IMAT Mathematics Master Handbook & Compendium

📌 Detailed Operational Derivation & Mathematical Proof: Decimal Conversion

Let $x = 0.a_1 a_2 \dots a_k \overline{p_1 p_2 \dots p_n}$ be a general mixed repeating decimal where the non-repeating part $A$ has $k$ digits and the repeating period $P$ has $n$ digits.

1. Multiply $x$ by $10^{k+n}$ to shift both the non-repeating antiperiod and one full repeating period to the left of the decimal point: $$10^{k+n} \cdot x = (a_1 \dots a_k p_1 \dots p_n) . \overline{p_1 \dots p_n}$$
2. Multiply $x$ by $10^k$ to shift only the non-repeating antiperiod to the left of the decimal point: $$10^k \cdot x = (a_1 \dots a_k) . \overline{p_1 \dots p_n}$$
3. Subtract equation (2) from equation (1). The infinite repeating decimal tails cancel out completely: $$(10^{k+n} - 10^k) x = (a_1 \dots a_k p_1 \dots p_n) - (a_1 \dots a_k)$$ $$10^k (10^n - 1) x = \text{Entire Digit Sequence} - \text{Non-repeating Digit Sequence}$$
4. Solve for $x$: Since $10^n - 1 = \underbrace{99\dots9}_{n \text{ times}}$ and multiplying by $10^k$ appends $k$ zeros, we get: $$x = \frac{\text{Entire Number} - \text{Non-Repeating Part}}{\underbrace{99\dots9}_{n \text{ nines}} \underbrace{00\dots0}_{k \text{ zeros}}}$$

📌 Detailed Operational Derivation & Proof: Double Radical Unnesting

We seek to express a nested radical $\sqrt{a \pm 2\sqrt{b}}$ in unnested form $\sqrt{x} \pm \sqrt{y}$ for positive real numbers $x, y$.

1. Square both sides of the assumed identity $\sqrt{a \pm 2\sqrt{b}} = \sqrt{x} \pm \sqrt{y}$: $$\left(\sqrt{a \pm 2\sqrt{b}}\right)^2 = \left(\sqrt{x} \pm \sqrt{y}\right)^2$$ $$a \pm 2\sqrt{b} = (\sqrt{x})^2 + (\sqrt{y})^2 \pm 2\sqrt{x}\sqrt{y}$$ $$a \pm 2\sqrt{b} = (x + y) \pm 2\sqrt{xy}$$
2. Equate rational and irrational parts on both sides: $$\begin{cases} x + y = a \\ xy = b \end{cases}$$
3. Construct quadratic equation for $t$ whose roots are $x$ and $y$: $$t^2 - (x+y)t + xy = 0 \implies t^2 - at + b = 0$$ $$t = \frac{a \pm \sqrt{a^2 - 4b}}{2}$$
4. Therefore, $x = \frac{a + \sqrt{a^2 - 4b}}{2}$ and $y = \frac{a - \sqrt{a^2 - 4b}}{2}$. Unnesting is rational if and only if $a^2 - 4b$ is a perfect square!

📌 Detailed Operational Derivation: Quadratic Formula & Vieta's Relations

1. Derivation of Quadratic Formula (Completing the Square):

2. Derivation of Vieta's Formulas:

📌 Detailed Operational Derivation: Parabola Vertex Coordinates

Method 1: Completing the Square (Standard Form conversion $y = a(x - h)^2 + k$):

Method 2: Differential Calculus (Critical Point $f'(x) = 0$):

📌 Detailed Operational Derivation: Base Change Theorem

Let $x = \log_a b$. We wish to express $x$ in terms of logarithms to a new base $c$.

1. Convert the logarithmic equation $x = \log_a b$ to its equivalent exponential form: $$a^x = b$$
2. Take the logarithm with base $c$ of both sides: $$\log_c(a^x) = \log_c b$$
3. Apply the power rule of logarithms $\log_c(a^x) = x \cdot \log_c a$: $$x \cdot \log_c a = \log_c b$$
4. Divide by $\log_c a$ (valid since $a \neq 1 \implies \log_c a \neq 0$): $$x = \frac{\log_c b}{\log_c a} \implies \log_a b = \frac{\log_c b}{\log_c a}$$

📌 Detailed Operational Derivation: Law of Cosines

Consider $\Delta ABC$ placed on Cartesian plane with vertex $A$ at origin $(0,0)$, side $b$ along positive $x$-axis so $C = (b,0)$.

