Master direct proof, mathematical induction, proof by cases, contradiction, contrapositive, and counter example — with visual flowcharts, step-marked solutions, MCQs, A-R, and PYQs.
Understanding mathematical proofs
In Classes IX, X and XI, we have learnt about the concepts of a statement, compound statement, negation, converse and contrapositive of a statement; axioms, conjectures, theorems and deductive reasoning.
"Proofs are to Mathematics what calligraphy is to poetry. Mathematical works do consist of proofs just as poems do consist of characters." — Vladimir Arnold
Here, we will discuss various methods of proving mathematical propositions.
Proof of a mathematical statement consists of a sequence of statements, each statement being justified with a definition or an axiom or a proposition that is previously established by the method of deduction using only the allowed logical rules.
Thus, each proof is a chain of deductive arguments, each of which has its premises and conclusions. Many a times, we prove a proposition directly from what is given in the proposition. But sometimes it is easier to prove an equivalent proposition rather than proving the proposition itself.
This leads to two ways of proving a proposition — directly or indirectly.
See proof methods in action
Straightforward approach, induction, and cases
It is a chain of arguments which leads 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.
Show that if $x^2 - 5x + 6 = 0$, then $x = 3$ or $x = 2$.
Explanation: Starting with $p$, we deduce $p \Rightarrow r \Rightarrow s \Rightarrow \ldots \Rightarrow q$. This proves $p \Rightarrow q$ is true.
Example: Prove that $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 2x + 5$ is one-one.
$f(x_1) = f(x_2) \Rightarrow x_1 = x_2$
$2x_1 + 5 = 2x_2 + 5$
$2x_1 = 2x_2 \Rightarrow x_1 = x_2$
Deductive proof strategy for natural numbers
For a given subset $S$ of $\mathbb{N}$, if
then $S = \mathbb{N}$.
According to the principle of mathematical induction, if a statement $P(n)$ is true for $n = 1$ (or for some starting point $j$), and if "$P(n)$ is true for $n = k$" implies that "$P(n)$ is true for $n = k+1$", then the statement is true for all positive integers $n \geq j$.
Show that if $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$, then $A^n = \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}$ for all $n \in \mathbb{N}$.
$A^1 = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ ✓
Assume $A^k = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$ is true.
$A^{k+1} = A^k \cdot A = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & k+1 \\ 0 & 1 \end{pmatrix}$ ✓
Examining every possible case
This method of proving $p \Rightarrow q$ is possible only when $p$ can be split into several cases $r, s, t$ (say) so that $p = r \vee s \vee t$.
If $r \Rightarrow q$, $s \Rightarrow q$, and $t \Rightarrow q$ are proved, then $(r \vee s \vee t) \Rightarrow q$ is proved, and so $p \Rightarrow q$ is proved.
Show that in any triangle ABC, $a = b\cos C + c\cos B$
Any triangle is either acute, obtuse, or right-angled. We prove the result for each case.
From right triangle ADB: $BD = c\cos B$
From right triangle ADC: $CD = b\cos C$
$a = BD + CD = c\cos B + b\cos C$ ✓
From right triangle ADB: $BD = c\cos B$
From right triangle ADC: $CD = -b\cos C$ (since $\cos(180° - C) = -\cos C$)
$a = BD - CD = c\cos B + b\cos C$ ✓
$a = c\cos B$ and $b\cos C = b\cos 90° = 0$
$a = c\cos B + 0 = c\cos B + b\cos C$ ✓
Contradiction, contrapositive, and counter example
Instead of proving the given proposition directly, we establish the proof of the proposition through proving a proposition which is equivalent to the given proposition.
Reductio Ad Absurdum
Here, we start with the assumption that the given statement is false. By rules of logic, we arrive at a conclusion contradicting the assumption and hence it is inferred that the assumption is wrong and hence the given statement is true.
Show that the set of all prime numbers is infinite.
Assume the set of all prime numbers is finite: $P_1, P_2, \ldots, P_k$.
Consider $N = (P_1 P_2 \ldots P_k) + 1$.
N is larger than any number in the list, so N is not in the list.
If N is prime, then there exists a prime not in the list — contradiction.
