If there is one prayer that you should pray/sing every day and every hour, it is the
LORD's prayer (Our FATHER in Heaven prayer)
- Samuel Dominic Chukwuemeka
It is the most powerful prayer.
A pure heart, a clean mind, and a clear conscience is necessary for it.
For in GOD we live, and move, and have our being.
- Acts 17:28
The Joy of a Teacher is the Success of his Students.
- Samuel Chukwuemeka
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 $d^{12}$
(2.) Prove that the sum of two even integers is 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.
(4.) Prove that the sum of three consecutive integers is a multiple of three.
(5.) Prove that for all integers $n$, if $n^2$ is even, then $n$ is even using proof by contrapositive.
(6.) Prove that the sum of four consecutive integers is an 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.
(8.) Prove or disprove that if $c$ and $f$ are even, then $k$ is even.
(9.) Prove that the difference between the cube and the square of an integer is always an even number.
(10.) Prove that the sum of the squares of two consecutive integers is an odd integer.
(11.) Prove by mathematical induction that the formula works for the set of positive integers.
$4 + 9 + 14 + ... + (5n - 1) = \dfrac{n}{2}(5n + 3)$
| 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 $4 + 9 + 14 + ... + (5k - 1) = \dfrac{k}{2}(5k + 3)...eqn.(1)$ |
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|
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:
\[ 1 + 2 + 3 + ... + n = \dfrac{n(n + 1)}{2} \]
| Steps | $LHS$ | $RHS$ |
|---|---|---|
| $1 + 2 + 3 + ... + n$ | $\dfrac{n(n + 1)}{2}$ | |
|
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: $1 + 2$ $3$ |
$\dfrac{2(2 + 1)}{2}$ $\dfrac{2(3)}{2}$ $1(3)$ $3$ |
|
Optional testing Test for $n = 3$ |
First term + Second term + Third term: $1 + 2 + 3$ $6$ |
$\dfrac{3(3 + 1)}{2}$ $\dfrac{3(4)}{2}$ $3(2)$ $6$ |
| 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:
\[ 1 + 2 + 3 + ... + n = \dfrac{n(n + 1)}{2} \]
| Steps | $LHS$ | $RHS$ |
|---|---|---|
| $1 + 2 + 3 + ... + n$ | $\dfrac{n(n + 1)}{2}$ | |
|
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: $1 + 2$ $3$ |
$\dfrac{2(2 + 1)}{2}$ $\dfrac{2(3)}{2}$ $1(3)$ $3$ |
|
Optional testing Test for $n = 3$ |
First term + Second term + Third term: $1 + 2 + 3$ $6$ |
$\dfrac{3(3 + 1)}{2}$ $\dfrac{3(4)}{2}$ $3(2)$ $6$ |
| 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. |
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(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:
\[ 1^3 + 2^3 + 3^3 + ... + n^3 = \left[\dfrac{n(n + 1)}{2}\right]^2 \]
| Steps | $LHS$ | $RHS$ |
|---|---|---|
| $1^3 + 2^3 + 3^3 + ... + n^3$ | $\left[\dfrac{n(n + 1)}{2}\right]^2$ | |
|
Initial Step Test for $n = 1$ |
First term: $1^3$ $1$ |
$\left[\dfrac{n(n + 1)}{2}\right]^2$ $\left[\dfrac{1(1 + 1)}{2}\right]^2$ $\left[\dfrac{1(2)}{2}\right]^2$ $\left[1(1)\right]^2$ $1^2$ $1$ |
| The initial step works | ||
|
Optional testing Test for $n = 2$ |
First term + Second term: $1^3 + 2^3$ $1 + 8$ $9$ |
$\left[\dfrac{2(2 + 1)}{2}\right]^2$ $\left[\dfrac{2(3)}{2}\right]^2$ $\left[1(3)\right]^2$ $3^2$ $9$ |
|
Optional testing Test for $n = 3$ |
First term + Second term + Third term: $1^3 + 2^3 + 3^3$ $1 + 8 + 27$ $36$ |
$\left[\dfrac{3(3 + 1)}{2}\right]^2$ $\left[\dfrac{3(4)}{2}\right]^2$ $\left[3(2)\right]^2$ $6^2$ $36$ |
| 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: $\color{darkblue}{\left[\dfrac{k(k + 1)}{2}\right]^2} + (k + 1)^3$ $\dfrac{k^2(k + 1)^2}{2^2} + \dfrac{(k + 1)^3}{1}$ $\dfrac{k^2(k + 1)^2}{4} + \dfrac{(k + 1)^3}{1}$ $\dfrac{k^2(k + 1)^2 + 4(k + 1)^3}{4}$ $\dfrac{(k + 1)^2[k^2 + 4(k + 1)]}{4}$ $\dfrac{(k + 1)^2[k^2 + 4k + 4]}{4}$ $\dfrac{(k + 1)^2[(k + 2)(k + 2)]}{4}$ $\dfrac{(k + 1)^2 (k + 2)^2}{4}$ |
$\left[\dfrac{(k + 1)[(k + 1) + 1]}{2}\right]^2$ $\left[\dfrac{(k + 1)[k + 1 + 1]}{2}\right]^2$ $\left[\dfrac{(k + 1)[k + 2]}{2}\right]^2$ $\left[\dfrac{(k + 1)(k + 2)}{2}\right]^2$ $\dfrac{(k + 1)^2 (k + 2)^2}{4}$ |
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$LHS = RHS$ The formula is proved. |
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(15.) Prove that the sum of the cubes of three consecutive positive integers is a multiple of $3$
(16.) Prove that the difference between the squares of two consecutive integers is the sum of the integers.
