Mathematics: Class 12 (Book 1) — Chapter 7: Proofs in Mathematics

A chain of arguments leading directly from what is given or assumed, with the help of axioms, definitions, or already proved theorems, to what is to be proved using rules of logic ($p \Rightarrow r \Rightarrow s \Rightarrow \dots \Rightarrow q$).

Mathematical Problem Statement: Show that if $x^2 - 5x + 6 = 0$, then $x = 3$ or $x = 2$.

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Step-by-Step Algorithmic Solution Trace:

1. Given: $x^2 - 5x + 6 = 0$.
2. Factorize the quadratic expression: $(x - 3)(x - 2) = 0$ (replacing an expression by an equivalent expression).
3. Apply the zero-product property: $x - 3 = 0$ or $x - 2 = 0$ (from the established theorem $ab = 0 \Rightarrow a = 0$ or $b = 0$ for $a, b \in \mathbf{R}$).
4. Add equal quantities to both sides: $x - 3 + 3 = 0 + 3$ or $x - 2 + 2 = 0 + 2$.
5. Simplify using additive identity: $x = 3$ or $x = 2$.

Final Evaluation Output: $x = 3$ or $x = 2$

Mathematical Problem Statement: Prove that the function $f : \mathbf{R} \to \mathbf{R}$ defined by $f(x) = 2x + 5$ is one-one.

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Step-by-Step Algorithmic Solution Trace:

1. Assume $f(x_1) = f(x_2)$ for $x_1, x_2 \in \mathbf{R}$.
2. Substitute the function definition: $2x_1 + 5 = 2x_2 + 5$.
3. Subtract $5$ from both sides: $2x_1 + 5 - 5 = 2x_2 + 5 - 5 \Rightarrow 2x_1 = 2x_2$.
4. Divide both sides by $2$: $\frac{2}{2}x_1 = \frac{2}{2}x_2 \Rightarrow x_1 = x_2$.

Final Evaluation Output: Since $f(x_1) = f(x_2) \Rightarrow x_1 = x_2$, the function $f$ is one-one.

A deductive strategy based on the following axiom for a subset $S \subset \mathbf{N}$:

• The natural number $1 \in S$.
• If $k \in S \Rightarrow k + 1 \in S$, then $S = \mathbf{N}$.

Mathematical Problem Statement: If $A = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$, then prove that $A^n = \begin{bmatrix} \cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{bmatrix}$ for all $n \in \mathbf{N}$.

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Step-by-Step Algorithmic Solution Trace:

1. Base Case ($n = 1$): $P(1): A^1 = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$, which is true.
2. Inductive Step: Assume $P(k)$ is true for some $k \in \mathbf{N}$:$$A^k = \begin{bmatrix} \cos k\theta & \sin k\theta \\ -\sin k\theta & \cos k\theta \end{bmatrix}$$
3. Prove $P(k+1)$ is true:$$A^{k+1} = A^k \cdot A = \begin{bmatrix} \cos k\theta & \sin k\theta \\ -\sin k\theta & \cos k\theta \end{bmatrix} \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$$
4. Apply matrix multiplication and trigonometric identities ($\cos A \cos B - \sin A \sin B = \cos(A+B)$):$$A^{k+1} = \begin{bmatrix} \cos k\theta \cos \theta - \sin k\theta \sin \theta & \cos k\theta \sin \theta + \sin k\theta \cos \theta \\ -\sin k\theta \cos \theta - \cos k\theta \sin \theta & -\sin k\theta \sin \theta + \cos k\theta \cos \theta \end{bmatrix}$$$$A^{k+1} = \begin{bmatrix} \cos(k+1)\theta & \sin(k+1)\theta \\ -\sin(k+1)\theta & \cos(k+1)\theta \end{bmatrix}$$

Final Evaluation Output: By the principle of mathematical induction, $P(n)$ is true for all $n \ge 1$.

This method is used when a statement $p \Rightarrow q$ can be split into several mutually exclusive and exhaustive cases $r, s, t$ such that $p = r \lor s \lor t$. If $r \Rightarrow q$, $s \Rightarrow q$, and $t \Rightarrow q$ are all proved, then $p \Rightarrow q$ is proved.

Mathematical Problem Statement: Show that in any triangle $ABC$, $a = b \cos C + c \cos B$.

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Step-by-Step Algorithmic Solution Trace:

1. Let $p$ be "ABC is any triangle" and $q$ be "$a = b \cos C + c \cos B$". We split $p$ into three cases based on the nature of $\angle C$:
2. Case (i): $\angle C$ is acute. Draw perpendicular $AD \perp BC$. In right-angled $\triangle ADB$, $BD = c \cos B$. In right-angled $\triangle ADC$, $CD = b \cos C$. Since $D$ lies on $BC$, $a = BD + CD = c \cos B + b \cos C$.
3. Case (ii): $\angle C$ is obtuse. Draw perpendicular $AD \perp BC$ produced. In right-angled $\triangle ADB$, $BD = c \cos B$. In right-angled $\triangle ADC$, $CD = b \cos(\angle ACD) = b \cos(180^\circ - C) = -b \cos C$. Since $C$ lies between $B$ and $D$, $a = BC = BD - CD = c \cos B - (-b \cos C) = c \cos B + b \cos C$.
4. Case (iii): $\angle C$ is a right angle. Here, $C = 90^\circ$. Thus, $b \cos C = b \cos 90^\circ = 0$. In right-angled $\triangle ACB$, $a = BC = c \cos B$. Thus, $a = 0 + c \cos B = b \cos C + c \cos B$.

