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# Solved Examples on Proofs

Prerequisites:
(1.) Factoring
(2.) Inequality
(3.) Pascal's Triangle
(4.) Exponents and Logarithms

Attempt all questions.
Identify the type of proof.
Show all work.

(1.) If $d, e, f$ are integers such that:
$e$ is a multiple of $d^3$ and
$f$ is a multiple of $e^4$
Prove that $f$ is a multiple of $d^{12}$

Direct Proof

$e\;\;is\;\;a\;\;multiple\;\;of\;\;d^3 \$3ex] \implies e = pd^3 \;\;where\;\;p\;\;is\;\;an\;\;integer...Definition\;\;of\;\;a\;\;multiple \\[3ex] f\;\;is\;\;a\;\;multiple\;\;of\;\;e^4 \\[3ex] \implies f = re^4 \;\;where\;\;r\;\;is\;\;an\;\;integer...Definition\;\;of\;\;a\;\;multiple \\[3ex] f = re^4 \\[3ex] But\;\; e = pd^3 \\[3ex] f = r(pd^3)^4 \\[3ex] f = rp^4d^{12} \\[3ex] Because\;\;p\;\;is\;\;an\;\;integer,\;\;p^4\;\;is\;\;also\;\;an\;\;integer \\[3ex] r * p^4\;\;is\;\;also\;\;an\;\;integer \\[3ex] \therefore f = some\;\;integer * d^{12} \\[3ex] \implies f\;\;is\;\;a\;\;multiple\;\;of\;\;d^{12} (2.) Prove that the sum of two even integers is even. Direct Proof Say\;\;c\;\;and\;\;d\;\;are\;\;even\;\;integers \\[3ex] For\;\;some\;\;integers:\;\;e\;\;and\;\;f \\[3ex] c = 2e...Definition\;\;of\;\;even \\[3ex] d = 2f...Definition\;\;of\;\;odd \\[3ex] c + d \\[3ex] = 2e + 2f \\[3ex] = 2(e + f) \\[3ex] Because\;\;e\;\;and\;\;f\;\;are\;\;integers,\;\;e + f\;\;is\;\;an\;\;integer \\[3ex] = 2 * some\;\;integer \\[3ex] = even (3.) Prove that the average of two numbers is greater than or equal to at least one of the numbers using proof by contradiction. Proof by Contradiction This question deals with Inequalities (greater than or equal to, less than) We shall use the Rules of Inequalities Statement: The average of two numbers is greater than or equal to at least one of the numbers. Contradiction: The average of two numbers is less than one of the numbers. Let\;\;c, d\;\;be\;\;the\;\;two\;\;numbers \\[3ex] Average\;\;of\;\;c\;\;and\;\;d = \dfrac{c + d}{2} \\[5ex] \underline{Statement} \\[3ex] \dfrac{c + d}{2} \ge c \\[5ex] \dfrac{c + d}{2} \ge d \\[5ex] \underline{Contradiction} \\[3ex] \dfrac{c + d}{2} \lt c...ineq.(1) \\[5ex] \dfrac{c + d}{2} \lt d...ineq.(2) \\[5ex] Rules\;\;of\;\;Inequality...Number(9.) \\[3ex] \dfrac{c + d}{2} + \dfrac{c + d}{2} \lt c + d \\[5ex] \dfrac{c + d + c + d}{2} \lt c + d \\[5ex] \dfrac{2c + 2d}{2} \lt c + d \\[5ex] \dfrac{2(c + d)}{2} \lt c + d \\[5ex] c + d \lt c + d \\[3ex] This\;\;is\;\;false \\[3ex] \therefore the\;\;statement\;\;is\;\;true (4.) Prove that the sum of three consecutive integers is a multiple of three. Direct Proof Let\;\;the\;\;first\;\;integer = n \\[3ex] Next\;\;consecutive\;\;integer = n + 1 \\[3ex] Next\;\;consecutive\;\;integer = n + 2 \\[3ex] Sum\;\;of\;\;three\;\;consecutive\;\;integers \\[3ex] = n + (n + 1) + (n + 2) \\[3ex] = n + n + 1 + n + 2 \\[3ex] = 3n + 3 \\[3ex] = 3(n + 1) \\[3ex] Because\;\;n\;\;is\;\;an\;\;integer \\[3ex] (n + 1) \;\;is\;\;also\;\;an\;\;integer \\[3ex] = 3 * some\;\;integer \\[3ex] = Definition\;\;of\;\;multiple\;\;of\;\;3 (5.) Prove that for all integers n, if n^2 is even, then n is even using proof by contrapositive. Proof by Contrapositive \underline{Statement} \\[3ex] If\;\;n^2\;\;is\;\;even,\;\;then\;\;n\;\;is\;\;even \\[3ex] \underline{Contrapositive} \\[3ex] If\;\;n\;\;is\;\;not\;\;even,\;\;then\;\;n^2\;\;is\;\;not\;\;even \\[3ex] This\;\;means\;\;that \\[3ex] If\;\;n\;\;is\;\;odd,\;\;then\;\;n^2\;\;is\;\;odd \\[3ex] n = 2e + 1\;\;for\;\;some\;\;integer\;\;e...Definition\;\;of\;\;odd \\[3ex] n^2 \\[3ex] = (2e + 1)^2 \\[3ex] = (2e + 1)(2e + 1) \\[3ex] = 4e^2 + 2e + 2e + 1 \\[3ex] = 4e^2 + 4e + 1 \\[3ex] = 4(e^2 + e) + 1 \\[3ex] = 2[2(e^2 + e)] + 1 \\[3ex] Because\;\;e\;\;is\;\;an\;\;integer \\[3ex] e^2\;\;is\;\;also\;\;an\;\;integer \\[3ex] e^2 + e\;\;is\;\;also\;\;an\;\;integer \\[3ex] 2(e^2 + e)\;\;is\;\;also\;\;an\;\;integer \\[3ex] = 2[some\;\;integer] + 1...Definition\;\;of\;\;odd \\[3ex] = odd (6.) Prove that the sum of four consecutive integers is an even integer. Direct Proof Let\;\;the\;\;first\;\;integer = n \\[3ex] Next\;\;consecutive\;\;integer = n + 1 \\[3ex] Next\;\;consecutive\;\;integer = n + 2 \\[3ex] Next\;\;consecutive\;\;integer = n + 3 \\[3ex] Sum\;\;of\;\;four\;\;consecutive\;\;integers \\[3ex] = n + (n + 1) + (n + 2) + (n + 3) \\[3ex] = n + n + 1 + n + 2 + n + 3 \\[3ex] = 4n + 6 \\[3ex] = 2(2n + 3) \\[3ex] Because\;\;n\;\;is\;\;an\;\;integer \\[3ex] 2n\;\;is\;\;also\;\;an\;\;integer \\[3ex] 2n + 3\;\;is\;\;also\;\;an\;\;integer \\[3ex] = 2 * some\;\;integer \\[3ex] = Definition\;\;of\;\;even\;\;integer For Questions (7.) and (8.): Assume: a = \dfrac{b}{c} \\[5ex] d = \dfrac{e}{f} \\[5ex] p = a - d \\[3ex] p = \dfrac{j}{k} \\[5ex] where: a, d, p are simplified rational numbers p \ne 0 (7.) Prove or disprove that if c and f are odd, then k is odd. Direct Proof Even and Odd Integers Proof Test\;\;to\;\;disprove \\[3ex] p = a - d \\[3ex] a - d = p \\[3ex] \dfrac{b}{c} - \dfrac{e}{f} = \dfrac{h}{k} \\[5ex] \dfrac{3}{5} - \dfrac{1}{5} = \dfrac{3 - 1}{5} = \dfrac{2}{5} \\[5ex] Simplified\;\;fractions \\[3ex] 5, 5, 5\:\:are\;\;odd \\[3ex] Test\;\;again\;\;to\;\;disprove \\[3ex] \dfrac{3}{7} - \dfrac{1}{5} = \dfrac{15 - 7}{35} = \dfrac{8}{35} \\[5ex] Simplified\;\;fractions \\[3ex] 7, 5, 