is √8 an irrational number? yes no

Explanation:

Step1: Simplify the square - root.

We know that $\sqrt{8}=\sqrt{4\times2}$. According to the property $\sqrt{ab}=\sqrt{a}\times\sqrt{b}$ ($a = 4$, $b = 2$ and $a\geq0$, $b\geq0$), so $\sqrt{8}=\sqrt{4}\times\sqrt{2}=2\sqrt{2}$.

Step2: Recall the definition of irrational numbers.

An irrational number is a number that cannot be written as a fraction $\frac{p}{q}$ where $p$ and $q$ are integers and $q\neq0$. Since $\sqrt{2}$ is an irrational number and $2\sqrt{2}$ is a non - zero multiple of $\sqrt{2}$, $2\sqrt{2}$ (i.e., $\sqrt{8}$) is also an irrational number.

Answer:

yes