Table of Content

• Methods of Proving Theorems
• 1. Direct Proof Method
• 2. Indirect Proof Method
• Mistakes in Proofs
• Solved Examples
• Important Mathematical Proofs – Practice Problems

S No.	Step	Reason
1.	a = b	Given
2.	a2 = ab	Multiply both sides of (1) by a
3.	a2 – b2 = ab – b2	Subtract b2 from both sides of (2)
4	(a-b) (a + b) = b(a-b)	Factor both sides of (3)
5.	a + b = b	Divide both sides of (4) by a – b
6.	2b = b	Replace a by b in (5) because a = b and simplify
7.	2 = 1	Divide both sides of (6) by b

Solution:

Every step is valid except for one , step 5 where we divided both sides by (a – b) . The error is that (a – b) equals zero; division of both sides of an equation by the same quantity is valid as long as this quantity is not zero.

2. Suppose you want to prove the assertion:

Let a, b, ∈ Z where a = 1 mod 3 and b = 2 mod 3. Then (a + b) = 0 mod 3.

Incorrect Proof:

Since a = 1 mod 3 there is an integer k in Z such that a = 3k + 1. Since b = 2 mod 3, we can write b = 3k + 2. Thus a + b = (3k + 1) + (3k + 2) = 6k + 3 = 3(2k + 1), so (a + b) = 0 mod 3.

Error in the Proof:

The attempted proof assumes that b = a + 1 since b − a = (3k + 2) − (3k + 1) = 1. So the proof is only valid for that limited set of choices for a and b.

Correct Proof:

Since a = 1 mod 3 there is an integer k in Z such that a = 3k+1. Since b = 2 mod 3, there is an integer n in Z such that b = 3n + 2. Therefore a + b = (3k + 1) + (3n + 2) = 3k + 3n + 3 = 3(k + n + 1), so (a + b) = 0 mod 3

Solved Examples

Example 1 : Proof of the Pythagorean Theorem:

Given: Right triangle ABC with right angle at C.

To Prove: a² + b² = c², where a, b are legs and c is hypotenuse.

Proof:

Construct squares on each side of the triangle.

Draw altitude from C to hypotenuse, creating two similar triangles.

Area of square on hypotenuse = c²

Areas of squares on legs = a² and b²

Using similarity of triangles, show that c² = a² + b²

Therefore, a² + b² = c²

Example 2 : Proof that √2 is Irrational:

Assume √2 is rational.

Then √2 = a/b, where a and b are integers with no common factors.

Square both sides: 2 = a²/b²

Multiply by b²: 2b² = a²

a² is even, so a must be even. Let a = 2k.

Substitute: 2b² = (2k)² = 4k²

Simplify: b² = 2k²

b² is even, so b must be even.

But this contradicts our assumption that a and b have no common factors.

Therefore, √2 cannot be rational.

Example 3 : Proof of the Fundamental Theorem of Algebra:

Statement: Every non-constant polynomial with complex coefficients has at least one complex root.

Outline of proof (this is a complex proof, so we’ll provide a sketch):

Assume f(z) is a polynomial of degree n ≥ 1.

Show that |f(z)| → ∞ as |z| → ∞.

Define g(z) = 1/f(z) for |z| > R, where R is large.

Show g(z) is continuous and bounded on |z| ≥ R.

Use Liouville’s theorem to show g must have a zero.

This zero corresponds to a root of f(z).

Example 4 :Proof of Euler’s Formula (e^(iπ) + 1 = 0):

Start with Euler’s formula: e^(ix) = cos x + i sin x

Let x = π:

e^(iπ) = cos π + i sin π

= -1 + 0i

= -1

Add 1 to both sides:

e^(iπ) + 1 = -1 + 1 = 0

Therefore, e^(iπ) + 1 = 0

Example 5 : Proof of the Infinitude of Primes:

Assume there are finitely many primes: p₁, p₂, …, pn.

Consider N = (p₁ × p₂ × … × pn) + 1.

N is either prime or not prime.

If N is prime, we have a prime larger than pn, contradicting our assumption.

If N is not prime, it must have a prime factor.

This prime factor cannot be any of p₁, p₂, …, pn (as dividing N by any of these leaves remainder 1).

So we have found a new prime, again contradicting our assumption.

Therefore, there must be infinitely many primes.

Example 6 :Proof of the Triangle Inequality:

To prove: |x + y| ≤ |x| + |y| for any real numbers x and y.

Proof:

(|x + y|)² = (x + y)(x + y) = x² + 2xy + y²

(|x| + |y|)² = |x|² + 2|x||y| + |y|² = x² + 2|x||y| + y²

Note that 2xy ≤ 2|x||y| for all x and y.