35\;\;are\;\;odd \\[3ex] Could\:\:not\:\:disprove \\[3ex] Prove \\[3ex] Let: \\[3ex] c = 2x + 1 ...Definition\:\:of\:\:odd\;\;integer \\[3ex] f = 2y + 1 ...Definition\;\;of\;\;odd\;\;integer \\[3ex] p = a - d \\[3ex] \dfrac{j}{k} = \dfrac{b}{c} - \dfrac{e}{f} \\[5ex] \dfrac{j}{k} = \dfrac{bf - ce}{ef} \\[5ex] \dfrac{j}{k} = \dfrac{b(2y + 1) - e(2x + 1)}{(2x + 1)(2y + 1)} \\[5ex] \implies k = (2x + 1)(2y + 1) \\[3ex] We need to show that k is odd Show that k is of the form: 2 * some\;\;value + 1...Definition of odd integer k = (2x + 1)(2y + 1) \\[3ex] = 4xy + 2x + 2y + 1 \\[3ex] = 2(xy + x + y) + 1 \\[3ex] = 2(some\;\;value) + 1 \\[3ex] \therefore k\;\;is\;\;odd \\[3ex] Proved. (8.) Prove or disprove that if c and f are even, then k is even. Direct Proof Even and Odd Integers Proof Test\;\;to\;\;disprove \\[3ex] p = a - d \\[3ex] a - d = p \\[3ex] \dfrac{b}{c} - \dfrac{e}{f} = \dfrac{h}{k} \\[5ex] \dfrac{3}{8} - \dfrac{1}{8} = \dfrac{3 - 1}{8} = \dfrac{2}{8} = \dfrac{1}{4} \\[5ex] Simplified\;\;fractions \\[3ex] 8, 8, 4\;\;are\;\;even \\[3ex] Test\;\;again\;\;to\;\;disprove \\[3ex] \dfrac{5}{6} - \dfrac{1}{6} = \dfrac{5 - 1}{6} = \dfrac{4}{6} = \dfrac{2}{3} \\[5ex] Simplified\;\;fractions \\[3ex] 6, 6\;\;are\;\;even \\[3ex] But\;\;3\;\;is\;\;odd \\[3ex] Disproved. (9.) Prove that the difference between the cube and the square of an integer is always an even number. Direct Proof We shall review all the definitions of an integer. Let\;\;p\;\;be\;\;an\;\;integer \\[3ex] \underline{Case\;1} \\[3ex] p = 0 \\[3ex] cube - square \\[3ex] 0^3 - 0^2 \\[3ex] = 0 \\[3ex] = even \\[5ex] \underline{Case\;2} \\[3ex] p\;\;is\;\;even \\[3ex] p = 2e\;\;where\;\;e\;\;is\;\;an\;\;integer...Definition\;\;of\;\;even \\[3ex] cube - square \\[3ex] (2e)^3 - (2e)^2 \\[3ex] = 8e^3 - 4e^2 \\[3ex] = 4e^2(2e - 1) \\[3ex] = 2[2e^2(2e - 1)] \\[3ex] Because\;\;e\;\;is\;\;an\;\;integer \\[3ex] 2e\;\;is\;\;also\;\;an\;\;integer \\[3ex] 2e - 1\;\;is\;\;also\;\;an\;\;integer \\[3ex] e^2\;\;is\;\;also\;\;an\;\;integer \\[3ex] 2e^2\;\;is\;\;also\;\;an\;\;integer \\[3ex] 2e^2(2e - 1)\;\;is\;\;also\;\;an\;\;integer \\[3ex] = 2[some\;\;integer]...Definition\;\;of\;\;even \\[3ex] = even \\[5ex] \underline{Case\;3} \\[3ex] p\;\;is\;\;odd \\[3ex] p = 2e + 1\;\;where\;\;e\;\;is\;\;an\;\;integer...Definition\;\;of\;\;odd \\[3ex] cube - square \\[3ex] (2e + 1)^3 - (2e + 1)^2 \\[3ex] (2e + 1)^2 \\[3ex] = (2e + 1)(2e + 1) \\[3ex] = (2e)^2 + 2e + 2e + 1 \\[3ex] = 4e^2 + 4e + 1 \\[3ex] (2e + 1)^3 \\[3ex] = (2e + 1)(2e + 1)^2 \\[3ex] = (2e + 1)(4e^2 + 4e + 1) \\[3ex] = 8e^3 + 8e^2 + 2e + 4e^2 + 4e + 1 \\[3ex] = 8e^3 + 12e^2 + 6e + 1 \\[3ex] cube - square \\[3ex] = (8e^3 + 12e^2 + 6e + 1) - (4e^2 + 4e + 1) \\[3ex] = 8e^3 + 12e^2 + 6e + 1 - 4e^2 - 4e - 1 \\[3ex] = 8e^3 + 8e^2 + 2e \\[3ex] = 2(4e^3 + 4e^2 + e) \\[3ex] Because\;\;e\;\;is\;\;an\;\;integer \\[3ex] 4e^2\;\;is\;\;also\;\;an\;\;integer \\[3ex] 4e^3\;\;is\;\;also\;\;an\;\;integer \\[3ex] 4e^3 + 4e^2 + e\;\;is\;\;also\;\;an\;\;integer \\[3ex] = 2[some\;\;integer]...Definition\;\;of\;\;even \\[3ex] = even (10.) Prove that the sum of the squares of two consecutive integers is an odd integer. Direct Proof Let\;\;the\;\;first\;\;integer = n \\[3ex] Next\;\;consecutive\;\;integer = n + 1 \\[3ex] Square\;\;n = n^2 \\[3ex] Square\;\;(n + 1) \\[3ex] = (n + 1)^2 \\[3ex] = (n + 1)(n + 1) \\[3ex] = n^2 + n + n + 1 \\[3ex] = n^2 + 2n + 1 \\[3ex] Sum\;\;n^2\;\;and\;\;(n + 1)^2 \\[3ex] = n^2 + n^2 + 2n + 1 \\[3ex] = 2n^2 + 2n + 1 \\[3ex] = 2(n^2 + n) + 1 \\[3ex] Because\;\;n\;is\;\;an\;\;integer \\[3ex] n^2\;\;is\;\;also\;\;an\;\;integer \\[3ex] n^2 + n \;\;is\;\;also\;\;an\;\;integer \\[3ex] = 2(some\;\;integer) + 1...Definition\;\;of\;\;odd \\[3ex] = odd (11.) Prove by mathematical induction that the formula works for the set of positive integers. 4 + 9 + 14 + ... + (5n - 1) = \dfrac{n}{2}(5n + 3) Proof by Mathematical Induction Steps LHS RHS 4 + 9 + 14 + ... + (5n - 1) \dfrac{n}{2}(5n + 3) Initial Step Test for n = 1 First term: 4 \dfrac{1}{2}[5(1) + 3] \dfrac{1}{2}(5 + 3) \dfrac{1}{2}(8) 4 The initial step works Optional testing Test for n = 2 First term + Second term: 4 + 9 13 \dfrac{2}{2}[5(2) + 3] 1(10 + 3) 1(13) 13 The optional testing works Optional testing Test for n = 3 First term + Second term + Third term: 4 + 9 + 14 27 \dfrac{3}{2}[5(3) + 3] \dfrac{3}{2}(15 + 3) \dfrac{3}{2}(18) 3(9) 27 The optional testing works Induction Hypothesis Assume n = k This is the kth term 4 + 9 + 14 + ... + (5k - 1) = \dfrac{k}{2}(5k + 3)...eqn.(1) Induction Step Assume n = k + 1 This is the (k + 1)st term 4 + 9 + 14 + ... + (5k - 1) + [5(k + 1) - 1] = \dfrac{k + 1}{2}[5(k + 1) + 3]...eqn.(2) Induction Step \color{darkblue}{4 + 9 + 14 + ... + (5k - 1)} + [5(k + 1) - 1] From \color{darkblue}{eqn. (1)}, substitute: \color{darkblue}{\dfrac{k}{2}(5k + 3)} + 5(k + 1) - 1 \dfrac{k}{2}(5k + 3) + 5k + 5 - 1 \dfrac{k}{2}(5k + 3) + 5k + 4 \dfrac{k(5k + 3)}{2} + \dfrac{5k + 4}{1} \dfrac{k(5k + 3) + 2(5k + 4)}{2} \dfrac{5k^2 + 3k + 10k + 8}{2} \dfrac{5k^2 + 13k + 8}{2} \dfrac{5k^2 + 5k + 8k + 8}{2} \dfrac{5k(k + 1) + 8(k + 1)}{2} \dfrac{(k + 1)(5k + 8)}{2} \dfrac{k + 1}{2}[5(k + 1) + 3] \dfrac{k + 1}{2}[5k + 5 + 3] \dfrac{k + 1}{2}(5k + 8) \dfrac{(k + 1)(5k + 8)}{2} LHS = RHS The formula is proved. (12.) Prove