Real Numbers

Hello students, welcome to today's mathematics lesson. I am so happy to see you all here, ready to learn about one of the most fascinating chapters in your Class 10 curriculum – Chapter 1: Real Numbers.

Now, before we begin, let me ask you something. Do you remember studying about real numbers in Class IX? You were introduced to the concept of irrational numbers, you learned that they exist, and you even learned how to locate some irrational numbers on the number line. But I am sure there was one thing that you might have wondered about – how do we actually prove that a number like √2 is irrational? We accepted it, we used it, but we never actually proved it. Well, students, that is exactly what we are going to do in this chapter. We are going to prove the irrationality of numbers like √2, √3, √5, and more. But to do that, we need some powerful tools, and those tools are the Fundamental Theorem of Arithmetic and something called Euclid's division algorithm.

Now, let me tell you something interesting. The name Euclid might sound familiar to you. Yes, that's right – Euclid was an ancient Greek mathematician, and his work forms the basis of much of the geometry we study. But today, we are going to learn about his division algorithm, which is a method for dividing positive integers. And then we will learn about the Fundamental Theorem of Arithmetic, which is all about writing numbers as products of primes. These two concepts might seem simple at first, but they have very deep and significant applications in mathematics. We will use them to prove that certain numbers are irrational, and also to understand when the decimal expansion of a rational number terminates and when it repeats.

So, students, let us begin our exploration of real numbers.

In your earlier classes, you have seen that any natural number can be written as a product of its prime factors. For instance, 2 equals 2, 4 equals 2 times 2, 253 equals 11 times 23, and so on. You have been doing this since your primary school days. Now, let us look at this from the opposite direction. That is, can any natural number be obtained by multiplying prime numbers together? Let us think about this.

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, we can produce infinitely many numbers this way. Let me show you a few examples. If we multiply 7 times 11 times 23, we get 1771. If we multiply 3 times 7 times 11 times 23, we get 5313. If we multiply 2 times 3 times 7 times 11 times 23, we get 10626. If we take 2 cubed times 3 times 7 cubed, we get 8232. And if we take 2 squared times 3 times 7 times 11 times 23, we get 21252. And we can go on and on, multiplying primes in different combinations to get infinitely many numbers.

Now, here is an interesting question. Your collection of primes includes all possible primes – how many primes do you think there are? Are there only a finite number of primes, or are there infinitely many? Well, students, there are infinitely many primes. This is a famous result in mathematics, proved long ago. So, if we combine all these infinitely many primes in all possible ways, we will get an infinite collection of numbers – all the primes themselves and all possible products of primes. The question is – can we produce all the composite numbers this way? Do you think there might be a composite number which is not the product of powers of primes? Before we answer this, let us actually factorise some positive integers, that is, do the opposite of what we have done so far.

We are going to use the factor tree, with which I am sure all of you are familiar. Let us take some large number, say 32760, and factorise it. We can break it down step by step. After factorising, we get 32760 equals 2 times 2 times 2 times 3 times 3 times 5 times 7 times 13. In other words, 32760 equals 2 cubed times 3 squared times 5 times 7 times 13, which is a product of powers of primes. Let us try another number, say 123456789. This can be written as 3 squared times 3803 times 3607. Of course, you would need to verify that 3803 and 3607 are prime numbers. I encourage you to try this out for several other natural numbers yourself.

This leads us to a very important observation. Every composite number can be written as the product of powers of primes. In fact, this statement is true, and it is called the Fundamental Theorem of Arithmetic. It is one of the most important theorems in the study of integers, which is why it is called the "Fundamental" Theorem of Arithmetic. Let me now formally state this theorem for you.

Theorem 1.1, the Fundamental Theorem of Arithmetic, states: Every composite number can be expressed, that is factorised, as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.

Now, let me explain what this means in simple words. The theorem says that every composite number can be broken down into prime numbers. But more importantly, it says that this breakdown is unique. What do I mean by unique? I mean that given any composite number, there is one and only one way to write it as a product of primes, except for the order in which we write the primes. For example, we can write 12 as 2 times 2 times 3, or as 2 times 3 times 2, or as 3 times 2 times 2. But these are essentially the same factorisation, just with the order changed. If we agree to write the primes in ascending order, then the factorisation is completely unique. So, for example, we would always write 12 as 2 squared times 3, and not as 3 times 2 squared.

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.