1. Coordinates of vertex $B$ are given by polar projection: $B = (c \cos A, c \sin A)$.
2. Apply distance formula for side length $a = |BC|$: $$a^2 = (c \cos A - b)^2 + (c \sin A - 0)^2$$
3. Expand both squared terms: $$a^2 = c^2 \cos^2 A - 2bc \cos A + b^2 + c^2 \sin^2 A$$
4. Group terms and apply Pythagorean identity $\cos^2 A + \sin^2 A = 1$: $$a^2 = b^2 + c^2 (\cos^2 A + \sin^2 A) - 2bc \cos A = b^2 + c^2 - 2bc \cos A \qquad \blacksquare$$

📌 Detailed Operational Proof: Inclusion-Exclusion Principle

Let $A$ and $B$ be finite sets. We count the elements in $A \cup B$:

1. The sum $|A| + |B|$ counts every element in $A \setminus B$ once, every element in $B \setminus A$ once, but counts every element in $A \cap B$ TWICE.
2. To ensure each element in $A \cup B$ is counted exactly once, subtract the double-counted intersection $|A \cap B|$: $$|A \cup B| = |A| + |B| - |A \cap B| \qquad \blacksquare$$

📌 Detailed Operational Proof: Contrapositive Equivalence via Truth Tables

Conclusion: Columns for $P \implies Q$ and $\neg Q \implies \neg P$ are identical for all truth values, proving logical equivalence.

📌 Detailed Operational Derivation: Bayes' Theorem

1. By definition of conditional probability for events $A$ and $B$: $$P(A \cap B) = P(A | B) \cdot P(B)$$
2. Symmetrically, conditioning $B$ on $A$: $$P(A \cap B) = P(B | A) \cdot P(A)$$
3. Equate both expressions for joint probability $P(A \cap B)$: $$P(A | B) \cdot P(B) = P(B | A) \cdot P(A)$$
4. Divide by $P(B)$ (where $P(B) > 0$): $$P(A | B) = \frac{P(B | A) \cdot P(A)}{P(B)} \qquad \blacksquare$$

📌 Detailed Operational Derivation: Geometric Series Sum Formulas

Let $S_n = a + ar + ar^2 + \dots + ar^{n-1}$.

1. Multiply $S_n$ by common ratio $r$: $$r S_n = ar + ar^2 + ar^3 + \dots + ar^n$$
2. Subtract $r S_n$ from $S_n$. All intermediate terms cancel in a telescoping sum: $$S_n - r S_n = a - ar^n \implies S_n (1 - r) = a (1 - r^n)$$ $$S_n = \frac{a(1 - r^n)}{1 - r} \quad (r \neq 1)$$
3. For infinite series ($n \to \infty$), if $|r| < 1$, then $\lim_{n \to \infty} r^n = 0$: $$S_\infty = \frac{a(1 - 0)}{1 - r} = \frac{a}{1 - r} \qquad \blacksquare$$

📌 Detailed Operational Derivation: Centroid 2:1 Median Division Ratio

Let $G$ be the intersection of medians $AD$ and $BE$ in $\Delta ABC$.

1. Connect midpoints $D$ and $E$ to form the mid-segment $DE$. By Midpoint Theorem, $DE \parallel AB$ and $DE = \frac{1}{2} AB$.
2. Since $DE \parallel AB$, opposite interior angles are equal: $\angle GDE = \angle GAB$ and $\angle GED = \angle GBA$.
3. Therefore, $\Delta GAB \sim \Delta GDE$ by AA similarity.
4. The ratio of corresponding sides equals the ratio of similarity scale factor: $$\frac{AG}{GD} = \frac{BG}{GE} = \frac{AB}{DE} = \frac{AB}{\frac{1}{2} AB} = 2$$
5. Thus, the centroid $G$ divides each median in a strict $2 : 1$ ratio from vertex to midpoint! $\blacksquare$