Introduction to Proofs

Last Updated : 11 Aug, 2025

Mathematical proof is an argument we give logically to validate a mathematical statement. To validate a statement, we consider two things: A statement and Logical operators.

A statement is either true or false but not both.
Logical operators are AND, OR, NOT, IF THEN, and IF AND ONLY IF.
Coupled with quantifiers like for all and there exists.
We apply operators to the statement to check its correctness.

A proof is a logically sound, step-by-step argument demonstrating that a specific statement (theorem, lemma, corollary) must be true, given a set of fundamental assumptions (axioms/postulates) and previously proven results. Proof transforms conjectures into established truths.

Types of Mathematical Proofs

Mathematicians use several styles of proof, depending on the nature of the statement being proven and the tools available. The most common types are:

Proof by Cases

In this method, we evaluate every case of the statement to conclude its truthiness.

Example: For every integer x, the integer x(x + 1) is even

Proof: If x is even, hence, x = 2k for some number k. now the statement becomes: 2k(2k + 1)
which is divisible by 2, hence it is even.

If x is odd, hence x = 2k + 1 for some number k, now the statement becomes: (2k+1)(2k+1+1) = (2k + 1) 2(k + 1)
which is again divisible by 2 and hence in both cases we proved that x(x+1) is even.

Used in algorithm analysis when different inputs cause the algorithm to behave differently, and you need to prove correctness for each scenario.

Proof by Contradiction

We assume the negation of the given statement and then proceed to conclude the proof.

Example: Prove that sqrt(2) is irrational.

Suppose sqrt(2) is rational.
sqrt(2) = a/b, for some integers a and b with b != 0.
Let us choose integers a and b with sqrt(2) = a/b, such that b is positive and as small as possible. (Well-Ordering Principle)
a² = 2b²
Since a² is even, it follows that a is even.
a = 2k for some integer k, so a² = 4k²
b² = 2k². Since b²is even, it follows that b is even.
Since a and b are both even, a/2 and b/2 are integers with b/2 > 0, and sqrt(2) = (a/2)/(b/2), because (a/2)/(b/2) = a/b.
But it contradicts our assumption b is as small as possible. Therefore sqrt(2) cannot be rational.

Common in theoretical computer science to prove impossibility results or limits.

Proof by Induction

The Principle of Mathematical Induction (PMI). Let P(n) be a statement about the positive integer n. If the following are true:

1. P(1),
2. (for all n there exists Z+) P(n) implies P(n + 1),

Then (for all n there exists Z+) P(n).

Example: For every positive integer n prove that , 1 + 2 +···+ n = n(n + 1)/ 2

Proof:
Base case: If n = 1,
1 + ··· + n = 1
n(n + 1)/2 = 1
Inductive step:
Suppose that for a given n there exists Z+,
1 + 2 +···+ n = n(n + 1)/ 2 ---- (i) (inductive hypothesis)
Our goal is to show that:
1 + 2 +···+ n + (n + 1) = [n + 1]([n + 1] + 1)/ 2
i.e. 1 + 2 +···+ n + (n + 1) = (n + 1)(n + 2) /2
Add n + 1 both sides to equation (i), we get,
1 + 2 +···+ n + (n + 1)
= n(n + 1)/ 2 + (n + 1)
= n(n + 1) /2 + 2(n + 1) /2
= (n + 2)(n + 1) /2
Hence Proved

Essential for proving the correctness of recursive algorithms, loop invariants, and formulas used in algorithms.

Direct Proof

When we want to prove a conditional statement p implies q, we assume that p is true, and follow implications to get to show that q is then true.
It is Mostly an application of hypothetical syllogism, [(p → r) ∧ (r → q)] → (p → q)]
We just have to find the propositions that lead us to q.

Theorem: If m is even and n is odd, then their sum is odd
Proof:
Since m is even, there is an integer j such that m = 2j.
Since n is odd, there is an integer k such that n = 2k+1. Then,
m+n = (2j)+(2k+1) = 2(j+k)+1
Since j+k is an integer, we see that m+n is odd.

