 Class 10

### The Fundamental Theorem of Arithmetic

In your earlier classes, you have seen that any natural number can be written as a product of its prime factors. For instance, 2 = 2, 4 = 2 × 2, 253 = 11 × 23, and so on.
Now, let us try and look at natural numbers from the other direction. That is, can any natural number be obtained by multiplying prime numbers? Let us see.

Take any collection of prime numbers, say 2, 3, 7, 11 and 23. If we multiply some or all of these numbers, allowing them to repeat as many times as we wish,
we can produce a large collection of positive integers (In fact, infinitely many). Let us list a few :

7 × 11 × 23 = 1771 \ \ \ \ \ \ \ \ \ \ 3 × 7 × 11 × 23 = 5313
2 × 3 × 7 × 11 × 23 = 10626 \ \ \ \ \ \ \ \ \ \ 2^3 × 3 × 73 = 8232
22 × 3 × 7 × 11 × 23 = 21252

and so on.
Now, let us suppose your collection of primes includes all the possible primes. What is your guess about the size of this collection? Does it contain only a finite number of integers, or infinitely many? Infact, there are infinitely many primes. So, if we combine all these primes in all possible ways, we will get an infinite collection of numbers, all the primes and all possible products of primes. The question is – can we produce all the composite numbers this way? What do you think? Do you think that there may be a composite number which is not the product of powers of primes? Before we answer this, let us factorise positive integers, that is, do the opposite of what we have done so far.
We are going to use the factor tree with which you are all familiar. Let us take
some large number, say, 32760, and factorise it as shown : So we have factorised 32760 as 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13 as a product of primes, i.e., 32760 = 23 × 32 × 5 × 7 × 13 as a product of powers of primes. Let us try another number, say, 123456789. This can be written as 32 × 3803 × 3607. Of course, you have to check that 3803 and 3607 are primes (Try it out for several other natural numbers yourself.) This leads us to a conjecture that every composite number can be written as the product of powers of primes. In fact, this statement is true, and is called the Fundamental Theorem of Arithmetic because of its basic crucial importance to the study of integers. Let us now formally state this theorem.

Theorem 1.2 (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.

The Fundamental Theorem of Arithmetic says that every composite number can be factorised as a product of primes. Actually it says more. It says that given any composite number it can be factorised as a product of prime numbers in a ‘unique’ way, except for the order in which the primes occur. That is, 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. So, for example, 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. This fact is also stated in the following form:

The prime factorisation of a natural number is unique, except for the order of its factors.

In general, given a composite number x, we factorise it as x = p_1p_2 ... p_n, where
p_1, p_2,..., p_n are primes and written in ascending order, i.e., p_1 ≤ p_2 ≤ . . . ≤ p_n. If we combine the same primes, we will get powers of primes. For example, 32760 = 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13 = 2^3 × 3^2 × 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. Let us look at some examples.
Q 3119756619 Consider the numbers 4^n, where n is a natural number. Check whether there is any value of n for which 4^n ends with the digit zero.
Class 12 Chapter 1 Example 5 Solution:

If the number 4^n, for any n, were to end with the digit zero, then it would be divisible by 5. That is, the prime factorisation of 4^n would contain the prime 5. This is not possible because 4^n = (2)2^n; so the only prime in the factorisation of 4n is 2. So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorisation of 4n. So, there is no natural number n for which 4^n ends with the digit zero.

You have already learnt how to find the HCF and LCM of two positive integers using the Fundamental Theorem of Arithmetic in earlier classes, without realising it This method is also called the prime factorisation method. Let us recall this method through an example.
Q 3139856712 Find the LCM and HCF of 6 and 20 by the prime factorisation method.
Class 10 Chapter 1 Example 6 Solution:

We have : 6 = 21 × 31 and 20 = 2 × 2 × 5 = 22 × 51.
You can find HCF (6, 20) = 2 and LCM (6, 20) = 2 × 2 × 3 × 5 = 60, as done in your
earlier classes.
Note that HCF (6, 20) = 21 = Product of the smallest power of each common prime factor in the numbers.
LCM (6, 20) = 22 × 31 × 51 = Product of the greatest power of each prime factor, involved in the numbers.
From the example above, you might have noticed that HCF (6, 20) × LCM(6, 20) = 6 × 20. In fact, we can verify that for any two positive integers a and b,
HCF (a, b) × LCM (a, b) = a × b. We can use this result to find the LCM of two positive integers, if we have already found the HCF of the two positive integers.
Q 3149856713 Find the HCF of 96 and 404 by the prime factorisation method. Hence, find their LCM.
Class 10 Chapter 1 Example 7 Solution:

The prime factorisation of 96 and 404 gives :

96 = 25 × 3, 404 = 22 × 101

Therefore, the HCF of these two integers is 2^2 = 4.

Also, LCM (96, 404) = ( 96xx404)/(HCF (96 , 404))= (96xx404)/4 = 9696
Q 3169856715 Find the HCF and LCM of 6, 72 and 120, using the prime factorisation method.
Class 10 Chapter 1 Example 8 Solution:

We have : 6 = 2 × 3, 72 = 2^3 × 3^2, 120 = 2^3 × 3 × 5

Here, 2^1 and 3^1 are the smallest powers of the common factors 2 and 3, respectively.

So, HCF (6, 72, 120) = 21 × 31 = 2 × 3 = 6

23, 32 and 51 are the greatest powers of the prime factors 2, 3 and 5 respectively involved in the three numbers.
So, LCM (6, 72, 120) = 23 × 32 × 51 = 360

Remark : Notice, 6 × 72 × 120 ≠ HCF (6, 72, 120) × LCM (6, 72, 120). So, the product of three numbers is not equal to the product of their HCF and LCM. 