An equivalent version of this theorem 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 book called Disquisitiones Arithmeticae. 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 lived from 1777 to 1855 and made fundamental contributions to both mathematics and science.

So, students, to summarise what we have learned so far: every composite number can be written as a product of prime numbers, and this way of writing it is unique, except for the order of the primes.

Now, let us see how we actually write a composite number in its prime factorised form. In general, given a composite number x, we factorise it as x equals p1 p2 ... pn, where p1, p2, ..., pn are primes. We usually write them in ascending order, that is, p1 is less than or equal to p2, and so on, up to pn. If we combine the same primes together, we will get powers of primes. For example, as we saw earlier, 32760 equals 2 times 2 times 2 times 3 times 3 times 5 times 7 times 13, which equals 2 cubed times 3 squared 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.

Now, the Fundamental Theorem of Arithmetic has many applications, both within mathematics and in other fields. Let us look at some examples to understand how we use this theorem.

Example 1: Consider the numbers 4 to the power n, where n is a natural number. Check whether there is any value of n for which 4 to the power n ends with the digit zero.

Now, students, think about this. When does a number end with the digit zero? A number ends with the digit zero if and only if it is divisible by 10. And 10 equals 2 times 5. So, for a number to end with zero, it must have both 2 and 5 as its prime factors. Now, let us look at 4 to the power n. 4 equals 2 squared, so 4 to the power n equals 2 to the power 2n. This means the only prime in the factorisation of 4 to the power n is 2. There is no factor of 5 at all. So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorisation of 4 to the power n. Therefore, it is impossible for 4 to the power n to be divisible by 5, and hence impossible for it to end with the digit zero. So, there is no natural number n for which 4 to the power n ends with the digit zero. This is a direct application of the Fundamental Theorem of Arithmetic.

Now, students, you have already learnt how to find the HCF and LCM of two positive integers using the Fundamental Theorem of Arithmetic in your earlier classes, although you might not have realised it at the time! 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.

Let us factorise 6 and 20 into primes. 6 equals 2 times 3, which we can write as 2 to the power 1 times 3 to the power 1. 20 equals 2 times 2 times 5, which is 2 squared times 5 to the power 1.

Now, to find the HCF, we take the product of the smallest power of each common prime factor. The only common prime factor between 6 and 20 is 2. The smallest power of 2 that appears in both factorisations is 2 to the power 1. So, HCF of 6 and 20 equals 2.

To find the LCM, we take the product of the greatest power of each prime factor that appears in either number. The prime factors involved are 2, 3, and 5. The greatest power of 2 is 2 squared, the greatest power of 3 is 3 to the power 1, and the greatest power of 5 is 5 to the power 1. So, LCM of 6 and 20 equals 2 squared times 3 times 5, which equals 4 times 3 times 5, equals 60.

Now, students, did you notice something interesting? HCF of 6 and 20 is 2, LCM is 60, and if we multiply HCF and LCM, we get 2 times 60 equals 120, which is exactly 6 times 20. This is not a coincidence. In fact, for any two positive integers a and b, HCF of a and b multiplied by LCM of a and b equals a times b. We can use this result to find the LCM of two positive integers if we have already found the HCF, or vice versa.

Let us look at another example to reinforce this concept.

Example 3: Find the HCF of 96 and 404 by the prime factorisation method. Hence, find their LCM.

First, let us factorise 96 and 404. 96 equals 2 to the power 5 times 3. 404 equals 2 squared times 101.

Now, the common prime factor is 2. The smallest power of 2 in both factorisations is 2 squared. So, HCF of 96 and 404 equals 2 squared, which is 4.

Now, to find the LCM, we can use the formula: LCM equals product of the two numbers divided by their HCF. So, LCM of 96 and 404 equals 96 times 404 divided by 4. Let me calculate that: 96 times 404 is 96 times 400 plus 96 times 4, which is 38400 plus 384, equals 38784. Dividing by 4 gives us 9696. So, LCM is 9696.

Now, let us try an example with three numbers.

Example 4: Find the HCF and LCM of 6, 72 and 120, using the prime factorisation method.

Let us factorise each number. 6 equals 2 times 3. 72 equals 2 cubed times 3 squared. 120 equals 2 cubed times 3 times 5.

Now, to find the HCF of three numbers, we look at the common prime factors among all three numbers. The common prime factors are 2 and 3. For 2, the smallest power that appears in all three numbers is 2 to the power 1. For 3, the smallest power is 3 to the power 1. So, HCF equals 2 times 3, which is 6.