Used in program verification to show that a function meets its specification directly from the code logic.

Indirect Proof

This method is also called as proof by contraposition.

In this method to prove a conditional statement p → q, we assume that ∼q is true and then follow the implication by applying knowledge and facts that we know to show that ∼p is true.
Since, we know that ∼q → ∼p is equivalent to p->q, by proving ∼q → ∼p, we'll indirectly be showing that p → q. Hence, the name "indirect proof".

Example: Prove that if n is an integer and 3n + 2 is even, then n is even.

Proof: Assume n is odd,
Hence n = 2k + 1, for some integer k
Now, 3n + 2 = 3(2k + 1) + 2
3n + 2 = 3(2k) + 3(1) + 2
3n + 2 = 2(3k) + 3 + 2
3n + 2 = 2(3k) + 5, where 3k is some integer
So, 3n + 2 is an odd integer.
From above, n is odd implies 3n + 2 is odd. Hence Proved.

Disproof by Counter Example

In this proof technique, a given statement is disproved by providing a counterexample.

Example: Prove or disprove the conjecture, "For every positive integer n, n! <= n².

Proof: 4 is a positive integer, but 4! is not less than or equal to 4². Therefore, given conjecture is false.

Used in testing and debugging to invalidate incorrect algorithms or claims.

Solved Examples

Statement: Prove that if n is an odd integer, then n²is odd.

Proof:

Let n be an odd integer, By definition, n can be written as n = 2k+1 for some integer k.
Now, square both sides: n² = (2k +1)²
Expand the square : n² = 4k² + 4k + 1
Factor out the 2 from the first two terms: n²= 2(2k² +2k) + 1
Since 2k² + 2k is an integer , we can write n² = 2m +1 , where m =2k² + 2k
Therefore , n² is odd

Statement: Prove that there is no smallest positive rational number.

Proof:

Assume , for the sake of contradiction, that there is a smallest positive rational number, say r.
Since r is rational , we can write r = p/q where p and q are positive and integers with no common factors.
Consider the number r/2 = p/2q
Since p/2q is positive and smaller than r, this contradicts our assumption that r is the smallest positive rational number.
Hence, there is no smallest positive rational number.

Statement: Prove that if n is not divisible by 3, then n² is not divisible by 3

Proof:

The contrapositive of the statement is "if n² is divisible by 3, then n is divisible by 3.
Assume that n2 is divisible by 3. This means that n² = 3k for some integers k.
if n² is divisible by 3, then n must be divisible by 3 (because if a square of number is divisible by a prime number, the number itself must be divisible by that prime number).
Therefore, if n² is divisible by 3, n must be divisible by 3.
Since the contrapositive is true, the original statement is true.

Introduction to Proofs - Practice Problems

Problem 1: Prove that the product of any two odd integers is odd.

Problem 2: Prove that there is no integer n such that n² = 2n +1

Problem 3: Prove that if n² is even, then n is even.

Problem 4: Prove that there are infinitely many prime numbers

Problem 5: Prove that there is no rational number that is equal to √3

Problem 6: Prove that there is no integer solution to the equation x² + y² = 3

Problem 7: Prove that if a triangle is not equilateral, then it does not have all equal angles.

Problem 8: Prove that 3ⁿ +2 is divisible by 4 for all natural numbers n.

Problem 9: Prove that the square of any even integer is divisible by 4.

Problem 10: Prove that for any integer n, if n is divisible by 6, then n is divisible by both 2 and 3.

Sonu Tiwari

Improve

Article Tags :

Mathematics

Introduction to Proofs

Types of Mathematical Proofs

Proof by Cases

Proof by Contradiction

Proof by Induction

Direct Proof

Indirect Proof

Disproof by Counter Example

Solved Examples

Introduction to Proofs - Practice Problems

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