FUNDAMENTAL THEOREM OF ARITHMETIC - EXPLANATION AND ITS APPLICATION WITH EXAMPLES

Please study
Prime Factorisation before Fundamental Theorem of Arithmetic,
if you have not already done so. It is a prerequisite here.

In Elementary Number Theory, it is proved that every natural number,
except 1, is a product of prime factors and this product is unique.

For example, 144 = 2 x 2 x 2 x 2 x 3 x 3
= 2⁴ x 3²;
and 1547 = 7 x 13 x 17.

Because multiplication is commutative and associative,
the order of factors is irrelevant.

In proving this theorem, mathematicians defined
an abnormal number to be a hypothetical number
that does not have a unique prime factorization.

They proved the theorem by proving that
there does not exist an abnormal number.

Unique Prime Factorisation Theorem (or) Fundamental Theorem of Arithmetic

Every natural number, except 1 can be expressed as
the product of Prime Factors in a unique way
except for the order in which the factors appear.

This is also known as Fundamental Theorem of Arithmetic.

Solved Example 1 of Fundamental Theorem of Arithmetic

Verify the truth of Unique Prime Factorisation Theorem
(Fundamental Theorem of Arithmetic)
by expressing the following numbers as the product of prime factors.

(i) 21 (ii) 36 (iii) 56 (iv) 72 (v) 50 (vi) 45

Solution :
(i) 21

3 | 21
  ----------
7 |  7
  ----------
     1 

Note : If you want explanation of the above vertical presentation, see
Prime Factorisation

(ii) 36

2 | 36
  ----------
2 | 18
  ----------
3 |  9
  ----------             
3 |  3
  ----------
     1 

(iii) 56

2 | 56
  ----------
2 | 28
  ----------
2 | 14
  ----------             
7 |  7
  ----------
     1 

(iv) 72

2 | 72
  ----------
2 | 36
  ----------
2 | 18
  ----------             
3 |  9
  ----------
3 |  3 
  ----------
     1

(v) 50

2 | 50
  ----------
5 | 25
  ----------
5 |  5
  ----------             
     1

(vi) 45

5 | 45
  ----------
3 |  9
  ----------
3 |  3
  ----------             
     1

Thus, all the six given numbers are expressed as
products of prime factors in a unique way except
for the order in which the factors appear.

This verifies the truth of Unique Prime Factorisation Theorem
(Fundamental Theorem of Arithmetic). Ans.

Application of Fundamental Theorem of Arithmetic

From the fundamental theorem of arithmetic, it is clear that
each natural number can be formed from the product of
Prime Numbers  in a unique way.

Finding the prime factorization of a natural number allows
derivation of all its factors, both prime and non-prime.

Number of Factors of a natural number and finding all the factors

If n = p^a.q^b.r^c.......
where n is a natural number ( ≠ 1 ),
p, q, r.... are distinct Prime Factors
and a, b, c....are their multiplicitys (exponents)
then, the number of Factors of n = (a + 1)(b + 1)(c + 1)...........

Different combinations of the products of
the distinct Prime Factors with different exponents
gives all the factors of the natural number.

The following example makes the concept clear.

Solved Example 2 of Fundamental Theorem of Arithmetic :

Find out the number of factors of the numbers in the previous Example
(Solved Example 1) and verify them by writing all the factors.

Solution :
(i) We have 21 = 3 x 7. = 3¹.7¹
Number of Factors of 21 = (1 + 1)(1 + 1) = 4.

The Factors of 21 = { 1, 3, 7, 21 } which are 4. (Verified.)

(ii) We have 36 = 2 x 2 x 3 x 3. = 2².3²
Number of Factors of 36 = (2 + 1)(2 + 1) = 9.

The Factors of 36 = { 1, 2, 3, 4, 6, 9, 12, 18, 36 } which are 9. (Verified.)

(iii) We have 56 = 2 x 2 x 2 x 7. = 2³.7¹
Number of Factors of 56 = (3 + 1)(1 + 1) = 8.

The Factors of 56 = { 1, 2, 4, 7, 8, 14, 28, 56 } which are 8. (Verified.)

(iv) We have 72 = 2 x 2 x 2 x 3 x 3. = 2³.3²
Number of Factors of 72 = (3 + 1)(2 + 1) = 12.

The Factors of 72 = {1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72} which are 12.(Verified.)

(v) We have 50 = 2 x 5 x 5. = 2¹.5²
Number of Factors of 50 = (1 + 1)(2 + 1) = 6.

The Factors of 50 = { 1, 2, 5, 10, 25, 50 } which are 6. (Verified.)

(vi) We have 45 = 5 x 3 x 3. = 5¹.3²
Number of Factors of 45 = (1 + 1)(2 + 1) = 6.

The Factors of 50 = { 1, 3, 5, 9, 15, 45 } which are 6. (Verified.)

Exercise 1 of Fundamental Theorem of Arithmetic

(1) Verify the truth of Unique Prime Factorisation Theorem
(Fundamental Theorem of Arithmetic)
by expressing the following numbers as the product of prime factors.

(i) 18 (ii) 30 (iii) 63 (iv) 75 (v) 84 (vi) 96

(2) Find out the number of factors of the numbers in the previous
question (question 1) and verify them by writing all the factors.

Answers to Exercise 1of Fundamental Theorem of Arithmetic

(1) (i) 2 x 3 x 3 (ii) 2 x 3 x 5 (iii) 3 x 3 x 7 (iv) 3 x 5 x 5
(v) 2 x 2 x 3 x 7 (vi) 2 x 2 x 2 x 2 x 2 x 3

(2) (i) 6; { 1, 2, 3, 6, 9, 18 } (ii) 8; { 1, 2, 3, 5, 6, 10, 15, 30 }
(iii) 6; { 1, 3, 7, 9, 21, 63 } (iv) 6; { 1, 3, 5, 15, 25, 75 }
(v) 12; { 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84 }
(vi) 12; { 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96 }