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Fundamental Theorem of Arithmetic

🎓 Class 10 Mathematics CBSE Theory Ch 1 — Real Numbers ⏱ ~35 min
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This MCQ module is based on: Fundamental Theorem of Arithmetic

This mathematics assessment will be based on: Fundamental Theorem of Arithmetic
Targeting Class 10 level in Number Theory, with Intermediate difficulty.

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1.1 Introduction

In Class IX, you began exploring the world of real numbers? and encountered irrational numbers. We continue our discussion on real numbers in this chapter. We begin with two very important properties of positive integers, namely Euclid's division algorithm and the Fundamental Theorem of Arithmetic?.

Euclid's division algorithm, as the name suggests, has to do with divisibility of integers. Stated simply, it says any positive integer \(a\) can be divided by another positive integer \(b\) in such a way that it leaves a remainder \(r\) that is smaller than \(b\). You probably recognise this as the usual long division process. Although it is quite easy to state and understand, it has many applications related to the divisibility properties of integers.

The Fundamental Theorem of Arithmetic, on the other hand, has to do with multiplication of positive integers. You already know that every composite number can be expressed as a product of primes in a unique way -- this important fact is the Fundamental Theorem of Arithmetic. We use it for two main applications:

  1. To prove the irrationality of numbers such as \(\sqrt{2}\), \(\sqrt{3}\) and \(\sqrt{5}\).
  2. To explore when exactly the decimal expansion of a rational number \(\frac{p}{q}\) (\(q \neq 0\)) is terminating and when it is non-terminating repeating.

1.2 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, \(2 = 2\), \(4 = 2 \times 2\), \(253 = 11 \times 23\), 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?

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:

\(7 \times 11 \times 23 = 1771\)   \(3 \times 7 \times 11 \times 23 = 5313\)
\(2 \times 3 \times 7 \times 11 = 10626\)   \(2^2 \times 3 \times 7^2 = 8232\)
\(2^2 \times 3 \times 7 \times 11 \times 23 = 21252\)

Now, let us suppose your collection of primes includes all the possible primes. Does it contain only a finite number of integers, or infinitely many? In fact, 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? Before we answer this, let us factorise a large number, say 32760, and factorise it as shown in the factor tree below.

32760 2 16380 2 8190 2 4095 3 1365 3 455 5 91 91 = 7 × 13 Primes (in gold): 2, 2, 2, 3, 3, 5, 7, 13
Factor tree for 32760: \(32760 = 2^3 \times 3^2 \times 5 \times 7 \times 13\)

So we have factorised 32760 as \(2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 7 \times 13\), which is a product of primes. That is, \(32760 = 2^3 \times 3^2 \times 5 \times 7 \times 13\), written as a product of powers of primes.

Let us try another number, say 123456790. This can be written as \(3^2 \times 3803 \times 3607\). (Of course, you have to check that 3803 and 3607 are primes!) 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.

Theorem 1.1 (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.
G

Carl Friedrich Gauss

(1777 -- 1855)

An equivalent version of Theorem 1.1 was probably first recorded as Proposition 14 of Book IX in Euclid's Elements, before it came to be known as the Fundamental Theorem of Arithmetic. However, the first correct proof was given by Carl Friedrich Gauss in his Disquisitiones Arithmeticae. 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.

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. So, for example, we regard \(2 \times 3 \times 5 \times 7\) as the same as \(3 \times 5 \times 7 \times 2\), or any other possible order in which these primes are written. This fact is also stated in the following form:

Uniqueness of Prime Factorisation
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 \(x = p_1 p_2 \ldots p_n\) where \(p_1, p_2, \ldots, p_n\) are primes and written in ascending order, i.e., \(p_1 \le p_2 \le \ldots \le p_n\). If we combine the same primes, we will get powers of primes. For example:

\(32760 = 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 7 \times 13 = 2^3 \times 3^2 \times 5 \times 7 \times 13\)

Once we have decided that the order will be ascending, then the way the number is factorised is unique.

HCF and LCM using the Fundamental Theorem of Arithmetic

The Fundamental Theorem of Arithmetic has many applications, both within mathematics and in other fields. Let us look at some examples.

Example 1: Can \(4^n\) end with the digit zero?

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.
Step 1: If a number ends with the digit zero, then it would be divisible by 5.

Step 2: That is, if \(4^n\) ends with the digit zero, then the prime factorisation of \(4^n\) would contain the prime 5.