For the LCM, we take the greatest power of each prime factor that appears in any of the numbers. The prime factors involved are 2, 3, and 5. The greatest power of 2 is 2 cubed, the greatest power of 3 is 3 squared, and the greatest power of 5 is 5 to the power 1. So, LCM equals 2 cubed times 3 squared times 5, which is 8 times 9 times 5, equals 72 times 5, equals 360.

Now, students, I want you to notice something important here. For two numbers, we have the relationship that HCF times LCM equals product of the two numbers. But for three numbers, this does not hold. Let me check: 6 times 72 times 120 equals 6 times 72 is 432, times 120 is 51840. But HCF times LCM is 6 times 360, which is 2160. These are not equal. So, the product of three numbers is not equal to the product of their HCF and LCM. This is an important point to remember.

Now, students, let us move on to the next section of the chapter, where we revisit irrational numbers.

In Class IX, you were introduced to irrational numbers and many of their properties. You studied about their existence and how the rationals and the irrationals together make up the real numbers. You even studied how to locate irrationals on the number line. However, you did not prove that they were irrationals. In this section, we will prove that √2, √3, √5 and, in general, √p is irrational, where p is a prime. One of the theorems we use in our proof is the Fundamental Theorem of Arithmetic, which we just learned.

First, let me remind you what an irrational number is. A number s is called irrational if it cannot be written in the form p divided by q, where p and q are integers and q is not equal to 0. Some examples of irrational numbers, with which you are already familiar, are: √2, √3, √15, π, negative √2 divided by √3, and 0.10110111011110... and so on.

Now, before we prove that √2 is irrational, we need to prove a theorem first. This theorem is crucial for our proof, and its proof is based on the Fundamental Theorem of Arithmetic.

Theorem 1.2 states: Let p be a prime number. If p divides a squared, then p divides a, where a is a positive integer.

Let me prove this theorem for you. Let the prime factorisation of a be as follows: a equals p1 p2 ... pn, where p1, p2, ..., pn are primes, not necessarily distinct. Therefore, a squared equals p1 squared times p2 squared times ... times pn squared. Now, we are given that p divides a squared. Therefore, from the Fundamental Theorem of Arithmetic, it follows that p is one of the prime factors of a squared. However, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realise that the only prime factors of a squared are p1, p2, ..., pn. So p is one of p1, p2, ..., pn. Now, since a equals p1 p2 ... pn, p divides a. This completes the proof.

Now, students, we are ready to prove that √2 is irrational. The proof is based on a technique called proof by contradiction. In this technique, we assume the opposite of what we want to prove, and then we show that this assumption leads to a contradiction. If we get a contradiction, then our original assumption must be false, and therefore, what we wanted to prove must be true.

Theorem 1.3: √2 is irrational.

Proof: Let us assume, to the contrary, that √2 is rational. So, we can find integers r and s, where s is not equal to 0, such that √2 equals r divided by s. Suppose r and s have a common factor other than 1. Then, we divide by the common factor to get √2 equals a divided by b, where a and b are coprime, that is, they have no common factors other than 1.

So, b√2 equals a. Squaring on both sides and rearranging, we get 2b squared equals a squared. Therefore, 2 divides a squared. Now, by Theorem 1.2, it follows that 2 divides a. So, we can write a equals 2c for some integer c.

Substituting for a, we get 2b squared equals 4c squared, that is, b squared equals 2c squared. This means that 2 divides b squared, and so 2 divides b (again using Theorem 1.2 with p equals 2). Therefore, a and b have at least 2 as a common factor. But this contradicts the fact that a and b have no common factors other than 1. This contradiction has arisen because of our incorrect assumption that √2 is rational. So, we conclude that √2 is irrational.

Now, students, let me explain this proof step by step. We start by assuming that √2 is rational, which means it can be written as a fraction in simplest form, with no common factors between the numerator and denominator. Then we square both sides to get 2b squared equals a squared. This tells us that a squared is even, which means a must be even (because if a were odd, a squared would also be odd). So, we can write a as 2c. Substituting this back, we get 2b squared equals 4c squared, which simplifies to b squared equals 2c squared. This means b squared is even, so b must be even. But if both a and b are even, then they have a common factor of 2, which contradicts our assumption that they are coprime. This contradiction proves that our initial assumption was wrong, and therefore √2 must be irrational.

Now, let us look at another example to prove that √3 is irrational.

