Identifying Irrational Number Between 3 and 5
The question asks for an irrational number that falls strictly between 3 and 5.
Analyzing the Options
We need to evaluate each option to determine if it is an irrational number and if it lies within the specified range (3, 5).
-
Option 1: $\sqrt{5}$
Since $2^2 = 4$ and $3^2 = 9$, we know that $2 < \sqrt{5} < 3$. This value is not between 3 and 5. -
Option 2: $\sqrt{27}$
Since $5^2 = 25$ and $6^2 = 36$, we know that $5 < \sqrt{27} < 6$. This value is not between 3 and 5. -
Option 3: $\sqrt{3}$
Since $1^2 = 1$ and $2^2 = 4$, we know that $1 < \sqrt{3} < 2$. This value is not between 3 and 5. -
Option 4: $\sqrt{17}$
Since $4^2 = 16$ and $5^2 = 25$, we know that $4 < \sqrt{17} < 5$. This value lies between 3 and 5.
Checking for Irrationality
An irrational number cannot be expressed as a simple fraction $p/q$. Square roots of non-perfect squares are irrational numbers.
- $\sqrt{5}$, $\sqrt{27}$, $\sqrt{3}$, and $\sqrt{17}$ are all irrational numbers because 5, 27, 3, and 17 are not perfect squares.
Conclusion
Out of the given options, only $\sqrt{17}$ is an irrational number that falls within the range of 3 to 5. Therefore, $\sqrt{17}$ is the correct answer.
Similar Questions
- $(\sqrt{5} + \sqrt{7})^2$ is a:
- Which of the following rational number lies between $\frac{1}{4}$ and $\frac{1}{2}$.
- Which of the following rational number lies between 9.2 and 10.5?
- The arrangement of rational numbers $-\frac{7}{10}, \frac{5}{-8}, \frac{2}{-3}$ in ascending order is:
Important Questions from Rational or Irrational Numbers
If \(\sqrt{1+\frac{\sqrt{3}}{2}}- \sqrt{1-\frac{\sqrt{3}}{2}}= c\) , then the value of c is:
If \(\frac{\sqrt{38-5\sqrt{3} } }{\sqrt{26+7\sqrt{3} } }= \frac{a+b\sqrt{3} }{23} \) , b > 0, then the value of (b – a) is:
If \( \frac{5}{4{\sqrt 2 }} + \frac{{3 + 2\sqrt 2 }}{{3 - 2\sqrt 2 }} - \frac{{3 - 2\sqrt 2 }}{{3 + 2\sqrt 2 }} = a + b\sqrt 2 \) , then what is the value of (3a + 4b)?
If \(\frac {8 + 2\sqrt 3}{3\sqrt 3 + 5} = a\sqrt 3 - b,\) then the value of a + b is equal to:
If \(\frac{\sqrt{26-7\sqrt{3} } }{\sqrt{14+5\sqrt{3} } } = \frac{b+a\sqrt{3} }{11}\) , b > 0, then what is the value of \(\sqrt{(b-a)} \) ?
