diff --git a/chapter_5/exercises.md b/chapter_5/exercises.md index 18eae68..6a52740 100644 --- a/chapter_5/exercises.md +++ b/chapter_5/exercises.md @@ -2831,17 +2831,170 @@ Page 320 show that any amount of money of at least 14¢ can be made up using 3¢ and 8¢ coins. +**Proof (by mathematical induction):** + +Let $P(n)$ be the sentence: + +$n$¢ can be obtained using $3$¢ and $8$¢ coins. + +_Basis Step:_ + +Prove $P(14)$: + +$P(14)$ is true because $14$¢ can be obtained using one $8$¢ coin and two $3$¢ +coins. + +_Inductive Step:_ + +Let $k$ be any integer where $k \geq 14$. + +Suppose $P(k)$ is true. That is: + +$k$¢ can be obtained using $3$¢ and $8$¢ coins. + +Prove $P(k + 1)$. That is: + +$k + 1$¢ can be obtained using $3$¢ and $8$¢ coins. + +_Case 1 (there is a $8$¢ coin among those used to make up $k$¢):_ + +In this case, replace the $8$¢ coin with three $3$¢ coins. The result will be +$k + 1$¢. + +_Case 2 (there is not a $8$¢ coin among those used to make up $k$¢):_ + +In this case, because $k \geq 14$, at least 5 $3$¢ coins must have been used. So +remove five $3$¢ coins and replace them with two $8$¢ coins. The result will be +$k + 1$¢. + +Therefore in either case $(k + 1)$¢ can be obtained using $3$¢ and $8$¢ coins. + +Q.E.D. + 2. Use mathematical induction to show that any postage of at least 12¢ can be obtained using 3¢ and 7¢ stamps. +**Proof (by mathematical induction):** + +Let $P(n)$ be the sentence: + +$n$¢ postage can be obtained using $3$¢ and $7$¢ stamps. + +_Basis Step:_ + +Prove $P(12)$. That is: + +$12$¢ postage can be obtained using $3$¢ and $7$¢ stamps. + +$12$¢ can be obtained using four $3$¢ stamps. Therefore $P(12)$ is true. + +_Inductive Step:_ + +Let $k$ be any integer where $k \geq 12$. + +Suppose $P(k)$. That is: + +$k$¢ postage can be obtained using $3$¢ and $7$¢ stamps. + +Prove $P(k + 1)$. That is: + +$(k + 1)$¢ postage can be obtained using $3$¢ and $7$¢ stamps. + +_Case 1 (at least one 2 $3$¢ stamps are used to make up $k$¢):_ + +Replace the two $3$¢ stamps with a $7$¢ stamp. This results in $(k + 1)$¢. + +_Case 2 (there are 1 or 0 $3$¢ stamps among those used to make up $k$¢):_ + +Replace two $7$¢ stamps with five $3$¢ stamps. This results in $(k + 1)$¢. + +Therefore, in both cases $(k + 1)$ postage can be obtained using $3$¢ and $7$¢ +stamps. + +Q.E.D. + 3. Stamps are sold in packages containing either 5 stamps or 8 stamps. a. Show that a person can obtain 5, 8, 10, 13, 15, 16, 20, 21, 24, or 25 stamps by buying a collection of 5-stamp packages and 8-stamp packages. +- 5 stamps can be obtained by purchasing one 5 stamp package. + +- 8 stamps can be obtained by purchasing one 8 stamp package. + +- 10 stamps can be obtained by purchasing two 5 stamp packages. + +- 13 stamps can be obtained by purchasing one 5 stamp package and one 8 stamp + package. + +- 15 stamps can be obtained by purchasing three 5 stamp packages. + +- 16 stamps can be obtained by purchasing two 8 stamp packages. + +- 20 stamps can be obtained by purchasing four 5 stamp packages. + +- 21 stamps can be obtained by purchasing two 8 stamp packages and one 5 stamp + package. + +- 24 stamps can be obtained by purchasing three 8 stamp packages. + +- 25 stamps can be obtained by purchasing five 5 stamp packages. + b. Use mathematical induction to show that any quantity of at least 28 stamps can be obtained by buying a collection of 5-stamp packages and 8-stamp packages. +**Proof (by mathematical induction):** + +Let $P(n)$ be the sentence: + +$n$ stamps can be obtained by buying a collection of 5-stamp packages and +8-stamp packages. + +_Basis Step:_ + +Prove $P(28)$. That is: + +$28$ stamps can be obtained by buying a collection of 5-stamp packages and +8-stamp packages. + +$28$ stamps can be obtained by buying four 5-stamp packages and one 8-stamp +package. + +Therefore $P(28)$ is true. + +_Inductive Step:_ + +Let $k$ be any integer where $k \geq 28$. + +Suppose $P(k)$. That is: + +$k$ stamps can be obtained by buying a collection of 5-stamp packages and +8-stamp packages. + +Prove $P(k + 1)$. That is: + +$(k + 1)$ stamps can be obtained by buying a collection of 5-stamp packages and +8-stamp packages. + +_Case 1 (at least three 5-stamp packages are used in obtaining $k$ stamps):_ + +Replace three 5-stamp packages with two 8-stamp packages. This results in +$(k + 1)$ stamps. + +_Case 2 (at most two 5-stamp packages are used in obtaining $k$ stamps):_ + +If there at most two 5-stamp packages, that means that $28-10=18$ must be made +up of 8-stamp packages. So at least 3 8-stamp packages must be used to exceed +the 28 minimum. + +Replace three 8-stamp packages with 5 5-stamp packages. This results in +$(k + 1)$ stamps. + +Therefore in both cases $(k + 1)$ stamps can be obtained by buying a collection +of 5-stamp packages and 8-stamp packages. + +Q.E.D. + 4. For each positive integer $n$, let $P(n)$ be the sentence that describes the following divisibility property: @@ -2849,26 +3002,58 @@ $$ 5^n - 1 \text{ is divisible by } 4 $$ a. Write $P(0)$. Is $P(0)$ true? +$$ 5^0 - 1 = 1 - 1 = 0 $$ + +$P(0)$ is true, as $0 = 0 \cdot 4$. + b. Write $P(k)$. +$$ P(k) = 5^k - 1 \text{ is divisible by } 4 $$ + c. Write $P(k + 1)$. +$$ P(k + 1) = 5^{k + 1} - 1 \text{ is divisible by } 4 $$ + d. In a proof by mathematical induction that this divisibility property holds for every integer $n \geq 0$, what must be shown in the inductive step? +It must be shown that supposing that $5^k - 1$ is divisible by $4$ for some +integer $k \geq 0$, that therefore $5^{k + 1} - 1$ is divisible by $4$. + 5. For each positive integer $n$, let $P(n)$ be the inequality $$ 2^n < (n + 1)! $$ a. Write $P(2)$. Is $P(2)$ true? +$$ P(2) = 2^2 < (2 + 1)! $$ + +$$ P(2) = 4 < (3)! $$ + +$$ P(2) = 4 < (3 \cdot 2 \cdot 1) $$ + +$$ P(2) = 4 < 6 $$ + +Yes, $P(2)$ is true because $4$ is less than $6$. + b. Write $P(k)$. +$$ P(k) = 2^k < (k + 1)! $$ + c. Write $P(k + 1)$. +$$ P(k + 1) = 2^{k + 1} < ((k + 1) + 1)! $$ + +Alternatively: + +$$ P(k + 1) = 2^{k + 1} < (k + 2)! $$ + d. In a proof by mathematical induction that this inequality holds for every integer $n \geq 2$, what must be shown in the inductive step? +It must be shown that supposing $2^k < (k + 1)!$ is true for any integer +$k \geq 2$, that therefore $2^{k + 1} < (k + 2)!$ is true. + 6. For each positive integer $n$, let $P(n)$ be the sentence Any checkerboard with dimensions $2 \times 3n$ can be completely covered with @@ -2876,13 +3061,30 @@ L-shaped trominoes. a. Write $P(1)$. Is $P(1)$ true? +Any checkerboard with dimensions $2 \times 3(1)$ can be completely covered with +L-shaped trominoes. + +Yes, this is true, a $2 \times 3$ dimension checkerboard can be completely +covered with L-shaped trominoes (2 in fact.) + b. Write $P(k)$. +Any checkerboard with dimensions $2 \times 3k$ can be completely covered with +L-shaped trominoes. + c. Write $P(k + 1)$. +Any checkerboard with dimensions $2 \times 3(k + 1)$ can be completely covered +with L-shaped trominoes. + d. In a proof by mathematical induction that $P(n)$ is true for each integer $n \geq 1$, what must be shown in the inductive step? +It must be shown that supposing any checkerboard with dimensions $2 \times 3k$ +can be completely covered with L-shaped trominoes for any integer $k \geq 1$, +that therefore any checkerboard with dimensions $2 \times 3(k + 1)$ can be +completely covered with L-shaped trominoes. + 7. For each positive integer $n$, let $P(n)$ be the sentence In any round-robin tournament involving $n$ teams, the teams can be labeled @@ -2891,54 +3093,1183 @@ $i = 1, 2, \dots, n$. a. Write $P(2)$. Is $P(2)$ true? +In any round-robin tournament involving $2$ teams, the teams can be labeled +$T_1$, $T_2$, so that $T_i$ beats $T_{i + 1}$ for every $i = 