Question

8.ns.a.1 : determine whether the following are rational or irrational numbers.

	r/irr		r/irr
$sqrt{16}$		$sqrt{(25/4)}$
$-\sqrt{36}$		$4.\overline{3}$
$3.7$		$4/9$
$sqrt{244}$		$4\pi/\pi$
$7/5$		$sqrt{6^2}$
$-2/3$		$sqrt{18}$
$\pi$		$21.989$
$sqrt{110}$		$sqrt{19}$
$-88$		$sqrt{230}$
$2/8$		$sqrt{121}$
$sqrt{37}$		$58$

© beulah jonnalagadda 2022

Explanation:

Step1: Recall definitions

A rational number is a number that can be expressed as $\frac{p}{q}$, where $p$ and $q$ are integers and $q
eq0$. An irrational number is a non - repeating, non - terminating decimal and cannot be expressed as a fraction of two integers. Also, the square root of a non - perfect square is irrational, and $\pi$ is irrational.

Step2: Analyze each number

• $\boldsymbol{\sqrt{2}}$: 2 is not a perfect square. So $\sqrt{2}$ is irrational (IRR).
• $\boldsymbol{2\pi}$: $\pi$ is irrational, and multiplying an irrational number by a non - zero rational number (2) gives an irrational number (IRR).
• $\boldsymbol{\sqrt{16}}$: $\sqrt{16} = 4=\frac{4}{1}$, so it is rational (R).
• $\boldsymbol{\sqrt{\frac{25}{4}}}$: $\sqrt{\frac{25}{4}}=\frac{5}{2}$, which is a fraction of two integers. So it is rational (R).
• $\boldsymbol{-\sqrt{36}}$: $-\sqrt{36}=- 6=\frac{-6}{1}$, so it is rational (R).
• $\boldsymbol{4.\overline{3}}$: It is a repeating decimal. Repeating decimals can be expressed as fractions. Let $x = 4.\overline{3}$, then $10x=43.\overline{3}$, $10x - x=43.\overline{3}-4.\overline{3}$, $9x = 39$, $x=\frac{13}{3}$. So it is rational (R).
• $\boldsymbol{3.7}$: $3.7=\frac{37}{10}$, so it is rational (R).
• $\boldsymbol{\frac{4}{9}}$: It is a fraction of two integers, so it is rational (R).
• $\boldsymbol{\sqrt{244}}$: 244 is not a perfect square ($15^2 = 225$, $16^2=256$), so $\sqrt{244}$ is irrational (IRR).
• $\boldsymbol{\frac{4\pi}{\pi}}$: $\frac{4\pi}{\pi}=4=\frac{4}{1}$ (after canceling $\pi$), so it is rational (R).
• $\boldsymbol{\frac{7}{5}}$: It is a fraction of two integers, so it is rational (R).
• $\boldsymbol{\sqrt{6^{2}}}$: $\sqrt{6^{2}} = 6=\frac{6}{1}$, so it is rational (R).
• $\boldsymbol{-\frac{2}{3}}$: It is a fraction of two integers, so it is rational (R).
• $\boldsymbol{\sqrt{18}}$: 18 is not a perfect square ($4^2 = 16$, $5^2 = 25$), so $\sqrt{18}$ is irrational (IRR).
• $\boldsymbol{\pi}$: $\pi$ is a non - repeating, non - terminating decimal, so it is irrational (IRR).
• $\boldsymbol{21.989}$: $21.989=\frac{21989}{1000}$, so it is rational (R).
• $\boldsymbol{\sqrt{110}}$: 110 is not a perfect square ($10^2 = 100$, $11^2 = 121$), so $\sqrt{110}$ is irrational (IRR).
• $\boldsymbol{\sqrt{19}}$: 19 is not a perfect square ($4^2 = 16$, $5^2 = 25$), so $\sqrt{19}$ is irrational (IRR).
• $\boldsymbol{-88}$: $-88=\frac{-88}{1}$, so it is rational (R).
• $\boldsymbol{\sqrt{230}}$: 230 is not a perfect square ($15^2 = 225$, $16^2 = 256$), so $\sqrt{230}$ is irrational (IRR).
• $\boldsymbol{\frac{2}{8}}$: $\frac{2}{8}=\frac{1}{4}$, which is a fraction of two integers, so it is rational (R).
• $\boldsymbol{\sqrt{121}}$: $\sqrt{121}=11=\frac{11}{1}$, so it is rational (R).
• $\boldsymbol{\sqrt{37}}$: 37 is not a perfect square ($6^2 = 36$, $7^2 = 49$), so $\sqrt{37}$ is irrational (IRR).
• $\boldsymbol{58}$: $58=\frac{58}{1}$, so it is rational (R).

Answer:

Number	R/IRR	Number	R/IRR
$\sqrt{16}$	R	$\sqrt{\frac{25}{4}}$	R
$-\sqrt{36}$	R	$4.\overline{3}$	R
$3.7$	R	$\frac{4}{9}$	R
$\sqrt{244}$	IRR	$\frac{4\pi}{\pi}$	R
$\frac{7}{5}$	R	$\sqrt{6^{2}}$	R
$-\frac{2}{3}$	R	$\sqrt{18}$	IRR
$\pi$	IRR	$21.989$	R
$\sqrt{110}$	IRR	$\sqrt{19}$	IRR
$-88$	R	$\sqrt{230}$	IRR
$\frac{2}{8}$	R	$\sqrt{121}$	R
$\sqrt{37}$	IRR	$58$	R