Introduction to Proofs (original) (raw)

Last Updated : 2 Jun, 2026

Mathematical proof is a logical argument used to show that a mathematical statement is true. To verify a statement, we use statements, logical operators, and quantifiers.

A proof is a clear, step-by-step explanation that shows why a statement (theorem, lemma, or corollary) is true using axioms, definitions, and previously proven results. Proofs help convert conjectures into established mathematical truths.

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**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.

**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 b2 is 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.

**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

**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.

**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.

Solved Examples

**Statement: Prove that if n is an odd integer, then n2 is 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.

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.