Question

the number, \\(\sqrt{3}\\) is
a whole number
an integer
rational
irrational

Explanation:

To determine the type of \(\sqrt{3}\), we recall the definitions:

• Whole numbers: Non - negative integers (\(0, 1, 2, \dots\)). \(\sqrt{3}\approx1.732\) is not a whole number.
• Integers: Positive/negative whole numbers or zero. \(\sqrt{3}\) is not an integer.
• Rational numbers: Can be expressed as \(\frac{p}{q}\) (\(p,q\) integers, \(q

eq0\)). Assume \(\sqrt{3}=\frac{p}{q}\) (in lowest terms), then \(3q^{2}=p^{2}\), so \(p\) is divisible by 3. Let \(p = 3k\), then \(3q^{2}=9k^{2}\Rightarrow q^{2}=3k^{2}\), so \(q\) is also divisible by 3, contradicting \(\frac{p}{q}\) in lowest terms. Thus, \(\sqrt{3}\) is not rational.

• Irrational numbers: Cannot be expressed as \(\frac{p}{q}\), and their decimal expansions are non - repeating, non - terminating. Since \(\sqrt{3}\) fails to be whole, integer, or rational, it is irrational.

Answer:

irrational