by mathematical induction that the sum of first n natural numbers beginning from 1 is given by the formula: \dfrac{n(n + 1)}{2} In other words, prove by mathematical induction that: \[ \sum_{p = 1}^{n} p = \dfrac{n(n + 1)}{2}$ In other words, prove by mathematical induction that: $1 + 2 + 3 + ... + n = \dfrac{n(n + 1)}{2}$ Proof by Mathematical Induction Steps$LHSRHS1 + 2 + 3 + ... + n\dfrac{n(n + 1)}{2}$Initial Step Test for$n = 1$First term:$1\dfrac{n(n + 1)}{2}\dfrac{1(1 + 1)}{2}\dfrac{1(2)}{2}1(1)1$The initial step works Optional testing Test for$n = 2$First term + Second term:$1 + 23\dfrac{2(2 + 1)}{2}\dfrac{2(3)}{2}1(3)3$Optional testing Test for$n = 3$First term + Second term + Third term:$1 + 2 + 36\dfrac{3(3 + 1)}{2}\dfrac{3(4)}{2}3(2)6$The optional testing works Induction Hypothesis Assume$n = k$This is the$kth$term$1 + 2 + 3 + ... + k = \dfrac{k(k + 1)}{2}...eqn.(1)$Induction Step Assume$n = k + 1$This is the$(k + 1)st$term$1 + 2 + 3 + ... + k + (k + 1) = \dfrac{(k + 1)[(k + 1) + 1]}{2}$Induction Step$\color{darkblue}{1 + 2 + 3 + ... + k} + (k + 1)$From$\color{darkblue}{eqn. (1)}$, substitute:$\color{darkblue}{\dfrac{k(k + 1)}{2}} + k + 1\dfrac{k(k + 1)}{2} + \dfrac{k + 1}{1}\dfrac{k(k + 1) + 2(k + 1)}{2}\dfrac{(k + 1)(k + 2)}{2}\dfrac{(k + 1)[(k + 1) + 1]}{2}\dfrac{(k + 1)[k + 1 + 1]}{2}\dfrac{(k + 1)[k + 2]}{2}\dfrac{(k + 1)(k + 2)}{2}LHS = RHS$The formula is proved. (13.) Prove by mathematical induction that the sum of first$n$natural numbers beginning from$1$is given by the formula:$\dfrac{n(n + 1)}{2}$In other words, prove by mathematical induction that: $\sum_{p = 1}^{n} p = \dfrac{n(n + 1)}{2}$ In other words, prove by mathematical induction that: $1 + 2 + 3 + ... + n = \dfrac{n(n + 1)}{2}$ Proof by Mathematical Induction Steps$LHSRHS1 + 2 + 3 + ... + n\dfrac{n(n + 1)}{2}$Initial Step Test for$n = 1$First term:$1\dfrac{n(n + 1)}{2}\dfrac{1(1 + 1)}{2}\dfrac{1(2)}{2}1(1)1$The initial step works Optional testing Test for$n = 2$First term + Second term:$1 + 23\dfrac{2(2 + 1)}{2}\dfrac{2(3)}{2}1(3)3$Optional testing Test for$n = 3$First term + Second term + Third term:$1 + 2 + 36\dfrac{3(3 + 1)}{2}\dfrac{3(4)}{2}3(2)6$The optional testing works Induction Hypothesis Assume$n = k$This is the$kth$term$1 + 2 + 3 + ... + k = \dfrac{k(k + 1)}{2}...eqn.