1, 2$. + +This is true, in a round-robin tournament involving only $2$ teams, one can +label the teams such that $T_2$ beats $T_1$. + b. Write $P(k)$. +In any round-robin tournament involving $k$ teams, the teams can be labeled +$T_1$, $T_2$, $T_3$, \dots, $T_k$, so that $T_i$ beats $T_{i + 1}$ for every +$i = 1, 2, \dots, k$. + c. Write $P(k + 1)$. +In any round-robin tournament involving $(k + 1)$ teams, the teams can be +labeled $T_1$, $T_2$, $T_3$, \dots, $T_{k + 1}$, so that $T_i$ beats $T_{i + 1}$ +for every $i = 1, 2, \dots, (k + 1)$. + d. In a proof by mathematical induction that $P(n)$ is true for each integer $n \geq 2$, what must be shown in the inductive step? +It must be shown that supposing in any round-robin tournament involving $k$ +teams, the teams can be labeled $T_1, T_2, T_3, \dots T_k$, so that $T_i$ beats +$T_{i + 1}$ for every $i = 1, 2, \dots k$ for any integer $k \geq 2$, then +therefore in any round-robin tournament involving $(k + 1)$ teams, the teams can +be labeled $T_1, T_2, T_3, \dots T_{k + 1}$ so that $T_i$ beats $T_{i + 1}$ for +every $i = 1, 2, \dots (k + 1)$. + Prove each statement in 8-23 by mathematical induction. 8. $5^n - 1$ is divisible by $4$, for every integer $n \geq 0$. +**Proof (by mathematical induction):** + +Let $P(n)$ be the sentence: + +$$ 5^n - 1 \text{ is divisible by } 4 $$ + +_Basis Step:_ + +Prove $P(0)$. That is: + +$$ 5^0 - 1 \text{ is divisible by } 4 $$ + +$$ 1 - 1 \text{ is divisible by } 4 $$ + +$$ 0 \text{ is divisible by } 4 $$ + +This sentence is true as $0 = 0 \cdot 4$, which shows that $0$ is divisible by +$4$ by the definition of divisibility. + +Therefore $P(0)$ is true. + +_Inductive Step:_ + +Let $k$ be any integer where $k \geq 0$. + +Suppose $P(k)$. That is: + +$$ 5^k - 1 \text{ is divisible by } 4 $$ + +This is the inductive hypothesis. + +Prove $P(k + 1)$. That is: + +$$ 5^{k + 1} - 1 \text{ is divisible by } 4 $$ + +$$ 5^{k + 1} - 1 $$ + +$$ = 5^k \cdot 5 - 1 $$ + +$$ = 5^k \cdot (4 + 1) - 1 $$ + +$$ = 5^k \cdot 4 + 5^k - 1 $$ + +Since we know by the inductive hypothesis that $5^k - 1$ is divisible by $4$. By +the definition of divisibility: + +$$ 5^k - 1 = 4r $$ + +for some integer $r$. Our equation now becomes: + +$$ = 5^k \cdot 4 + 4r $$ + +$$ = 4(5^k + r) $$ + +Now, we know that $5^k + r$ is an integer by the sum and product of integers. +Therefore, by the definition of divisibility, $5^{k + 1} - 1$ is divisible by +$4$. + +Q.E.D. + 9. $7^n - 1$ is divisible by $6$, for every integer $n \geq 0$. +**Proof (by mathematical induction):** + +Let $P(n)$ be the sentence: + +$$ 7^n - 1 \text{ is divisible by } 6 $$ + +_Basis Step:_ + +Prove $P(0)$. That is: + +$$ 7^0 - 1 \text{ is divisible by } 6 $$ + +$$ 7^0 - 1 $$ + +$$ = 1 - 1 $$ + +$$ = 0 $$ + +$0$ is divisible by $6$ because $0 = 0 \cdot 6$. + +Therefore $P(0)$ is true. + +_Inductive Step:_ + +Let $k$ be any integer where $k \geq 0$. + +Suppose $P(k)$. That is: + +$$ 7^k - 1 \text{ is divisible by } 6 $$ + +Prove $P(k + 1)$. That is: + +$$ 7^{k + 1} - 1 \text{ is divisible by } 6 $$ + +$$ 7^{k + 1} - 1 $$ + +$$ = 7^k \cdot 7 - 1 $$ + +$$ = 7^k \cdot (6 + 1) - 1 $$ + +$$ = 7^k \cdot 6 + (7^k - 1) $$ + +By the inductive hypothesis and by the definition of divisibility: + +$$ = 7^k \cdot 6 + 6r $$ + +for some integer $r$. + +$$ = 6(7^k + r) $$ + +Now, we know that $7^k + r$ is an integer by the sum and product of integers. +Therefore, by the definition of divisibility, $7^{k + 1} - 1$ is divisible by +$6$. + +Q.E.D. + 10. $n^3 - 7n + 3$ is divisible by $3$, for each integer $n \geq 0$. +**Proof (by mathematical induction):** + +Let $P(n)$ be the sentence: + +$$ n^3 - 7n + 3 \text{ is divisible by } 3 $$ + +_Basis Step:_ + +Prove $P(0)$. That is: + +$$ (0)^3 - 7(0) + 3 \text{ is divisible by } 3 $$ + +$$ (0)^3 - 7(0) + 3 $$ + +$$ = 0 - 0 + 3 $$ + +$$ = 3 $$ + +By the definition of divisibility, $3 \mid 3$, as $3 = 1 \cdot 3$. + +Therefore, $P(0)$ is true. + +_Inductive Step:_ + +Let $k$ be any integer where $k \geq 0$. + +Suppose $P(k)$. That is: + +$$ k^3 - 7k + 3 \text{ is divisible by } 3 $$ + +Prove $P(k + 1)$. That is: + +$$ (k + 1)^3 - 7(k + 1) + 3 \text{ is divisible by } 3 $$ + +$$ (k + 1)^3 - 7(k + 1) + 3 $$ + +$$ = (k + 1)(k + 1)(k + 1) - 7k - 7 + 3 $$ + +$$ = (k^2 + 2k + 1)(k + 1) - 7k - 7 + 3 $$ + +$$ = (k^2(k + 1) + 2k(k + 1) + 1(k + 1)) - 7k - 7 + 3 $$ + +$$ = (k^3 + k^2 + 2k^2 + 2k + k + 1) - 7k - 7 + 3 $$ + +$$ = (k^3 + 3k^2 + 3k + 1) - 7k - 7 + 3 $$ + +$$ = (k^3 - 7k + 3) + 3k^2 + 3k + 1 - 7 $$ + +$$ = (k^3 - 7k + 3) + 3k^2 + 3k - 6 $$ + +By the inductive hypothesis and definition of divisibility: + +$$ = (3r) + 3k^2 + 3k - 6 $$ + +for some integer $r$. + +$$ = 3(r + k^2 + k - 2) $$ + +Now, we know that $r + k^2 + k - 2$ is an integer by the product and sum of +integers. Thus, by the definition of divisibility, $(k + 1)^3 - 7(k + 1) + 3$ is +divisible by $3$. + +Therefore $P(k + 1)$ is true. + +Q.E.D. + 11. $3^{2n} - 1$ is divisible by $8$, for every integer $n \geq 0$. +**Proof (by mathematical induction):** + +Let $P(n)$ be the sentence: + +$$ 3^{2n} - 1 \text{ is divisible by } 8 $$ + +_Basis Step:_ + +Prove $P(0)$. That is: + +$$ 3^{2(0)} - 1 \text{ is divisible by } 8 $$ + +$$ 3^{2(0)} - 1 $$ + +$$ = 3^0 - 1 $$ + +$$ = 1 - 1 $$ + +$$ = 0 $$ + +$0$ is divisible by $8$ as $0 = 0 \cdot 8$. + +Therefore $P(0)$ is true. + +_Inductive Step:_ + +Let $k$ be any integer where $k \geq 0$. + +Suppose $P(k)$. That is: + +$$ 3^{2k} - 1 \text{ is divisible by } 8 $$ + +This is the inductive hypothesis. + +Prove $P(k + 1)$. That is: + +$$ 3^{2(k + 1)} - 1 \text{ is divisible by } 8 $$ + +$$ 3^{2(k + 1)} - 1 $$ + +$$ = 3^{2k + 2} - 1 $$ + +$$ = 3^{2k} \cdot 3^2 - 1 $$ + +$$ = 3^{2k} \cdot 9 - 1 $$ + +$$ = 3^{2k} \cdot (8 + 1) - 1 $$ + +$$ = 3^{2k} \cdot 8 + (3^{2k} - 1) $$ + +By the inductive hypothesis and the definition of divisibility: + +$$ = 3^{2k} \cdot 8 + 8r $$ + +for some integer $r$. + +$$ = 8(3^{2k} + r) $$ + +Now, $3^{2k} + r$ is an integer by the sum and product of integers. Thus +$3^{2(k + 1)} - 1$ is divisible by $8$ by the definition of divisibility. + +Therefore $P(k + 1)$ is true. + +Q.E.D. + 12. For any integer $n \geq 0$, $7^n - 2^n$ is divisible by $5$. +**Proof (by mathematical induction):** + +Let $P(n)$ be the sentence: + +$$ 7^n - 2^n \text{ is divisible by } 5 $$ + +_Basis Step:_ + +Prove $P(0)$. That is: + +$$ 7^0 - 2^0 \text{ is divisible by } 5 $$ + +$$ 7^0 - 2^0 $$ + +$$ = 1 - 1 $$ + +$$ = 0 $$ + +$0$ is divisible by $5$ as $0 = 0 \cdot 5$. + +Therefore $P(0)$ is true. + +_Inductive Step:_ + +Let $k$ be any integer where $k \geq 0$. + +Suppose $P(k)$. That is: + +$$ 7^k - 2^k \text{ is divisible by } 5 $$ + +This is the inductive hypothesis. + +Prove $P(k + 1)$. That is: + +$$ 7^{k + 1} - 2^{k + 1} \text{ is divisible by } 5 $$ + +$$ 7^{k + 1} - 2^{k + 1} $$ + +$$ = 7^k \cdot 7^1 - 2^k \cdot 2^1 $$ + +$$ = 7^k \cdot (5 + 2) - 2^k \cdot 2^1 $$ + +$$ = 7^k \cdot 5 + (2)7^k - 2^k \cdot 2^1 $$ + +$$ = 7^k \cdot 5 + 2(7^k - 2^k) $$ + +By the inductive hypothesis and the definition of divisibility: + +$$ = 7^k \cdot 5 + 2(5r) $$ + +For some integer $r$. + +$$ = 5(7^k + 2r) $$ + +Now, $7^k + 2r$ is an integer by the sum and product of integers. Thus +$7^{k + 1} - 2^{k + 1}$ is divisible by $5$ by the definition of divisibility. + +Therefore $P(k + 1)$ is true. + +Q.E.D. + 13. For any integer $n \geq 0$, $x^n -y^n$ is divisible by $x - y$, where $x$ and $y$ are any integers with $x \neq y$. +**Proof (by mathematical induction):** + +Suppose $x$ and $y$ are any integers with $x \neq y$. + +Let $P(n)$ be the sentence: + +$$ x^n - y^n \text{ is divisible by } x - y $$ + +_Basis Step:_ + +Prove $P(0)$. That is: + +$$ x^0 - y^0 \text{ is divisible by } x - y $$ + +$$ x^0 - y^0 $$ + +$$ = 1 - 1 $$ + +$$ = 0 $$ + +$0$ is divisible by $(x - y)$ as $0 = 0 \cdot (x - y)$. + +Therefore $P(0)$ is true. + +_Inductive Step:_ + +Let $k$ be any integer where $k \geq 0$. + +Suppose $P(k)$. That is: + +$$ x^k - y^k \text{ is divisible by } x - y $$ + +This is the inductive hypothesis. + +Prove $P(k + 1)$. That is: + +$$ x^{k + 1} - y^{k + 1} \text{ is divisible by } x - y $$ + +$$ x^{k + 1} - y^{k + 1} $$ + +$$ = x^k(x) - y^k(y) $$ + +$$ = x^k(x) - xy^k + xy^k - y^k(y) $$ + +$$ = x(x^k - y^k) + y^k(x - y) $$ + +By the inductive hypothesis: + +$$ = x(r(x - y)) + y^k(x - y) $$ + +for some integer $r$. + +$$ = (x - y)(xr + y^k) $$ + +We know $xr + y^k$ is an integer by the sum and product of integers. By the +definition of divisibility, $x^{k + 1} - y^{k + 1}$ is divisible by $x - y$. + +Therefore $P(k + 1)$ is true. + +Q.E.D. + 14. $n^3 - n$ is divisible by $6$, for each integer $n \geq 0$. +**Proof (by mathematical induction):** + +Let $P(n)$ be the sentence: + +$$ n^3 - n \text{ is divisible by } 6 $$ + +_Basis Step:_ + +Prove $P(0)$. That is: + +$$ 0^3 - 0 \text{ is divisible by } 6 $$ + +$$ 0^3 - 0 $$ + +$$ = 0 - 0 $$ + +$$ = 0 $$ + +$0$ is divisible by $6$ because $0 = 0 \cdot 6$. + +Therefore $P(0)$ is true. + +_Inductive Step:_ + +Let $k$ be any integer where $k \geq 0$. + +Suppose $P(k)$. That is: + +$$ k^3 - k \text{ is divisible by } 6 $$ + +This is the inductive hypothesis. + +Prove $P(k + 1)$. That is: + +$$ (k + 1)^3 - (k + 1) \text{ is divisible by } 6 $$ + +$$ (k + 1)^3 - (k + 1) $$ + +$$ = (k + 1)(k + 1)(k + 1) - (k + 1) $$ + +$$ = (k^2 + 2k + 1)(k + 1) - (k + 1) $$ + +$$ = (k^3 + k^2 + 2k^2 + 2k + k + 1) - (k + 1) $$ + +$$ = (k^3 + 3k^2 + 3k + 1) - (k + 1) $$ + +$$ = k^3 + 3k^2 + 3k + 1 - k - 1 $$ + +$$ = k^3 + 3k^2 + 2k $$ + +$$ = (k^3 - k) + 3k^2 + 3k $$ + +$$ = (k^3 - k) + 3k(k + 1) $$ + +By the inductive hypothesis and definition of divisibility: + +$$ = 6r + 3k(k + 1) $$ + +for some integer $r$. + +By Theorem 4.5.2, the product of any two consecutive integers must be even. + +$$ = 6r + 3(2m) $$ + +for some integer $m$. + +$$ = 6r + 6m $$ + +$$ = 6(r + m) $$ + +Now, $r + m$ is an integer by the sum of integers. + +Therefore $(k + 1)^3 - (k + 1)$ is divisible by $6$ by the definition of +divisibility. + +Therefore $P(k + 1)$ is true. + +Q.E.D. + 15. $n(n^2 + 5)$ is divisible by $6$, for each integer $n \geq 0$. +**Proof (by mathematical induction):** + +Let $P(n)$ be the sentence: + +$$ n(n^2 + 5) \text{ is divisible by } 6 $$ + +_Basis Step:_ + +Prove $P(0)$. That is: + +$$ 0(0^2 + 5) \text{ is divisible by } 6 $$ + +$$ 0(0^2 + 5) $$ + +$$ = 0(0 + 5) $$ + +$$ = 0(5) $$ + +$$ = 0 $$ + +$0$ is divisible by $6$ as $0 = 0 \cdot 6$. + +Therefore $P(0)$ is true. + +_Inductive Step:_ + +Let $k$ be any integer where $k \geq 0$. + +Suppose $P(k)$. That is: + +$$ k(k^2 + 5) \text{ is divisible by } 6 $$ + +This is the inductive hypothesis. + +Prove $P(k + 1)$. That is: + +$$ (k + 1)((k + 1)^2 + 5) \text{ is divisible by } 6 $$ + +$$ (k + 1)((k + 1)^2 + 5) $$ + +$$ = (k + 1)((k + 1)(k + 1) + 5) $$ + +$$ = (k + 1)(k^2 + 2k + 6) $$ + +$$ = k^3 + k^2 + 2k^2 + 2k + 6k + 6 $$ + +$$ = k^3 + 3k^2 + 8k + 6 $$ + +$$ = k^3 + 3k^2 + 5k + 3k + 6 $$ + +$$ = (k^3 + 5k) + 3k^2 + 3k + 6 $$ + +$$ = k(k^2 + 5) + 3k^2 + 3k + 6 $$ + +$$ = k(k^2 + 5) + 3(k^2 + k + 2) $$ + +By the inductive hypothesis and definition of divisibility: + +$$ = 6r + 3(k^2 + k + 2) $$ + +for some integer $r$. + +$$ = 6r + 3(k(k + 1) + 2) $$ + +By Theorem 4.5.2 $k(k + 1)$ is always even: + +$$ = 6r + 3(2m + 2) $$ + +for some integer $m$. + +$$ = 6r + 6m + 6 $$ + +$$ = 6(r + m + 1) $$ + +Now, $r + m + 1$ is an integer by the sum of integers. Thus +$(k + 1)((k + 1)^2 + 5)$ is divisible by $6$ by the definition of divisibility. + +Therefore $P(k + 1)$ is true. + +Q.E.D. + 16. $2^n < (n + 1)!