Final Evaluation Output: In all three cases, $a = b \cos C + c \cos B$ holds true.

To prove a statement $p$ is true, we start by assuming its negation $\sim p$ is true. We then logically deduce a conclusion that contradicts an established fact or our initial assumption. This forces the realization that $\sim p$ is false, hence $p$ must be true.

Mathematical Problem Statement: Prove that the set of all prime numbers is infinite.

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Step-by-Step Algorithmic Solution Trace:

1. Assume the negation: The set of all prime numbers is finite, say $P = \{P_1, P_2, P_3, \dots, P_k\}$.
2. Construct a new integer: $N = (P_1 P_2 P_3 \dots P_k) + 1$.
3. Analyze $N$: Since $N > P_i$ for all $i$, $N$ is not in the list of primes. If $N$ is prime, we have found a prime not in our finite list, which is a contradiction.
4. If $N$ is composite, it must have a prime factor $p$. If $p$ belongs to our list, then $p$ divides the product $P_1 P_2 \dots P_k$. Since $p$ also divides $N$, it must divide their difference $N - (P_1 P_2 \dots P_k) = 1$, which is impossible.

Final Evaluation Output: Our assumption is false; the set of prime numbers is infinite.

Instead of proving $p \Rightarrow q$, we prove its logically equivalent contrapositive statement: $\sim q \Rightarrow \sim p$.

Mathematical Problem Statement: Prove that if $f(x) = 2x + 5$ is one-one, then $x_1 \neq x_2 \Rightarrow f(x_1) \neq f(x_2)$.

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Step-by-Step Algorithmic Solution Trace:

1. We want to prove $p \Rightarrow q$, where $p: f(x_1) = f(x_2) \Rightarrow x_1 = x_2$.
2. The contrapositive is $\sim q \Rightarrow \sim p$, which is: "If $x_1 \neq x_2$, then $f(x_1) \neq f(x_2)$".
3. Start with $x_1 \neq x_2$. Multiply by $2$: $2x_1 \neq 2x_2$. Add $5$: $2x_1 + 5 \neq 2x_2 + 5$. This yields $f(x_1) \neq f(x_2)$.

Final Evaluation Output: Since $\sim q \Rightarrow \sim p$ is proved, the equivalent statement $p \Rightarrow q$ is true.

Mathematical Problem Statement: Show that "if a matrix $A$ is invertible, then $A$ is non-singular".

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Step-by-Step Algorithmic Solution Trace:

1. Let $p$: "$A$ is invertible" and $q$: "$A$ is non-singular".
2. We prove the contrapositive: $\sim q \Rightarrow \sim p$ ("If $A$ is singular, then $A$ is not invertible").
3. Assume $A$ is singular, which means $|A| = 0$. The inverse formula is $A^{-1} = \frac{\text{adj } A}{|A|}$.
4. Since $|A| = 0$, the division is undefined, and $A^{-1}$ does not exist. Thus, $A$ is not invertible.

Final Evaluation Output: The contrapositive is true; hence, if $A$ is invertible, then $A$ is non-singular.

To disprove a universal mathematical conjecture or statement (e.g., "All $X$ have property $Y$"), finding a single specific instance where the statement fails is mathematically sufficient. This single instance is called a counterexample.

Mathematical Problem Statement: Disprove the conjecture: "For each $n \in \mathbf{N}$, $2^{2^n} + 1$ is a prime."

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Step-by-Step Algorithmic Solution Trace:

1. Test $n = 1$: $2^{2^1} + 1 = 5$ (Prime).
2. Test $n = 2$: $2^{2^2} + 1 = 17$ (Prime).
3. Test $n = 3$: $2^{2^3} + 1 = 257$ (Prime).
4. Test $n = 5$: $2^{2^5} + 1 = 2^{32} + 1 = 4294967297$.
5. Factorize: $4294967297 = 641 \times 6700417$. Since it has factors other than $1$ and itself, it is composite.

Final Evaluation Output: The value $n = 5$ serves as a counterexample, disproving the general statement.

Mathematical Problem Statement: Disprove the statement: "Every continuous function is differentiable."

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Step-by-Step Algorithmic Solution Trace:

1. Consider the absolute value function $\phi(x) = |x|$.
2. This function is continuous everywhere, including at $x = 0$.
3. Evaluate differentiability at $x=0$: Left-hand derivative is $-1$, right-hand derivative is $+1$. Since they do not match, the function is not differentiable at $x=0$.

Final Evaluation Output: The function $\phi(x) = |x|$ is a counterexample that disproves the statement.