Integers and rational numbers

Integers, fractions, ordering, decimals, repeating decimals and irrational numbers. 8 problems with step-by-step solutions.

Solve each problem in your notebook first, then open the solution.

Problem 1 Integers

a) \(-3 < m < 3\), \(m \in \mathbb{Z}\). Find the greatest value of \(m^2 - 2m + 5\).

b) Evaluate \(4 - (4 - (4 - (4 - 3)))\).

c) Evaluate \(1 - 2 + 3 - 4 + \dots - 120\).

d) Evaluate \(-[25 + (-13)]\cdot[(-8) + 3]\).

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a) \(m^2 - 2m + 5 = (m - 1)^2 + 4\), largest at \(m = -2\): \(9 + 4 = 13\).

b) \(4 - 3 = 1\), \(4 - 1 = 3\), \(4 - 3 = 1\), \(4 - 1 = 3\). Answer: 3.

c) 60 pairs \((1 - 2), (3 - 4), \dots\) each equal \(-1\): \(-60\).

d) \(-12\cdot(-5) = 60\).

Problem 2 Integer values

a) \(\dfrac{3a + 14}{a + 2}\) is an integer. Find all integers \(a\).

b) \(x\) and \(y\) are integers with \(y = \dfrac{12}{x - 1}\). Find the sum of all possible \(x\).

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a) \(\frac{3a + 14}{a + 2} = 3 + \frac{8}{a + 2}\). \(a + 2 \in \{\pm1, \pm2, \pm4, \pm8\}\), so \(a \in \{-1, -3, 0, -4, 2, -6, 6, -10\}\).

b) \(x - 1\) divides 12: 12 divisors \(\pm1, \pm2, \pm3, \pm4, \pm6, \pm12\), they add up to 0. So the sum of the \(x\) values is \(12\cdot 1 + 0 = 12\).

Problem 3 Fractions

Calculate.

a) \(\frac23 + \frac56 - \frac34\) b) \(\left(1 - \frac14\right)\left(1 - \frac19\right)\left(1 - \frac{1}{16}\right)\) c) \(\dfrac{\frac35 - \frac12}{\frac{1}{4} + \frac{1}{20}}\) d) \(2\frac13\cdot 1\frac27 : 3\)

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a) \(\frac{8 + 10 - 9}{12} = \frac{9}{12} = \frac34\)
b) \(\frac34\cdot\frac89\cdot\frac{15}{16} = \frac{360}{576} = \frac58\)
c) \(\frac{1/10}{6/20} = \frac{1}{10}\cdot\frac{20}{6} = \frac13\)
d) \(\frac73\cdot\frac97\cdot\frac13 = 1\)

Problem 4 Telescoping

a) Evaluate \(\dfrac{1}{1\cdot 2} + \dfrac{1}{2\cdot 3} + \dots + \dfrac{1}{19\cdot 20}\).

b) Evaluate \(A = \left(1 + \frac12\right)\left(1 + \frac13\right)\cdots\left(1 + \frac{1}{49}\right)\).

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a) \(\frac{1}{k(k + 1)} = \frac1k - \frac{1}{k + 1}\). The sum is \(1 - \frac{1}{20} = \frac{19}{20}\).

b) \(\frac32\cdot\frac43\cdots\frac{50}{49} = \frac{50}{2} = 25\).

Problem 5 Ordering fractions

a) Order \(\frac{5}{8}, \frac{7}{12}, \frac{2}{3}, \frac{11}{18}\) from smallest to largest.

b) Find three fractions between \(\frac25\) and \(\frac12\).

c) \(2 < a < b < 30\) are integers. Find the greatest value of \(\frac{a}{b}\) and of \(\frac{b}{a}\).

Show solution

a) Common denominator 72: \(\frac{45}{72}, \frac{42}{72}, \frac{48}{72}, \frac{44}{72}\). Order: \(\frac{7}{12} < \frac{11}{18} < \frac58 < \frac23\).

b) \(\frac25 = \frac{16}{40}\), \(\frac12 = \frac{20}{40}\): \(\frac{17}{40}, \frac{18}{40}, \frac{19}{40}\).

c) \(\frac{a}{b}\) largest for \(a = 28\), \(b = 29\): \(\frac{28}{29}\). \(\frac{b}{a}\) largest for \(a = 3\), \(b = 29\): \(\frac{29}{3}\).

Problem 6 Decimals

a) Convert to decimals: \(\frac38\), \(-\frac{7}{20}\), \(\frac{5}{6}\).

b) Convert to fractions: \(-0.36\), \(2.625\), \(4.08\).

c) Evaluate \(\dfrac{0.003}{0.06} + \dfrac{0.08}{0.002}\).

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a) \(0.375\), \(-0.35\), \(0.8\overline{3}\).

b) \(-\frac{9}{25}\), \(\frac{21}{8}\), \(\frac{102}{25}\).

c) \(0.05 + 40 = 40.05\).

Problem 7 Repeating decimals

a) Convert to fractions: \(0.\overline{7}\), \(0.\overline{36}\), \(1.2\overline{5}\), \(0.4\overline{18}\).

b) Evaluate \(0.\overline{3} + 0.\overline{6}\) and \(0.\overline{1} + 0.\overline{2} + \dots + 0.\overline{8}\).

Show solution

Rule: (number without the point − non-repeating part) over (as many 9s as repeating digits, then as many 0s as non-repeating decimals).

a) \(\frac79\); \(\frac{36}{99} = \frac{4}{11}\); \(\frac{125 - 12}{90} = \frac{113}{90}\); \(\frac{418 - 4}{990} = \frac{414}{990} = \frac{23}{55}\).

b) \(\frac39 + \frac69 = 1\). \(\frac{1 + 2 + \dots + 8}{9} = \frac{36}{9} = 4\).

Problem 8 Irrational numbers

a) Prove that \(\sqrt3\) is irrational.

b) Is \(\sqrt{2}\cdot\sqrt{8}\) rational? Is \(\sqrt2 + \sqrt8\)?

Show solution

a) Suppose \(\sqrt3 = \frac{p}{q}\) in lowest terms. Then \(p^2 = 3q^2\), so 3 divides \(p\); write \(p = 3k\). Then \(9k^2 = 3q^2\), \(q^2 = 3k^2\), so 3 divides \(q\) too. Contradiction with lowest terms.

b) \(\sqrt{16} = 4\): rational. \(\sqrt2 + 2\sqrt2 = 3\sqrt2\): irrational.