$, for every integer $n \geq 2$. +**Proof (by mathematical induction):** + +Let $P(n)$ be the sentence: + +$$ 2^n < (n + 1)! $$ + +_Basis Step:_ + +Prove $P(2)$. That is: + +$$ 2^(2) < (2 + 1)! $$ + +$$ 4 < (3)! $$ + +$$ 4 < (3 \cdot 2 \cdot 1) $$ + +$$ 4 < 6 $$ + +$4$ is less than $6$. + +Therefore $P(2)$ is true. + +_Inductive Step:_ + +Let $k$ be any integer where $k \geq 2$. + +Suppose $P(k)$. That is: + +$$ 2^k < (k + 1)! $$ + +This is the inductive hypothesis. + +Prove $P(k + 1)$. That is: + +$$ 2^{k + 1} < ((k + 1) + 1)! $$ + +Alternatively: + +$$ 2^{k + 1} < (k + 2)! $$ + +By the inductive hypothesis and the laws of exponents: + +$$ = 2^{k} \cdot 2 < 2(k + 1)! $$ + +Since $k \geq 2$, then $2 < k + 2$, and so: + +$$ 2(k + 1)! < (k + 2)(k + 1)! = (k + 2)! $$ + +Combining these inequalities shows: + +$$ 2^{k + 1} < (k + 2)! $$ + +As was to be shown. + +Q.E.D. + 17. $1 + 3n \leq 4^n$, for every integer $n \geq 0$. +**Proof (by mathematical induction):** + +Let $P(n)$ be the inequality: + +$$ 1 + 3n \leq 4^n $$ + +_Basis Step:_ + +Prove $P(0)$. That is: + +$$ 1 + 3(0) \leq 4^0 $$ + +$$ = 1 + 0 \leq 1 $$ + +$$ = 1 \leq 1 $$ + +Since $1 = 1$, $1 \leq 1$ is a true statement. + +Therefore $P(0)$ is true. + +_Inductive Step:_ + +Let $k$ be any integer where $k \geq 0$. + +Suppose $P(k)$. That is: + +$$ 1 + 3k \leq 4^k $$ + +This is the inductive hypothesis. + +Prove $P(k + 1)$. That is: + +$$ 1 + 3(k + 1) \leq 4^{k + 1} $$ + +$$ (1 + 3k) + 3 \leq 4^k \cdot 4 $$ + +By the inductive hypothesis: + +$$ (1 + 3k) + 3 \leq 4^k + 3 $$ + +Now show: + +$$ 4^k + 3 \leq 4^{k + 1} $$ + +Since: + +$$ 4^{k + 1} = 4^k \cdot 4 $$ + +it is enough to show: + +$$ 3 \leq 3 \cdot 4^k $$ + +which is true for all $k \geq 0$. + +So: + +$$ 1 + 3(k + 1) \leq 4^k + 3 \leq 4^{k + 1} $$ + +$$ 1 + 3(k + 1) \leq 4^{k + 1} $$ + +Q.E.D. + 18. $5^n + 9 < 6^n$, for each integer $n \geq 2$. +**Proof (by mathematical induction):** + +Let $P(n)$ be the inequality: + +$$ 5^n + 9 < 6^n $$ + +_Basis Step:_ + +Prove $P(2)$. That is: + +$$ 5^2 + 9 < 6^2 $$ + +$$ 25 + 9 < 36 $$ + +$$ 34 < 36 $$ + +Since $34$ is less than $36$, this inequality is true. + +Therefore $P(2)$ is true. + +_Inductive Step:_ + +Let $k$ be any integer where $k \geq 2$. + +Suppose $P(k)$. That is: + +$$ 5^k + 9 < 6^k $$ + +This is the inductive hypothesis. + +Prove $P(k + 1)$. That is: + +$$ 5^{k + 1} + 9 < 6^{k + 1} $$ + +If we multiply the inductive hypothesis by 5: + +$$ 5(5^k + 9) < 5(6^k) $$ + +$$ 5^{k + 1} + 45 < 5(6^k) $$ + +$$ 5^{k + 1} + 45 < 5(6^k) < 6^{k + 1} $$ + +$$ 5^{k + 1} + 45 < 5(6^k) < 6^k \cdot 6 = 6^{k + 1} $$ + +Note that: + +$$ 5^{k + 1} + 9 < 5^{k + 1} + 45 < 5(6^k) < 6^k \cdot 6 = 6^{k + 1} $$ + +Therefore: + +$$ 5^{k + 1} + 9 < 6^{k + 1} $$ + +As was to be shown. + +Q.E.D. + 19. $n^2 < 2^n$, for every integer $n \geq 5$. +**Proof (by mathematical induction):** + +Let $P(n)$ be the inequality: + +$$ n^2 < 2^n $$ + +_Basis Step:_ + +Prove $P(5)$. That is: + +$$ 5^2 < 2^5 $$ + +$$ 25 < 32 $$ + +Since $25$ is less than $32$, this is a true statement. + +Therefore $P(5)$ is true. + +_Inductive Step:_ + +Let $k$ be any integer where $k \geq 5$. + +Suppose $P(k)$. That is: + +$$ k^2 < 2^k $$ + +This is the inductive hypothesis. + +Prove $P(k + 1)$. That is: + +$$ (k + 1)^2 < 2^{k + 1} $$ + +Now, expanding out the left-hand side: + +$$ (k + 1)^2 = k^2 + 2k + 1 $$ + +Consider the inductive hypothesis: + +$$ k^2 < 2^k $$ + +It follows that: + +$$ k^2 + 2k + 1 < 2^k + 2k + 1 $$ + +By proposition 5.3.2, $2k + 1 < 2^k$ since $k \geq 5 \geq 3$. + +Hence: + +$$ (k + 1)^2 = k^2 + 2k + 1 < 2^k + 2k + 1 < 2^k + 2^k = 2^{k + 1} $$ + +$$ (k + 1)^2 < 2^{k + 1} $$ + +As was to be shown. + +Q.E.D. + 20. $2^n < (n + 2)!