The Fundamental Theorem of Arithmetic

Mathematical Statement (or Proposition) 

A mathematical statement is a sentence which is either true or false . It can not be ambiguous.

Conjecture 

It is a conclusion or proposition based on incomplete information, for which no proof has been found.

Axiom 

A statement or proposition which is regarded as being established, accepted, or self-evidently true. It need not to be proved.

Theorem 

It is a statement that has been proved on the basis of previously established statements, such as other theorems, and generally accepted statements, such as axioms. A theorem is a logical consequence of the axioms.

Proof 

The process which can establish the truth of a mathematical statement based purely on logical arguments is called a mathematical proof.
The Fundamental Theorem of Arithmetic was first recorded as Proposition 14 of Book IX in Euclid’s Elements, before it came to be known as such. 

It's first correct proof was given by Carl Friedrich Gauss (1777-1855) in his Disquisitiones Arithmeticae (1801).
Carl Friedrich Gauss is often referred to as the ‘Prince of Mathematicians’ and is considered one of the three greatest mathematicians of all time, along with Archimedes and Newton. He has made fundamental contributions to both mathematics and science.

The Fundamental Theorem of Arithmetic

"Every composite number can be expressed ( factorised ) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur."

➧ Given any composite number, there is one and only one way to write it as a product of primes, as long as we are not particular about the order in which the primes occur. e.g., we regard 2 × 3 × 5 × 7 as the same as 3 × 5 × 7 × 2, or any other possible order in which these primes are written.
➧ In general, given a composite number x, we factorise it as x = p₁p₂ ... p_n, where p₁p₂ ... p_n are primes and written in ascending order, i.e. p₁≤ p₂ ≤ . . .≤ p_n.
➢ If we combine the same primes, we will get powers of primes. e.g.,
     32760 = 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13 = 2³ × 3² × 5 × 7 × 13
➢ Once we have decided that the order will be ascending, then the way the number is factorised, is unique.

➧ The Fundamental Theorem of Arithmetic has many applications, both within mathematics and in other fields
➢ One of its application is in Prime factorisation method for finding HCF & LCM of two or more integers.
➢ It is applied to prove the irrationality of many irrational numbers.
➢ It is applied to explore when exactly the decimal expansion of rational number is terminating and when it is non-terminating repeating.

Finding HCF & LCM of Integers

➢ For given  two positive integers p & q,

HCF(p,q) × LCM(p,q) = p × q

➢ For given  three positive integers p, q & r,

HCF(p,q,r)	=	p×q×r×LCM(p,q,r)
		LCM(p,q)×LCM(q,r)×LCM(r,p)

LCM(p,q,r)	=	p×q×r×HCF(p,q,r)
		HCF(p,q)×HCF(q,r)×HCF(r,p)

(Q) Find the HCF and LCM of 6, 72 and 120, using the prime factorisation method.
(A) 6 = 2 × 3, 72 = 2³ × 3², 120 = 2³ × 3 × 5
     HCF (6, 72 and 120) = 2 × 3  (common prime factors with lowest power in common)
     LCM (6, 72 and 120) = 2³ × 3²× 5 ( all prime factors with highest power)

Proving that a given number is irrational

To prove that a given number is irrational, we need the following theorem, whose proof is based on the Fundamental Theorem of Arithmetic.

Theorem : Let p be a prime number. If p divides a², then p divides a, where a is a positive integer.

Proof : 
Let a = p₁p₂...p_n , where p₁p₂...p_n are primes not necessarily distinct.
⇒ a² = p₁²p₂²...p_n²
⇒  p₁p₂...p_n  are prime factors of a²    
p divides a²                                                        (given)
⇒ p is one of the prime factor of a²      (from fundamental theorem of arithmetic)
But Prime factorisation of any number is unique
⇒ p is one of p₁p₂...p_n
∵a = p₁p₂...p_n
⇒ p divides a.

Theorem : √2 is irrational.

Proof : 
Let us assume that √2 is rational, such that for integers a and b (b≠0), √2 = ^a/_{b ,} where a and b are co-primes, i.e. ratio is in standard form.
⇒ b√2 = a
Squaring both sides, we get
(b√2)² = (a)²
⇒ 2b² = a²
⇒ 2 divides  a²
⇒ 2 divides  a
So, we can write a =2c for some integer c,
2b² = 4c²         (∵2b² = a² )
⇒ b² = 2c²
⇒ 2 divides  b²
⇒ 2 divides  b
∴  a and b have 2 as a common factor.
But this contradicts the fact that a & b are co-primes.
Hence, this contradicts our assumption that √2 is rational.
So, we conclude that √2 is irrational.

➤ The technique we used above is known as Proof by Contradiction.
➤ Rational number and Irrational number
   ➢ Sum and Difference of a rational number and a irrational number is irrational.
   ➣ The product and quotient of a non-zero rational number and irrational number is irrational.

