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.