evaluate:

- \(9^2 = 81\)
- \(10^2 = 100\)

Since \(99\) is greater than \(81\) but less than \(100\), we can conclude that:

\[
9 < \sqrt{99} < 10
\]

Thus, \(\sqrt{99}\) is greater than 9 and less than 10.

If we were looking for an example of an irrational number between 9 and 11, both \(\sqrt{99}\) (approximately 9.95) and any other number like \(\pi + 7\) (approximately 10.14) would qualify as they are irrational and fit within the specified range.

### Summary

- **Irrationality**: \(\sqrt{99}\) cannot be expressed as a fraction of two integers.
- **Range**: \(\sqrt{99}\) is greater than 9 and less than 10, placing it comfortably between 9 and 11.

Hence, \(\sqrt{99}\) is indeed a valid example of an irrational number between 9 and 11.

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