If N is composite, it must have a prime divisor. But none of $P_1, \ldots, P_k$ divides N (all leave remainder 1). So the prime divisor is not in the list — contradiction.
In both cases, we get a contradiction. Hence, our assumption is false.
Proving ~q ⇒ ~p instead of p ⇒ q
Instead of proving the conditional $p \Rightarrow q$, we prove its equivalent, i.e., $\sim q \Rightarrow \sim p$.
The contrapositive of a conditional can be formed by interchanging the conclusion and the hypothesis and negating both.
Prove that $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 2x + 5$ is one-one (using contrapositive).
$p \Rightarrow q$: If $f(x_1) = f(x_2)$, then $x_1 = x_2$.
$\sim q \Rightarrow \sim p$: If $x_1 \neq x_2$, then $f(x_1) \neq f(x_2)$.
Assume $x_1 \neq x_2$.
Then $2x_1 \neq 2x_2$ (multiplying by 2 preserves inequality).
Then $2x_1 + 5 \neq 2x_2 + 5$ (adding 5 preserves inequality).
So $f(x_1) \neq f(x_2)$. ✓
Show that "if a matrix A is invertible, then A is non-singular".
Contrapositive: If A is singular ($|A| = 0$), then A is not invertible.
If $|A| = 0$, then $A^{-1} = \frac{\text{adj } A}{|A|}$ does not exist.
Hence A is not invertible. ✓
Disproving a universal statement
A counter example is an example that disproves a statement. Since the disproof of a proposition $p \Rightarrow q$ is merely a proof of the proposition $\sim(p \Rightarrow q)$, this is also a method of proof.
Statement: For each $n \in \mathbb{N}$, $2^{2^n} + 1$ is a prime.
Check:
But for $n=5$: $2^{32} + 1 = 4294967297 = 641 \times 6700417$ (not prime)
Statement: Every continuous function is differentiable.
Counter Example: $f(x) = |x|$ is continuous at $x=0$ but not differentiable at $x=0$.
Speed techniques for proofs
Ask: Can I start from p and reach q directly?
p ⇒ q ≡ ~q ⇒ ~p
Assume the opposite, derive absurdity
One counter example disproves a universal statement
Base case + Inductive step = Done
Split into all possible cases
Avoid these:
Converse: q ⇒ p (not equivalent!)
Contrapositive: ~q ⇒ ~p (equivalent!)
Check every possibility
Always verify:
Click "Show Answer" to reveal step-by-step solutions with marks
Q1. Show that if $x^2 - 5x + 6 = 0$, then $x = 3$ or $x = 2$.
$x^2 - 5x + 6 = (x-3)(x-2) = 0$
If $ab = 0$, then $a = 0$ or $b = 0$
So $x - 3 = 0$ or $x - 2 = 0$
$x = 3$ or $x = 2$
Q2. Prove that $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 2x + 5$ is one-one.
$f$ is one-one if $f(x_1) = f(x_2) \Rightarrow x_1 = x_2$
$2x_1 + 5 = 2x_2 + 5$
$2x_1 = 2x_2 \Rightarrow x_1 = x_2$
Hence $f$ is one-one. ✓
Q3. Prove that the set of all prime numbers is infinite.
Assume primes are finite: $P_1, P_2, \ldots, P_k$.
$N = P_1 P_2 \ldots P_k + 1$
If N is prime → new prime, contradiction.
If N is composite → prime divisor not in list, contradiction.
Assumption false → primes are infinite. ✓
Q4. Show that in any triangle ABC, $a = b\cos C + c\cos B$.
Any triangle is acute, obtuse, or right-angled.
$BD = c\cos B$, $CD = b\cos C$
$a = BD + CD = c\cos B + b\cos C$ ✓
$BD = c\cos B$, $CD = -b\cos C$
$a = BD - CD = c\cos B + b\cos C$ ✓
$a = c\cos B$ and $b\cos C = 0$
$a = c\cos B + 0$ ✓
Q5. Prove that $f(x) = 2x + 5$ is one-one using contrapositive.
$p \Rightarrow q$: If $f(x_1) = f(x_2)$, then $x_1 = x_2$.
$\sim q \Rightarrow \sim p$: If $x_1 \neq x_2$, then $f(x_1) \neq f(x_2)$.
Assume $x_1 \neq x_2$.
