To classify [tex]\(\sqrt{32}\)[/tex], let's examine its properties step-by-step.
### Step 1: Simplification and Calculation
Firstly, we simplify and calculate [tex]\(\sqrt{32}\)[/tex]:
[tex]\[
\sqrt{32} = \sqrt{16 \cdot 2} = \sqrt{16} \cdot \sqrt{2} = 4\sqrt{2}
\][/tex]
We know that [tex]\(\sqrt{2}\)[/tex] is an irrational number. Therefore, [tex]\(4\sqrt{2}\)[/tex] is also irrational because the product of a rational number (4) and an irrational number ([tex]\(\sqrt{2}\)[/tex]) is irrational.
### Step 2: Determining the Type of Decimal Representation
Next, we need to determine whether [tex]\(\sqrt{32}\)[/tex] is a terminating or non-terminating decimal.
An irrational number cannot be expressed as a fraction of two integers, which implies that its decimal representation is non-terminating and non-repeating.
### Conclusion
Based on the properties of irrational numbers and the decimal characteristics we've discussed:
- [tex]\(\sqrt{32}\)[/tex] is an irrational number.
- Its decimal representation is non-terminating and non-repeating.
Therefore, the correct classification of [tex]\(\sqrt{32}\)[/tex] is:
[tex]\[
\boxed{\text{irrational number, nonrepeating decimal}}
\][/tex]
