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 = 2^{4} x 3^{2}; 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

Thus 21 = 3 x 7. Ans.

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

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.

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Number of Factors of a natural number and finding all the factors : Fundamental Theorem of Arithmetic

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. = 3^{1}.7^{1} 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. = 2^{2}.3^{2} 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. = 2^{3}.7^{1} 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. = 2^{3}.3^{2} 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. = 2^{1}.5^{2} 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. = 5^{1}.3^{2} Number of Factors of 45 = (1 + 1)(2 + 1) = 6.

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

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Do the following problems on 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

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