Therefore, (|x + y|)² ≤ (|x| + |y|)²

Taking the square root of both sides (which preserves inequality as both sides are non-negative):

|x + y| ≤ |x| + |y|

Example 7 :Proof of the Intermediate Value Theorem:

Statement: If f is a continuous function on [a,b] and k is between f(a) and f(b), then there exists c in [a,b] such that f(c) = k.

Proof (by contradiction):

Assume f is continuous on [a,b], f(a) < k < f(b), but there’s no c in [a,b] with f(c) = k.

Let A = {x in [a,b] : f(x) < k} and B = {x in [a,b] : f(x) > k}.

A and B are non-empty (a is in A, b is in B) and their union is [a,b].

Let s = sup(A). Then s is in [a,b].

If f(s) < k, then by continuity, f(x) < k for x near s, contradicting s = sup(A).

If f(s) > k, then by continuity, f(x) > k for x near s, contradicting that s is an upper bound of A.

Therefore, f(s) must equal k, contradicting our assumption.

Thus, there must exist c in [a,b] such that f(c) = k.

Example 8 :Proof of the Cauchy-Schwarz Inequality:

To prove: |⟨x,y⟩|² ≤ ⟨x,x⟩ ⟨y,y⟩ for vectors x and y in an inner product space.

Proof:

For any real t, consider the non-negative quantity:

⟨x – ty, x – ty⟩ ≥ 0

Expanding:

⟨x,x⟩ – 2t⟨x,y⟩ + t²⟨y,y⟩ ≥ 0

This quadratic in t is always non-negative, so its discriminant must be non-positive:

(2⟨x,y⟩)² – 4⟨x,x⟩⟨y,y⟩ ≤ 0

Simplifying:

⟨x,y⟩² ≤ ⟨x,x⟩⟨y,y⟩

Taking absolute values:

|⟨x,y⟩|² ≤ ⟨x,x⟩⟨y,y⟩

Example 9 :Proof of the Law of Quadratic Reciprocity:

Statement: For odd primes p and q, (p/q)(q/p) = (-1)^((p-1)(q-1)/4), where (p/q) is the Legendre symbol.

This proof is quite complex, so we’ll outline Gauss’s lemma approach:

Consider the set S = {1, 2, …, (p-1)/2} mod p.

Count the number of elements in S that become negative when multiplied by q.

Show this count is related to (q/p).

Use symmetry to relate (p/q) and (q/p).

Conclude the quadratic reciprocity law.

Example 10 : Proof of Fermat’s Little Theorem:

Statement: If p is prime and a is not divisible by p, then a^(p-1) ≡ 1 (mod p).

Proof:

Consider the numbers a, 2a, 3a, …, (p-1)a mod p.

These are all distinct mod p (if ia ≡ ja (mod p), then i ≡ j (mod p) as a is not divisible by p).

Therefore, these numbers are a permutation of 1, 2, …, p-1 mod p.

Multiply them together:

a × 2a × 3a × … × (p-1)a ≡ 1 × 2 × 3 × … × (p-1) (mod p)

a^(p-1) × (p-1)! ≡ (p-1)! (mod p)

Cancel (p-1)! on both sides (as p is prime, (p-1)! is not divisible by p):

a^(p-1) ≡ 1 (mod p)

Important Mathematical Proofs – Practice Problems

1).Prove that the sum of the angles in a triangle is 180°.

2).Prove that there are infinitely many Pythagorean triples.

3).Prove the Binomial Theorem using mathematical induction.

4).Prove that e is irrational.

5).Prove the Fundamental Theorem of Calculus.

6).Prove the Unique Factorization Theorem for integers.

7).Prove that the set of rational numbers is countable.

8).Prove the Intermediate Value Theorem for continuous functions.

9).Prove the Mean Value Theorem for derivatives.

10).Prove the Extreme Value Theorem for continuous functions on closed intervals.

Conclusion – Important Mathematical Proofs

Understanding different methods of proof is fundamental in mathematics. Direct proofs, indirect proofs, and proof by contradiction each provide unique ways to demonstrate the validity of mathematical statements. By mastering these techniques, students and mathematicians can construct clear and compelling arguments. This foundational knowledge not only strengthens mathematical reasoning but also enhances problem-solving skills across various domains.

FAQs on Important Mathematical Proofs

What is a direct proof?

A direct proof starts with the assumption that the hypothesis is true and uses logical steps to arrive at the conclusion. It involves using definitions, theorems, and axioms to show that if P(x) is true, then Q(x) must also be true.

How does an indirect proof differ from a direct proof?

An indirect proof, such as proof by contrapositive or contradiction, starts by assuming the negation of the conclusion. It then shows that this assumption leads to a contradiction, thereby proving the original statement.

What are common mistakes in mathematical proofs?

Common mistakes include incorrect assumptions, dividing by zero, and misapplying logical steps. Ensuring each step follows logically from the previous one and carefully checking each assumption can help avoid these errors.