In Class IX, you began your exploration of 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 in Sections 1.2 and 1.3, namely the 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. Many of you probably recognise this as the usual long division process. Although this result is quite easy to state and understand, it has many applications related to the divisibility properties of integers. We touch upon a few of them, and use it mainly to compute the HCF of two positive integers. The Fundamental Theorem of Arithmetic, on the other hand, has to do something 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. Again, while it is a result that is easy to state and understand, it has some very deep and significant applications in the field of mathematics. We use the Fundamental Theorem of Arithmetic for two main applications. First, we use it to prove the irrationality of many of the numbers you studied in Class IX, such as 2 , 3 and 5 . Second, we apply this theorem to explore when exactly the decimal p expansion of a rational number, say (q ≠ 0) , is terminating and when it is nonq terminating repeating. We do so by looking at the prime factorisation of the denominator q of p . You will see that the prime factorisation of q will completely reveal the nature q of the decimal expansion of p . q So let us begin our exploration.
1.2 Euclid’s Division Lemma
Consider the following folk puzzle*. A trader was moving along a road selling eggs. An idler who didn’t have much work to do, started to get the trader into a wordy duel. This grew into a fight, he pulled the basket with eggs and dashed it on the floor. The eggs broke. The trader requested the Panchayat to ask the idler to pay for the broken eggs. The Panchayat asked the trader how many eggs were broken. He gave the following response: If counted in pairs, one will remain; If counted in threes, two will remain; If counted in fours, three will remain; If counted in fives, four will remain; If counted in sixes, five will remain; If counted in sevens, nothing will remain; My basket cannot accomodate more than 150 eggs. So, how many eggs were there? Let us try and solve the puzzle. Let the number of eggs be a. Then working backwards, we see that a is less than or equal to 150: If counted in sevens, nothing will remain, which translates to a = 7p + 0, for some natural number p. If counted in sixes, a = 6 q + 5, for some natural number q. If counted in fives, four will remain. It translates to a = 5w + 4, for some natural number w. If counted in fours, three will remain. It translates to a = 4s + 3, for some natural number s. If counted in threes, two will remain. It translates to a = 3t + 2, for some natural number t. If counted in pairs, one will remain. It translates to a = 2u + 1, for some natural number u. That is, in each case, we have a and a positive integer b (in our example, b takes values 7, 6, 5, 4, 3 and 2, respectively) which divides a and leaves a remainder r (in our case, r is 0, 5, 4, 3, 2 and 1, respectively), that is smaller than b. The * This is modified form of a puzzle given in ‘Numeracy Counts!’ by A. Rampal, and others.
moment we write down such equations we are using Euclid’s division lemma, which is given in Theorem 1.1. Getting back to our puzzle, do you have any idea how you will solve it? Yes! You...