Example 5: Prove that √3 is irrational.

Solution: Let us assume, to the contrary, that √3 is rational. That is, we can find integers a and b, where b is not equal to 0, such that √3 equals a divided by b. Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.

So, b√3 equals a. Squaring on both sides, and rearranging, we get 3b squared equals a squared. Therefore, a squared is divisible by 3, and by Theorem 1.2, it follows that a is also divisible by 3. So, we can write a equals 3c for some integer c.

Substituting for a, we get 3b squared equals 9c squared, that is, b squared equals 3c squared. This means that b squared is divisible by 3, and so b is also divisible by 3 (using Theorem 1.2 with p equals 3). Therefore, a and b have at least 3 as a common factor. But this contradicts the fact that a and b are coprime. This contradiction has arisen because of our incorrect assumption that √3 is rational. So, we conclude that √3 is irrational.

Now, students, you might be wondering why we keep using the same proof technique. Actually, the proof for √5 would be very similar – we would assume √5 is rational, square both sides, get 5b squared equals a squared, conclude that 5 divides a, write a as 5c, substitute back, get b squared equals 5c squared, conclude that 5 divides b, and then get a contradiction because a and b would have a common factor of 5. This same method can be extended to prove that √p is irrational for any prime p.

Now, in Class IX, we mentioned some properties about rational and irrational numbers. Specifically, we mentioned that the sum or difference of a rational and an irrational number is irrational, and the product and quotient of a non-zero rational and irrational number is irrational. Let us prove some particular cases here.

Example 6: Show that 5 minus √3 is irrational.

Solution: Let us assume, to the contrary, that 5 minus √3 is rational. That is, we can find coprime a and b, where b is not equal to 0, such that 5 minus √3 equals a divided by b. Therefore, 5 minus a divided by b equals √3. Rearranging this equation, we get √3 equals 5 minus a divided by b, which equals 5b minus a, all divided by b. Since a and b are integers, 5b minus a divided by b is rational. So, √3 is rational. But this contradicts the fact that √3 is irrational. This contradiction has arisen because of our incorrect assumption that 5 minus √3 is rational. So, we conclude that 5 minus √3 is irrational.

Let me explain this proof. We want to prove that 5 minus √3 is irrational. We use the method of contradiction. We assume that 5 minus √3 is rational. If it is rational, then we can write it as a fraction a/b in simplest form. Then we rearrange the equation to isolate √3. We get √3 equals 5 minus a/b, which is a rational number (since the difference of two rational numbers is rational). But this would mean √3 is rational, which we know is false. Therefore, our assumption must be wrong, and 5 minus √3 must be irrational.

Example 7: Show that 3√2 is irrational.

Solution: Let us assume, to the contrary, that 3√2 is rational. That is, we can find coprime a and b, where b is not equal to 0, such that 3√2 equals a divided by b. Rearranging, we get √2 equals a divided by 3b. Since 3, a and b are integers, a divided by 3b is rational, and so √2 is rational. But this contradicts the fact that √2 is irrational. So, we conclude that 3√2 is irrational.

Now, students, notice the pattern in these proofs. In both cases, we assumed the opposite of what we wanted to prove, and then we showed that this assumption would lead to a situation where a known irrational number would become rational, which is impossible. This is a very powerful proof technique.

Now, let me give you a brief summary of everything we have learned in this chapter.

In this chapter, you have studied the following important points:

First, we learned about the Fundamental Theorem of Arithmetic, which states that every composite number can be expressed, that is factorised, as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur. This is a very fundamental result that has many applications.

Second, we learned that if p is a prime and p divides a squared, then p divides a, where a is a positive integer. This theorem is crucial for proving the irrationality of square roots.

Third, we learned how to prove that √2 and √3 are irrational using the method of proof by contradiction. We also learned that the sum or difference of a rational and an irrational number is irrational, and the product of a non-zero rational and irrational number is irrational.

We also reviewed how to find the HCF and LCM of numbers using the prime factorisation method, and we verified the relationship that for any two positive integers a and b, HCF of a and b times LCM of a and b equals a times b.

Students, this is a very important chapter because it lays the foundation for understanding the real number system. The concepts you have learned here, especially the Fundamental Theorem of Arithmetic and the method of proving irrationality, will be useful in many other areas of mathematics. Make sure you understand these concepts thoroughly and practice as many problems as you can.

That brings us to the end of today's lesson. Thank you for your attention, and I will see you in the next lesson.