Step 3: This is not possible because \(4^n = (2^2)^n = 2^{2n}\); so the only prime in the factorisation of \(4^n\) is 2.

Step 4: So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorisation of \(4^n\).

Conclusion: So, there is no natural number \(n\) for which \(4^n\) ends with the digit zero.
Think about it: Can \(6^n\) end with the digit zero for any natural number \(n\)? Since \(6^n = (2 \times 3)^n = 2^n \times 3^n\), and 5 is not a factor, \(6^n\) can never end 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.

Example 2: Find the LCM and HCF of 6 and 20 by the prime factorisation method.

Find the HCF and LCM of 6 and 20 by prime factorisation.
Step 1: Write the prime factorisations.
\(6 = 2^1 \times 3^1\)
\(20 = 2^2 \times 5^1\)

Step 2: Find HCF.
HCF = Product of the smallest power of each common prime factor.
Common prime factor: only 2.
\(\text{HCF}(6, 20) = 2^1 = 2\)

Step 3: Find LCM.
LCM = Product of the greatest power of each prime factor involved.
Prime factors involved: 2, 3, 5.
\(\text{LCM}(6, 20) = 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60\)

Verification: \(\text{HCF}(6, 20) \times \text{LCM}(6, 20) = 2 \times 60 = 120\) and \(6 \times 20 = 120\). ✓
Key Rule
HCF = Product of the smallest power of each common prime factor in the numbers.
LCM = Product of the greatest power of each prime factor, involved in the numbers.

Example 3: Find the HCF and LCM of 96 and 404.

Find the HCF and LCM of 96 and 404 by the prime factorisation method. Also verify that \(\text{HCF} \times \text{LCM} = \text{Product of the two numbers}\).
Step 1: Prime factorisations:
\(96 = 2^5 \times 3\)
\(404 = 2^2 \times 101\)

Step 2: HCF = product of common primes with smallest powers.
Common prime: 2. Smallest power: \(2^2 = 4\).
\(\text{HCF}(96, 404) = 2^2 = 4\)

Step 3: LCM = product of all primes with greatest powers.
\(\text{LCM}(96, 404) = 2^5 \times 3 \times 101 = 32 \times 3 \times 101 = 9696\)

Verification:
\(\text{HCF} \times \text{LCM} = 4 \times 9696 = 38784\)
\(96 \times 404 = 38784\) ✓
Both are equal. Hence verified.

Example 4: Find the HCF and LCM of 6, 72 and 120.

Find the HCF and LCM of 6, 72 and 120, using the prime factorisation method.
Step 1: Prime factorisations:
\(6 = 2 \times 3\)
\(72 = 2^3 \times 3^2\)
\(120 = 2^3 \times 3 \times 5\)

Step 2: HCF = product of smallest powers of common primes.
Common primes: 2 and 3.
Smallest power of 2: \(2^1\). Smallest power of 3: \(3^1\).
\(\text{HCF}(6, 72, 120) = 2^1 \times 3^1 = 2 \times 3 = 6\)

Step 3: LCM = product of greatest powers of all primes.
Greatest power of 2: \(2^3\). Greatest power of 3: \(3^2\). Greatest power of 5: \(5^1\).
\(\text{LCM}(6, 72, 120) = 2^3 \times 3^2 \times 5^1 = 8 \times 9 \times 5 = 360\)
Remark
Notice that \(6 \times 72 \times 120 = 51840\), but \(\text{HCF}(6, 72, 120) \times \text{LCM}(6, 72, 120) = 6 \times 360 = 2160\). So, the product of three numbers is not equal to the product of their HCF and LCM. The property \(\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b\) holds only for two positive integers.
Interactive: Prime Factorisation & HCF/LCM Calculator
Enter two numbers to see their prime factorisations, HCF, and LCM
Enter two numbers and click Compute.
Activity: Exploring the Factor Tree
L3 Apply
Predict: Will you always get the same set of prime factors for a number, no matter how you start the factor tree?
  1. Pick any composite number between 100 and 500, for example 360.
  2. Draw two different factor trees for it. In the first, start by dividing by the smallest prime. In the second, start by dividing by the largest factor you can think of.
  3. List all the prime factors at the leaves of each tree.
  4. Compare the two lists. Are they identical (ignoring order)?