$, for each integer $n \geq 0$. +**Proof (by mathematical induction):** + +Let $P(n)$ be the inequality: + +$$ 2^n < (n + 2)! $$ + +_Basis Step:_ + +Prove $P(0)$. That is: + +$$ 2^0 < (0 + 2)! $$ + +$$ 1 < (2)! $$ + +$$ 1 < 2 $$ + +Since $1$ is less than $2$. This is a true statement. + +Therefore $P(0)$ is true. + +_Inductive Step:_ + +Let $k$ be any integer such that $k \geq 0$. + +Suppose $P(k)$. That is: + +$$ 2^k < (k + 2)! $$ + +Prove $P(k + 1)$. That is: + +$$ 2^{k + 1} < ((k + 1) + 2)! $$ + +Alternatively: + +$$ 2^{k + 1} < (k + 3)! $$ + +Expanding out the left-hand side: + +$$ 2^{k + 1} = 2^k \cdot 2 $$ + +Consider the inductive hypothesis: + +$$ 2^k < (k + 2)! $$ + +Multiple both sides by $2$: + +$$ 2(2^k) < 2(k + 2)! $$ + +$$ 2^{k + 1} < 2(k + 2)! $$ + +Now, expanding out the right-hand side: + +$$ (k + 3)! = (k + 3)(k + 2)! $$ + +Since $k \geq 0$, it follows that $k + 3 \geq 3 \geq 2$. Putting out +inequalities together then, we get: + +$$ 2^{k + 1} < 2(k + 2)! < (k + 3)(k + 2)! = (k + 3)! $$ + +And now simplified: + +$$ 2^{k + 1} < (k + 3)! $$ + +As was to be shown. + +Q.E.D. + 21. $\sqrt{n} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \dots + \dfrac{1}{\sqrt{n}}$, for every integer $n \geq 2$. +**Proof (by mathematical induction):** + +Let $P(n)$ be the inequality: + +$$ \sqrt{n} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \dots + \dfrac{1}{\sqrt{n}} $$ + +_Basis Step:_ + +Prove $P(2)$. That is: + +$$ \sqrt{2} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \dots + \dfrac{1}{\sqrt{2}} $$ + +$\dots \dfrac{1}{\sqrt{2}}$ just ends at term, $\dfrac{1}{\sqrt{2}}$. + +$$ \sqrt{2} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} $$ + +$$ \sqrt{2} < 1 + \dfrac{1}{\sqrt{2}} $$ + +$$ \sqrt{2} < \frac{\sqrt{2}}{\sqrt{2}} + \dfrac{1}{\sqrt{2}} $$ + +$$ \sqrt{2} < \dfrac{\sqrt{2} + 1}{\sqrt{2}} $$ + +$$ (\sqrt{2})(\sqrt{2}) < \left(\dfrac{\sqrt{2} + 1}{\sqrt{2}}\right)(\sqrt{2}) $$ + +$$ 2 < \sqrt{2} + 1 \approx 2.414213562 $$ + +This statement is true. Therefore $P(2)$ is true. + +_Inductive Step:_ + +Let $k$ be any integer where $k \geq 2$. + +Suppose $P(k)$. That is: + +$$ \sqrt{k} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \dots + \dfrac{1}{\sqrt{k}} $$ + +This is the inductive hypothesis. + +Prove $P(k + 1)$. That is: + +$$ \sqrt{k + 1} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \dots + \dfrac{1}{\sqrt{k + 1}} $$ + +From the inductive hypothesis: + +$$ \sqrt{k} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \dots + \dfrac{1}{\sqrt{k}} $$ + +Add $\dfrac{1}{\sqrt{k + 1}}$ to both sides: + +$$ \sqrt{k} + \frac{1}{\sqrt{k + 1}} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \dots + \dfrac{1}{\sqrt{k}} + \frac{1}{\sqrt{k + 1}} $$ + +From here, it is enough to show: + +$$ \sqrt{k + 1} \leq \sqrt{k} + \frac{1}{\sqrt{k + 1}} $$ + +$$ \sqrt{k + 1} - \sqrt{k} \leq \frac{1}{\sqrt{k + 1}} $$ + +$$ \left(\sqrt{k + 1} - \sqrt{k}\right)\left(\frac{\sqrt{k + 1} + \sqrt{k}}{\sqrt{k + 1} + \sqrt{k}}\right) \leq \frac{1}{\sqrt{k + 1}} $$ + +$$ \frac{(k + 1) - k}{\sqrt{k + 1} + \sqrt{k}} \leq \frac{1}{\sqrt{k + 1}} $$ + +$$ \frac{1}{\sqrt{k + 1} + \sqrt{k}} \leq \frac{1}{\sqrt{k + 1}} $$ + +Since $\sqrt{k + 1} + \sqrt{k} > \sqrt{k + 1}$, this inequality holds. + +Simplified, our inequality becomes: + +$$ \sqrt{k + 1} < \sqrt{k} + \frac{1}{\sqrt{k + 1}} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \dots + \dfrac{1}{\sqrt{k + 1}} $$ + +$$ \sqrt{k + 1} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \dots + \dfrac{1}{\sqrt{k + 1}} $$ + +As was to be shown. + +Q.E.D. + 22. $1 + nx \leq (1 + x)^n$, for every real number $x > -1$ and every integer $n \geq 2$. +**Proof (by mathematical induction):** + +Suppose $x$ is any real number where $x > -1$. + +Let $P(n)$ be the sentence: + +$$ 1 + nx \leq (1 + x)^n $$ + +_Basis Step:_ + +Prove $P(2)$. That is: + +$$ 1 + 2x \leq (1 + x)^2 $$ + +$$ 1 + 2x \leq 1 + 2x + x^2 $$ + +$$ 0 \leq x^2 $$ + +This inequality always holds. + +Therefore $P(2)$ is true. + +_Inductive Step:_ + +Let $k$ be any integer where $k \geq 2$. + +Suppose $P(k)$. That is: + +$$ 1 + kx \leq (1 + x)^k $$ + +This is the inductive hypothesis. + +Prove $P(k + 1)$. That is: + +$$ 1 + (k + 1)x \leq (1 + x)^{k + 1} $$ + +Consider the inductive hypothesis: + +$$ 1 + kx \leq (1 + x)^k $$ + +Multiply each side by $(1 + x)$: + +$$ (1 + x)(1 + kx) \leq ((1 + x)^k)(1 + x) $$ + +$$ 1 + x + kx + kx^2 \leq (1 + x)^{k + 1} $$ + +Now it is enough to show that the left hand side of $P(k + 1)$ is less than or +equal to the left-hand side of $(1 + x)(P(k))$: + +$$ 1 + (k + 1)x \leq 1 + x + kx + kx^2 $$ + +$$ 1 + kx + x \leq 1 + x + kx + kx^2 $$ + +$$ 1 + x + kx \leq 1 + x + kx + kx^2 $$ + +$$ 0 \leq kx^2 $$ + +Since $k \geq 2$, this inequality will always hold. + +Simplified, our inequality is: + +$$ 1 + (k + 1)x \leq 1 + x + kx + kx^2 \leq (1 + x)^{k + 1} $$ + +$$ 1 + (k + 1)x \leq (1 + x)^{k + 1} $$ + +As was to be shown. + +Q.E.D. + 23. a. $n^3 > 2n + 1$, for each integer $n \geq 2$. +**Proof (by mathematical induction):** + +Let $P(n)$ be the inequality: + +$$ n^3 > 2n + 1 $$ + +_Basis Step:_ + +Prove $P(2)$. That is: + +$$ (2)^3 > 2(2) + 1 $$ + +$$ 8 > 4 + 1 $$ + +$$ 8 > 5 $$ + +Since $8$ is greater than $5$, this statement is true. + +Therefore $P(2)$ is true. + +_Inductive Step:_ + +Let $k$ be any integer where $k \geq 2$. + +Suppose $P(k)$. That is: + +$$ k^3 > 2k + 1 $$ + +This is the inductive hypothesis. + +Prove $P(k + 1)$. That is: + +$$ (k + 1)^3 > 2(k + 1) + 1 $$ + +Alternatively: + +$$ (k + 1)^3 > 2k + 2 + 1 $$ + +$$ (k + 1)^3 > 2k + 3 $$ + +Consider the inductive hypothesis: + +$$ k^3 > 2k + 1 $$ + +Add $2$ to both sides: + +$$ k^3 + 2 > 2k + 1 + 2 $$ + +$$ k^3 + 2 > 2k + 3 $$ + +Now it is enough to prove that the left-hand side of this inequality is less +than the left-hand side of the $P(k + 1)$ inequality: + +$$ (k + 1)^3 > k^3 + 2 $$ + +$$ (k + 1)(k + 1)(k + 1) > k^3 + 2 $$ + +$$ (k^2 + 2k + 1)(k + 1) > k^3 + 2 $$ + +$$ k^3 + k^2 + 2k^2 + 2k + k + 1 > k^3 + 2 $$ + +$$ k^3 + 3k^2 + 3k + 1 > k^3 + 2 $$ + +$$ 3k^2 + 3k > 1 $$ + +Since $k \geq 2$, this inequality will always hold. + +Simplified: + +$$ (k + 1)^3 > k^3 + 2 > 2k + 3 $$ + +$$ (k + 1)^3 > 2k + 3 $$ + +As was to be shown. + +Q.E.D. + b. $n! > n^2$, for each integer $n \geq 4$. +**Proof (by mathematical induction):** + +Let $P(n)$ be the inequality: + +$$ n! > n^2 $$ + +_Basis Step:_ + +Prove $P(4)$. That is: + +$$ 4! > 4^2 $$ + +$$ (4 \cdot 3 \cdot 2 \cdot 1) > 16 $$ + +$$ 24 > 16 $$ + +Since $24$ is greater than $16$, this statement is true. + +Therefore $P(4)$ is true. + +_Inductive Step:_ + +Let $k$ be any integer where $k \geq 4$. + +Suppose $P(k)$. That is: + +$$ k! > k^2 $$ + +This is the inductive hypothesis. + +Prove $P(k + 1)$. That is: + +$$ (k + 1)! > (k + 1)^2 $$ + +Take the inductive hypothesis: + +$$ k! > k^2 $$ + +And multiply each side by $(k + 1)$: + +$$ (k + 1)k! > k^2(k + 1) $$ + +$$ (k + 1)! > k^2(k + 1) $$ + +Now it is enough to show: + +$$ k^2(k + 1) > (k + 1)^2 $$ + +$$ k^2 > k + 1 $$ + +And this inequality holds for all $k \geq 4$. + +Simplified: + +$$ (k + 1)! > k^2(k + 1) > (k + 1)^2 $$ + +$$ (k + 1)! > (k + 1)^2 $$ + +As was to be shown. + +Q.E.D. + 24. A sequence $a_1, a_2, a_3, \dots$ is defined by letting $a_1 = 3$ and $a_k = 7a_{k - 1}$ for each integer $k \geq 2$. Show that $a_n = 3 \cdot 7^{n - 1}$ for every integer $n \geq 1$. diff --git a/chapter_5/test_yourself.md b/chapter_5/test_yourself.md index 6012e4d..414881e 100644 --- a/chapter_5/test_yourself.md +++ b/chapter_5/test_yourself.md @@ -63,5 +63,9 @@ Page 320 1. Mathematical induction differs from the kind of induction used in the natural sciences because it is actually a form of _____ reasoning. +deductive + 2. Mathematical induction can be used to _____ conjectures that have been made using inductive reasoning. + +prove diff --git a/leftoff.txt b/leftoff.txt index a1e0432..3ae0b93 100644 --- a/leftoff.txt +++ b/leftoff.txt @@ -1 +1 @@ -312 +321