(1)$Induction Step Assume$n = k + 1$This is the$(k + 1)st$term$1 + 2 + 3 + ... + k + (k + 1) = \dfrac{(k + 1)[(k + 1) + 1]}{2}$Induction Step$\color{darkblue}{1 + 2 + 3 + ... + k} + (k + 1)$From$\color{darkblue}{eqn. (1)}$, substitute:$\color{darkblue}{\dfrac{k(k + 1)}{2}} + k + 1\dfrac{k(k + 1)}{2} + \dfrac{k + 1}{1}\dfrac{k(k + 1) + 2(k + 1)}{2}\dfrac{(k + 1)(k + 2)}{2}\dfrac{(k + 1)[(k + 1) + 1]}{2}\dfrac{(k + 1)[k + 1 + 1]}{2}\dfrac{(k + 1)[k + 2]}{2}\dfrac{(k + 1)(k + 2)}{2}LHS = RHS$The formula is proved. (14.) Prove by mathematical induction that the sum of the cubes of the first$n$natural numbers beginning from$1$is given by the formula:$\left[\dfrac{n(n + 1)}{2}\right]^2$In other words, prove by mathematical induction that: $\sum_{p = 1}^{n} p^3 = \left[\dfrac{n(n + 1)}{2}\right]^2$ In other words, prove by mathematical induction that: $1^3 + 2^3 + 3^3 + ... + n^3 = \left[\dfrac{n(n + 1)}{2}\right]^2$ Proof by Mathematical Induction Steps$LHSRHS1^3 + 2^3 + 3^3 + ... + n^3\left[\dfrac{n(n + 1)}{2}\right]^2$Initial Step Test for$n = 1$First term:$1^31\left[\dfrac{n(n + 1)}{2}\right]^2\left[\dfrac{1(1 + 1)}{2}\right]^2\left[\dfrac{1(2)}{2}\right]^2\left[1(1)\right]^21^21$The initial step works Optional testing Test for$n = 2$First term + Second term:$1^3 + 2^31 + 89\left[\dfrac{2(2 + 1)}{2}\right]^2\left[\dfrac{2(3)}{2}\right]^2\left[1(3)\right]^23^29$Optional testing Test for$n = 3$First term + Second term + Third term:$1^3 + 2^3 + 3^31 + 8 + 2736\left[\dfrac{3(3 + 1)}{2}\right]^2\left[\dfrac{3(4)}{2}\right]^2\left[3(2)\right]^26^236$The optional testing works Induction Hypothesis Assume$n = k$This is the$kth$term$1^3 + 2^3 + 3^3 + ... + k^3 = \left[\dfrac{k(k + 1)}{2}\right]^2...eqn.(1)$Induction Step Assume$n = k + 1$This is the$(k + 1)st$term$1^3 + 2^3 + 3^3 + ... + k^3 + (k + 1)^3 = \left[\dfrac{(k + 1)[(k + 1) + 1]}{2}\right]^2$Induction Step$\color{darkblue}{1^3 + 2^3 + 3^3 + ... + k^3} + (k + 1)^3$From$\color{darkblue}{eqn. (1)}$, substitute:$\color{darkblue}{\left[\dfrac{k(k + 1)}{2}\right]^2} + (k + 1)^3\dfrac{k^2(k + 1)^2}{2^2} + \dfrac{(k + 1)^3}{1}\dfrac{k^2(k + 1)^2}{4} + \dfrac{(k + 1)^3}{1}\dfrac{k^2(k + 1)^2 + 4(k + 1)^3}{4}\dfrac{(k + 1)^2[k^2 + 4(k + 1)]}{4}\dfrac{(k + 1)^2[k^2 + 4k + 4]}{4}\dfrac{(k + 1)^2[(k + 2)(k + 2)]}{4}\dfrac{(k + 1)^2 (k + 2)^2}{4}\left[\dfrac{(k + 1)[(k + 1) + 1]}{2}\right]^2\left[\dfrac{(k + 1)[k + 1 + 1]}{2}\right]^2\left[\dfrac{(k + 1)[k + 2]}{2}\right]^2\left[\dfrac{(k + 1)(k + 2)}{2}\right]^2\dfrac{(k + 1)^2 (k + 2)^2}{4}LHS = RHS$The formula is proved. (15.) Prove that the sum of the cubes of three consecutive positive integers is a multiple of$3$Direct Proof I shall use Pascal's Triangle to cube the expressions. That is my preference to save time. However, you can use the normal expansion if you prefer.