Observation: No matter how you construct the factor tree, the set of prime factors (with multiplicities) is always the same. This illustrates the uniqueness part of the Fundamental Theorem of Arithmetic. For example, \(360 = 2^3 \times 3^2 \times 5\), whether you start dividing by 2 or by 10 or by any other factor.

Competency-Based Questions

Scenario: A school is organising a cultural event. There are 96 students in the dance group and 404 students in the drama group. The organisers want to divide each group into teams of equal size, where each team has only dance students or only drama students. They also want to know when both groups will perform again together if dance performs every 96 minutes and drama every 404 minutes.
Q1. What is the maximum number of students that can be in each team?
L3 Apply
  • (a) 2
  • (b) 4
  • (c) 8
  • (d) 12
Answer: (b) 4. The maximum equal team size is \(\text{HCF}(96, 404) = 4\).
Q2. After how many minutes will both groups perform together again?
L3 Apply
Answer: They will perform together after \(\text{LCM}(96, 404) = 9696\) minutes, which is approximately 6 days 17 hours 36 minutes.
Q3. Analyse why HCF gives us the maximum team size and not LCM.
L4 Analyse
Answer: HCF is the largest number that divides both 96 and 404 exactly. For equal teams, the team size must divide both group sizes with no remainder. Hence HCF (the highest common factor) gives the maximum team size. LCM, on the other hand, is the smallest number divisible by both, which is useful for scheduling (when events coincide again).
Q4. Evaluate: If a third group of 120 students joins, will the relationship \(\text{HCF} \times \text{LCM} = \text{product of numbers}\) still hold for all three groups?
L5 Evaluate
Answer: No. For two numbers, \(\text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b\). But for three numbers, this relationship does NOT hold. \(\text{HCF}(96, 404, 120) = 4\) and \(\text{LCM}(96, 404, 120)\) would need to include all prime factors at their highest powers: \(2^5 \times 3 \times 5 \times 101 = 48480\). Product = \(4 \times 48480 = 193920\), but \(96 \times 404 \times 120 = 4{,}654{,}080\). They are not equal.

Assertion--Reason Questions

Assertion (A): \(4^n\) can never end with the digit 0 for any natural number \(n\).
Reason (R): The prime factorisation of \(4^n\) is \(2^{2n}\), which does not contain 5 as a factor.
(a) Both true, R explains A.
(b) Both true, R does not explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a) -- Both are true. A number ending in 0 must be divisible by 5 (and 2). Since \(4^n = 2^{2n}\) has no factor of 5, it can never end in 0. R correctly explains A.
Assertion (A): \(\text{HCF}(6, 20) \times \text{LCM}(6, 20) = 6 \times 20\).
Reason (R): For any two positive integers \(a\) and \(b\), \(\text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b\).
(a) Both true, R explains A.
(b) Both true, R does not explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a) -- Both are true. \(\text{HCF}(6,20) = 2\), \(\text{LCM}(6,20) = 60\). \(2 \times 60 = 120 = 6 \times 20\). R states the general property that explains A.
Assertion (A): The prime factorisation of 140 is \(2^2 \times 5 \times 7\).
Reason (R): Every composite number has a unique prime factorisation (apart from the order of factors).
(a) Both true, R explains A.
(b) Both true, R does not explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a) -- \(140 = 2 \times 70 = 2 \times 2 \times 35 = 2^2 \times 5 \times 7\). A is true. R is the FTA which guarantees this unique factorisation. R explains A.

Frequently Asked Questions — Real Numbers

What is Fundamental Theorem of Arithmetic in NCERT Class 10 Mathematics?

Fundamental Theorem of Arithmetic is a key concept covered in NCERT Class 10 Mathematics, Chapter 1: Real Numbers. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.

How do I solve problems on Fundamental Theorem of Arithmetic step by step?

To solve problems on Fundamental Theorem of Arithmetic, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 10 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.

What are the most important formulas for Chapter 1: Real Numbers?

The essential formulas of Chapter 1 (Real Numbers) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.

Is Fundamental Theorem of Arithmetic important for the Class 10 board exam?

Fundamental Theorem of Arithmetic is part of the NCERT Class 10 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.

What mistakes should students avoid in Fundamental Theorem of Arithmetic?

Common mistakes in Fundamental Theorem of Arithmetic include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.

Where can I find more NCERT practice questions on Fundamental Theorem of Arithmetic?

End-of-chapter NCERT exercises for Fundamental Theorem of Arithmetic cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 1, and solve at least one previous-year board paper to consolidate your understanding.

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