Show that:2−√3  is irrational.
Proof : Let us assume that 2-√3 is rational, such that for co-primes a and b (b≠0), 2-√3 =  ^a/_b .
⇒ 2 − ^a/_b = √3_.
⇒ √3_.= 2 − ^a/_b = ^2b−a/_b
∵ a & b are integers
⇒ ^2b−a/_b is rational
⇒ √3 is rational
But this contradicts the fact that √3 is irrational.
So, we conclude that 2−√3  is irrational.

Rational Numbers & Their Decimal Expansions

➤ We know that decimal expansions of Rational numbers can be either
→ terminating or
→ non-terminating reoccurring (repeating).
➤ Can we predict by looking at denominator when the given rational is terminating and when it is non-terminating reoccurring.

Rational numbers whose decimal expansions are terminating

↬ Fraction whose denominator is power of 10 will terminate in its decimal expansion.
     e.g., ³⁷⁵/₁₀₀ = 3.75, ²³/₁₀₀₀ = 0.023 etc.
∵ Prime factorisation of 10 = 2×5
⇒ Prime factorisation of 10^k= 2^k×5^k   , where k is any positive integer.
⇒ Denominator of a fraction which is a power of 10 can be expressed as 2^k×5^k.

If we simplify these fraction into a standard rational number, where p & q are co-primes then some powers of two's or five's or both in the numerator & the denominator will cancel each other out, then Prime factorisation of q will have the form 2ⁿ×5^m, where m and n are some non-negative integers (can be 0).
e.g, 0.375 = ³⁷⁵/₁₀₀₀  = ^3×53/_10³ = ^3×53/₂³_×5³= ³/₂³

here, q = 2³×5⁰

↬ Fraction whose denominator is power of 2 only will terminating in its decimal expansion.
e.g., ²⁷/₂ = 13.5, ¹¹/₄ = 2.75, ⁷/₈ = 0.875 etc.
↬ Fraction whose denominator is power of 5 only will terminate in its decimal expansion. 
e.g., ¹/₅ = 0.2, ²⁰⁰/₁₂₅ = 1.6, etc.

Theorem: Let x = ^p/_q be a rational number, such that the prime factorisation of q is of the form  2ⁿ×5^m, where n, m are non-negative integers. Then x has a decimal expansion which terminates.

Example: Among following rational numbers, select which will terminate.
(i) ²⁷/₃₂₀ (ii) ⁸¹/₉₀ (iii) ⁴⁵/₂₁₀ (iv) ¹²⁷/₇₅₀
Solution:
(i) ²⁷/₃₂₀ = ³³/₂⁶_×5 , 
    Here, q = 2⁶×5¹ (= 2ⁿ×5^m)          
    ⇒ ²⁷/₃₂₀ will terminate and, 
    = 0.084375   
(ii) ⁸¹/₉₀ = ³⁴/_2×3²×5 = ³²/_2×5
    Here, q = 2¹×5¹ (= 2ⁿ×5^m, n=m)     
    ⇒ ⁸¹/₉₀ will terminate and,
    = 0.9         
(iii) ⁴⁵/₂₁₀ = ^32×5/_2×3×5×7 = ³/_2×7
    Here, q = 2¹× 7 (≠ 2ⁿ×5^m) 
    ⇒ ⁴⁵/₂₁₀ will not terminate.
(iv) ¹²⁷/₇₅₀ = ¹²⁷/_2×3×5³
    Here, q = 2×3×5³ (≠ 2ⁿ×5^m) 
    ⇒ ¹²⁷/₇₅₀ will not terminate.            

Theorem : Let x be a rational number whose decimal expansion terminates. Then x can be expressed in the form ^p/_q ,where p and q are co-prime, and the prime factorisation of q is of the form  2ⁿ×5^m, where n, m are non-negative integers.

e.g, ³⁷⁵/₁₀₀₀ (= 0.375)
       dividing by 125 (HCF), we get
       = ³/₈ = ³/_2³
      (= ^p/_q and q = 2³×5⁰)

We can convert the rational number in form of  ^p/_q to equivalent rational number of form  ^a/_b , where b is power of 10 .We can then easily write its decimal expansion which is terminating.
e.g,  ³/₈ = ³/_2³   (form of  ^p/_q)
= ^3×53/₂³_×5³ (multiplying both the numerator & denominator with 5³)
= ³⁷⁵/₁₀³ = 0.375

Rational numbers whose decimal expansions are non-terminating recurring (repeating)

Rational numbers in the form ^p/_q where q is not of the form  2ⁿ×5^m, will not have terminating decimal expansion.
e.g., ¹/₇ = 0.142857

Theorem: Let x = ^p/_q be a rational number, such that the prime factorisation of q is not of the form  2ⁿ×5^m, where n, m are non-negative integers. Then x has a decimal expansion which is non-terminating recurring (repeating).

e.g.,  ⁴⁵/₂₁₀ = ^32×5/_2×3×5×7 = ³/_2×7
    Here, q = 2¹× 7 (≠ 2ⁿ×5^m) 
    ⇒ ⁴⁵/₂₁₀ will not terminate and,
    =0.169333... = 0.1693

Euclid's Division Algorithm