$ Let\;\;the\;\;first\;\;positive\;\;integer = n \\[3ex] Next\;\;consecutive\;\;positive\;\;integer = n + 1 \\[3ex] Next\;\;consecutive\;\;positive\;\;integer = n + 2 \\[3ex] Cube\;\;each\;\;of\;\;them \\[3ex] 1st:\;\; n^3 \\[3ex] 2nd:\;\; (n + 1)^3 = n^3 + 3n^2 + 3n + 1...using\;\;Pascal's\;\;Triangle \\[3ex] 3rd:\;\; (n + 2)^3 \\[3ex] = n^3 + 3 * n^2 * (2) + 3 * n * (2)^2 + 2^3...using\;\;Pascal's\;\;Triangle \\[3ex] = n^3 + 6n^2 + 12n + 8 \\[3ex] Sum\;\;of\;\;cubes \\[3ex] = n^3 + (n^3 + 3n^2 + 3n + 1) + (n^3 + 6n^2 + 12n + 8) \\[3ex] = n^3 + n^3 + 3n^2 + 3n + 1 + n^3 + 6n^2 + 12n + 8 \\[3ex] = 3n^3 + 9n^2 + 15n + 9 \\[3ex] = 3(n^3 + 3n^2 + 5n + 3) \\[3ex] = 3 * some\;\;integer...Definition\;\;of\;\;multiple\;\;of\;\;3 \\[3ex] = multiple\;\;of\;\;3 $(16.) Prove that the difference between the squares of two consecutive integers is the sum of the integers. Direct Proof$ Let\;\;the\;\;first\;\;integer = n \\[3ex] Next\;\;consecutive\;\;integer = n + 1 \\[3ex] Sum\;\;of\;\;the\;\;integers \\[3ex] = n + (n + 1) \\[3ex] = n + n + 1 \\[3ex] = 2n + 1 \\[3ex] $To get$2n + 1$, we need to find the difference between$(n + 1)^2$and$n^2$rather than the difference between$n^2$and$(n + 1)^2$This is important so we can prove the statement.$ Square\;\;n = n^2 \\[3ex] Square\;\;(n + 1) \\[3ex] = (n + 1)^2 \\[3ex] = (n + 1)(n + 1) \\[3ex] = n^2 + n + n + 1 \\[3ex] = n^2 + 2n + 1 \\[3ex] Difference\;\;between\;\;(n + 1)^2\;\;and\;\;n^2 \\[3ex] = (n + 1)^2 - n^2 \\[3ex] = (n^2 + 2n + 1) - n^2 \\[3ex] = n^2 + 2n + 1 - n^2 \\[3ex] = 2n + 1 $(17.) Direct Proof I shall use Pascal's Triangle to cube the expressions. That is my preference to save time. However, you can use the normal expansion if you prefer.$ Let\;\;the\;\;first\;\;positive\;\;integer = n \\[3ex] Next\;\;consecutive\;\;positive\;\;integer = n + 1 \\[3ex] Next\;\;consecutive\;\;positive\;\;integer = n + 2 \\[3ex] Cube\;\;each\;\;of\;\;them \\[3ex] 1st:\;\; n^3 \\[3ex] 2nd:\;\; (n + 1)^3 = n^3 + 3n^2 + 3n + 1...using\;\;Pascal's\;\;Triangle \\[3ex] 3rd:\;\; (n + 2)^3 \\[3ex] = n^3 + 3 * n^2 * (2) + 3 * n * (2)^2 + 2^3...using\;\;Pascal's\;\;Triangle \\[3ex] = n^3 + 6n^2 + 12n + 8 \\[3ex] Sum\;\;of\;\;cubes \\[3ex] = n^3 + (n^3 + 3n^2 + 3n + 1) + (n^3 + 6n^2 + 12n + 8) \\[3ex] = n^3 + n^3 + 3n^2 + 3n + 1 + n^3 + 6n^2 + 12n + 8 \\[3ex] = 3n^3 + 9n^2 + 15n + 9 \\[3ex] = 3(n^3 + 3n^2 + 5n + 3) \\[3ex] = 3 * some\;\;integer...Definition\;\;of\;\;multiple\;\;of\;\;3 \\[3ex] = multiple\;\;of\;\;3 $(18.) Prove by contradiction that for any positive two real numbers,$m$and$n$; if$m * n \ge 100$, then either$m \gt 9$or$n \gt 9$Proof by Contradiction$ \underline{Statement} \\[3ex] (m * n \ge 100) \rightarrow [(m \gt 9) \lor (n \gt 9)] \\[3ex] \underline{Contradiction} \\[3ex] (m * n \ge 100) \rightarrow \neg[(m \gt 9) \lor (n \gt 