Then $2x_1 \neq 2x_2$ and $2x_1 + 5 \neq 2x_2 + 5$.
So $f(x_1) \neq f(x_2)$. ✓
Since $\sim q \Rightarrow \sim p$, $p \Rightarrow q$ is proved.
Q6. Disprove: "Every continuous function is differentiable."
Consider $f(x) = |x|$.
$f$ is continuous at $x = 0$ (LHL = RHL = f(0) = 0).
$Lf'(0) = -1$, $Rf'(0) = +1$
Since $Lf'(0) \neq Rf'(0)$, $f$ is not differentiable at 0.
Q7. Show that if $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$, then $A^n = \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}$ for all $n \in \mathbb{N}$.
$n=1$: $A^1 = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ ✓
Assume $A^k = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$
$A^{k+1} = A^k \cdot A = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & k+1 \\ 0 & 1 \end{pmatrix}$ ✓
By PMI, true for all $n \in \mathbb{N}$. ✓
Select your answer and click "Check Answers"
Each question has an Assertion (A) and a Reason (R). Choose the correct option.
(A) Both A and R are true, and R is the correct explanation of A.
(B) Both A and R are true, but R is NOT the correct explanation of A.
(C) A is true, but R is false.
(D) A is false, but R is true.
Frequently asked questions in board exams
Q. Show that if $x^2 - 5x + 6 = 0$, then $x = 3$ or $x = 2$.
$x^2 - 5x + 6 = (x-3)(x-2) = 0$
$x - 3 = 0$ or $x - 2 = 0$
$x = 3$ or $x = 2$
Q. Prove that $f(x) = 2x + 5$ is one-one.
$f(x_1) = f(x_2) \Rightarrow x_1 = x_2$
$2x_1 + 5 = 2x_2 + 5$
$2x_1 = 2x_2 \Rightarrow x_1 = x_2$
$f$ is one-one ✓
Q. Prove that the set of all prime numbers is infinite.
Assume primes are finite: $P_1, \ldots, P_k$.
$N = P_1 P_2 \ldots P_k + 1$
If N prime → new prime. If N composite → prime divisor not in list.
Both cases contradict. Assumption false.
Q. Show that in any triangle ABC, $a = b\cos C + c\cos B$.
Triangle is acute, obtuse, or right-angled.
$a = BD + CD = c\cos B + b\cos C$ ✓
$a = BD - CD = c\cos B + b\cos C$ ✓
$a = c\cos B + 0$ ✓
Q. Prove that $f(x) = 2x + 5$ is one-one using contrapositive.
$p \Rightarrow q$: $f(x_1)=f(x_2) \Rightarrow x_1=x_2$
$\sim q \Rightarrow \sim p$: $x_1 \neq x_2 \Rightarrow f(x_1) \neq f(x_2)$
$x_1 \neq x_2 \Rightarrow 2x_1 \neq 2x_2 \Rightarrow 2x_1+5 \neq 2x_2+5$
$\Rightarrow f(x_1) \neq f(x_2)$ ✓
$\sim q \Rightarrow \sim p$ proved → $p \Rightarrow q$ proved.
Q. Disprove: "Every continuous function is differentiable" using a counter example.
$f(x) = |x|$ is continuous at $x = 0$.
$Lf'(0) = -1$, $Rf'(0) = +1$
$Lf'(0) \neq Rf'(0)$ → not differentiable.
Q. Show that if $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$, then $A^n = \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}$ for all $n \in \mathbb{N}$.
$n=1$: $A^1 = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ ✓
$A^k = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$
$A^{k+1} = A^k \cdot A = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & k+1 \\ 0 & 1 \end{pmatrix}$ ✓
By PMI, true for all $n \in \mathbb{N}$. ✓
Q. Prove that the matrix A is invertible ⇒ A is non-singular (using contrapositive).
$p \Rightarrow q$: A invertible $\Rightarrow$ A non-singular
$\sim q \Rightarrow \sim p$: A singular $\Rightarrow$ A not invertible
If A is singular, $|A| = 0$.
$A^{-1} = \frac{\text{adj } A}{|A|}$ does not exist.
So A is not invertible. ✓
$\sim q \Rightarrow \sim p$ proved → $p \Rightarrow q$ proved.