(17.)
(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 $n \gt 9$
(19.) Prove by mathematical induction that
| Steps | $LHS$ | $RHS$ |
|---|---|---|
| $(1 + x)^n$ | $1 + nx$ | |
|
Initial Step Test for $n = 1, x = 1$ |
$(1 + x)^n$ $(1 + 1)^1$ $2^1$ $2$ |
$1 + nx$ $1 + 1(1)$ $1 + 1$ $2$ |
|
$2 \ge 2$ The initial step works |
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|
Optional testing Test for $n = 1, x = 0$ |
$(1 + x)^n$ $(1 + 0)^1$ $1^1$ $1$ |
$1 + nx$ $1 + 1(0)$ $1 + 0$ $1$ |
|
$1 \ge 1$ The optional testing works |
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|
Optional testing Test for $n = 2, x = 1$ |
$(1 + x)^n$ $(1 + 1)^2$ $2^2$ $4$ |
$1 + nx$ $1 + 2(1)$ $1 + 2$ $3$ |
|
$4 \ge 3$ The optional testing works |
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|
Induction Hypothesis Assume $n = k$ $(1 + x)^k \ge 1 + kx...eqn.(1)$ |
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|
Induction Step Assume $n = k + 1$ $(1 + x)^{k + 1} \ge 1 + (k + 1)x...eqn.(2)$ |
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| 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: $\color{darkblue}{1 + kx} * (1 + x)$ $(1 + kx)(1 + x)$ $1 + x + kx + kx^2$ $\color{black}{1 + x + kx} + kx^2$ |
$1 + (k + 1)x$ $1 + x(k + 1)$ $1 + kx + x$ $1 + x + kx$ $\color{black}{1 + x + kx}$ |
|
$x \gt -1$ $\therefore x^2 \ge 0$ $n \ge 1$ $n = k + 1$ $k = n - 1$ $\therefore k \ge 0$ $\implies kx^2 \ge 0$ $LHS \ge RHS$ The formula is proved. |
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(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
$
| Steps | $LHS$ | $RHS$ |
|---|---|---|
| $(1 + x)^n$ | $1 + nx$ | |
|
Initial Step Test for $n = 1, x = 1$ |
$(1 + x)^n$ $(1 + 1)^1$ $2^1$ $2$ |
$1 + nx$ $1 + 1(1)$ $1 + 1$ $2$ |
|
$2 \ge 2$ The initial step works |
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|
Optional testing Test for $n = 1, x = 0$ |
$(1 + x)^n$ $(1 + 0)^1$ $1^1$ $1$ |
$1 + nx$ $1 + 1(0)$ $1 + 0$ $1$ |
|
$1 \ge 1$ The optional testing works |
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|
Optional testing Test for $n = 2, x = 1$ |
$(1 + x)^n$ $(1 + 1)^2$ $2^2$ $4$ |
$1 + nx$ $1 + 2(1)$ $1 + 2$ $3$ |
|
$4 \ge 3$ The optional testing works |
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Induction Hypothesis Assume $n = k$ $(1 + x)^k \ge 1 + kx...eqn.(1)$ |
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Induction Step Assume $n = k + 1$ $(1 + x)^{k + 1} \ge 1 + (k + 1)x...eqn.(2)$ |
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| 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: $\color{darkblue}{1 + kx} * (1 + x)$ $(1 + kx)(1 + x)$ $1 + x + kx + kx^2$ $\color{black}{1 + x + kx} + kx^2$ |
$1 + (k + 1)x$ $1 + x(k + 1)$ $1 + kx + x$ $1 + x + kx$ $\color{black}{1 + x + kx}$ |
|
$x \gt -1$ $\therefore x^2 \ge 0$ $n \ge 1$ $n = k + 1$ $k = n - 1$ $\therefore k \ge 0$ $\implies kx^2 \ge 0$ $LHS \ge RHS$ The formula is proved. |
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