9)] \\[3ex] (m * n \ge 100) \rightarrow [(m \le 9) \land (n \le 9)] \\[3ex] Notice\;\;De'Morgan's\;\;Law\;\;from\;\;OR\;\;to\;\;AND \\[3ex] \underline{Proof} \\[3ex] \underline{RHS} \\[3ex] (m \le 9) \land (n \ge 9) \\[3ex] m * n \le 9 * 9 \\[3ex] m * n \le 81 \\[3ex] \underline{LHS} \\[3ex] m * n \ge 100 \\[3ex] RHS \ne LHS \\[3ex] This\;\;is\;\;false \\[3ex] \therefore the\;\;statement\;\;is\;\;true $(19.) Prove by mathematical induction that Proof by Mathematical Induction Steps$LHSRHS(1 + x)^n1 + nx$Initial Step Test for$n = 1, x = 1(1 + x)^n(1 + 1)^12^121 + nx1 + 1(1)1 + 122 \ge 2$The initial step works Optional testing Test for$n = 1, x = 0(1 + x)^n(1 + 0)^11^111 + nx1 + 1(0)1 + 011 \ge 1$The optional testing works Optional testing Test for$n = 2, x = 1(1 + x)^n(1 + 1)^22^241 + nx1 + 2(1)1 + 234 \ge 3$The optional testing works Induction Hypothesis Assume$n = k(1 + x)^k \ge 1 + kx...eqn.(1)$Induction Step Assume$n = k + 1(1 + x)^{k + 1} \ge 1 + (k + 1)x...eqn.(2)$Induction Step$(1 + x)^{k + 1}(1 + x)^{k + 1} = (1 + x)^k * (1 + x)^1...Law\;1...Exp\implies (1 + x)^k * (1 + x)^1\color{darkblue}{(1 + x)^k} * (1 + x)^1$From$\color{darkblue}{eqn. (1)}$, substitute:$\color{darkblue}{1 + kx} * (1 + x)(1 + kx)(1 + x)1 + x + kx + kx^2\color{black}{1 + x + kx} + kx^21 + (k + 1)x1 + x(k + 1)1 + kx + x1 + x + kx\color{black}{1 + x + kx}x \gt -1\therefore x^2 \ge 0n \ge 1n = k + 1k = n - 1\therefore k \ge 0\implies kx^2 \ge 0LHS \ge RHS$The formula is proved. (20.) Prove by mathematical induction that for all natural numbers:$n$and for all real numbers:$\{x| x \ge -1\} (1 + x)^n \ge 1 + nx $Proof by Mathematical Induction Steps$LHSRHS(1 + x)^n1 + nx$Initial Step Test for$n = 1, x = 1(1 + x)^n(1 + 1)^12^121 + nx1 + 1(1)1 + 122 \ge 2$The initial step works Optional testing Test for$n = 1, x = 0(1 + x)^n(1 + 0)^11^111 + nx1 + 1(0)1 + 011 \ge 1$The optional testing works Optional testing Test for$n = 2, x = 1(1 + x)^n(1 + 1)^22^241 + nx1 + 2(1)1 + 234 \ge 3$The optional testing works Induction Hypothesis Assume$n = k(1 + x)^k \ge 1 + kx...eqn.(1)$Induction Step Assume$n = k + 1(1 + x)^{k + 1} \ge 1 + (k + 1)x...eqn.(2)$Induction Step$(1 + x)^{k + 1}(1 + x)^{k + 1} = (1 + x)^k * (1 + x)^1...Law\;1...Exp\implies (1 + x)^k * (1 + x)^1\color{darkblue}{(1 + x)^k} * (1 + x)^1$From$\color{darkblue}{eqn. (1)}$, substitute:$\color{darkblue}{1 + kx} * (1 + x)(1 + kx)(1 + x)1 + x + kx + kx^2\color{black}{1 + x + kx} + kx^21 + (k + 1)x1 + x(k + 1)1 + kx + x1 + x + kx\color{black}{1 + x + kx}x \gt -1\therefore x^2 \ge 0n \ge 1n = k + 1k = n - 1\therefore k \ge 0\implies kx^2 \ge 0LHS \ge RHS\$
The formula is proved.