Question

classify the numbers as rational or irrational. \\(\sqrt{29}\\) \\(-2\\) \\(\sqrt{29} - -2\\) \\(\sqrt{29} + -2\\) \
rational \
irrational

Explanation:

Step1: Recall definitions

A rational number is a number that can be expressed as $\frac{p}{q}$ where $p,q$ are integers and $q
eq0$. An irrational number is a non - repeating, non - terminating decimal, and square roots of non - perfect squares are irrational.

Step2: Analyze $\sqrt{29}$

Since 29 is not a perfect square (because $5^2 = 25$ and $6^2=36$, and 29 is between them), $\sqrt{29}$ is an irrational number.

Step3: Analyze $- 2$

$-2$ can be written as $\frac{-2}{1}$, where $p=-2$ and $q = 1$ are integers with $q
eq0$. So, $-2$ is a rational number.

Step4: Analyze $\sqrt{29}-(-2)=\sqrt{29} + 2$

The sum of an irrational number ($\sqrt{29}$) and a rational number (2) is an irrational number. Because if we assume $\sqrt{29}+2=\frac{p}{q}$ (rational), then $\sqrt{29}=\frac{p}{q}-2=\frac{p - 2q}{q}$, which would imply $\sqrt{29}$ is rational, a contradiction.

Step5: Analyze $\sqrt{29}+(-2)=\sqrt{29}-2$

The difference of an irrational number ($\sqrt{29}$) and a rational number (2) is an irrational number. Because if we assume $\sqrt{29}-2=\frac{p}{q}$ (rational), then $\sqrt{29}=\frac{p}{q}+2=\frac{p + 2q}{q}$, which would imply $\sqrt{29}$ is rational, a contradiction.

Answer:

• Rational: $-2$
• Irrational: $\sqrt{29}$, $\sqrt{29}-(-2)$, $\sqrt{29}+(-2)$