The Fundamental Theorem of Airthmetic
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 x2 , 253, 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.
7x11x23 = 1771 3x7x11x23 = 5313
2x3x7x11x23 = 10626 23 x 3x73= 8232
22 x3x7x11x23 = 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 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 x2x 2x3x3x5x7 x13 as a product of primes ie, 32760 = 23 x32 x5 x 7 x 13 as a product of powers of primes. Let us try another number , say , 123456789. This can be written as 32 x 3803 x 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 Airthemetic because of its basic crucial importance to the study of integers. Let us now formally state this theorem.
Wednesday, October 21, 2009
Wednesday, October 7, 2009
Real numbers- Example 4
Example 4 : A sweetseller has 420 kaju barfis and 130 badam barfis. She wants to stack them in such a way that each stack has the same number , and they take up the least area of the trya. What is the maximum number of barfis that can be placed in each stack for this purpose ?
Solution : this can be done by trial and error. But to do it systematically , we find HCF (420,130) . Then this number will give the maximum number of barfis in each stack and the number of stacks will then be the least. The area of the tray that is used up will be the least.
Now, let us use Euclid’s algorithm to find their HCF : We have :
420 = 130 x 3 + 30
130 =30x 4 +10
30 = 10 x 3 + 0
So, the HCF of 420 and 130 is 10.
Therefore, the sweetseller can make stacks of 10 for both kinds of barfis.
Solution : this can be done by trial and error. But to do it systematically , we find HCF (420,130) . Then this number will give the maximum number of barfis in each stack and the number of stacks will then be the least. The area of the tray that is used up will be the least.
Now, let us use Euclid’s algorithm to find their HCF : We have :
420 = 130 x 3 + 30
130 =30x 4 +10
30 = 10 x 3 + 0
So, the HCF of 420 and 130 is 10.
Therefore, the sweetseller can make stacks of 10 for both kinds of barfis.
Wednesday, September 23, 2009
Real numbers- Example 2
Example 2 : Show that every positive even integer is of the form 2q, and that every positive odd integer is of the form 2q + 1, where q is some integer.
Solution : Let a be any positive integer and b = 2. Then , by Euclid’s algorithm, a = 2q +r, for some integer q 0, and r =0 or r= 1, because 0r<2. So a= 2q or 2q +1.
If a is of the form 2q, then a ids an even integer . Also, a positive integer can be either even or odd. Therefore, any positive odd integer I sof the form 2q +1.
Solution : Let a be any positive integer and b = 2. Then , by Euclid’s algorithm, a = 2q +r, for some integer q 0, and r =0 or r= 1, because 0r<2. So a= 2q or 2q +1.
If a is of the form 2q, then a ids an even integer . Also, a positive integer can be either even or odd. Therefore, any positive odd integer I sof the form 2q +1.
Thursday, September 10, 2009
Real Numbers -Introduction
1.1 Introduction
In Class IX, you began your exploration of the world of real nubers 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 Airthemetic.
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 recognize 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 properities 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 Airthemetic , 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 Airthemetic . 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 Airthemetic for two main applications. First, we use it prove the irrationality of many of the nubers you studied in Class IX, such as , and . Second , we apply this theorem to explore when exactly the decimal expansion of a rational number , say p/q (q0), is terminating and when it is non-terminating repeating. We do so by looking at the prime factorization of the denominator q of p/q. You will see that the prime factorization of q will completely reveal the nature of the decimal expansion of p/q. So let us begin our exploration.
In Class IX, you began your exploration of the world of real nubers 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 Airthemetic.
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 recognize 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 properities 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 Airthemetic , 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 Airthemetic . 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 Airthemetic for two main applications. First, we use it prove the irrationality of many of the nubers you studied in Class IX, such as , and . Second , we apply this theorem to explore when exactly the decimal expansion of a rational number , say p/q (q0), is terminating and when it is non-terminating repeating. We do so by looking at the prime factorization of the denominator q of p/q. You will see that the prime factorization of q will completely reveal the nature of the decimal expansion of p/q. So let us begin our exploration.
Monday, September 7, 2009
Real numbers - Example 1
Example 1: Use Euclid’s algorithm to find the HCF of 4052 and 12576.
Solution :
Step 1 : Since 12576 > 4052, we apply the division lemma to 12576 and 4052, to get
12576 + 4052 x 3 + 420
Step 2 : Since the remainder 420 0, we apply the division lemma to 4052 and 420, to get
4052 = 420 x9 + 272
Step 3 : We consider the new divisor 420 and the new remainder 272, and apply the division lemma to get
420 = 272 x 1 + 148
We consider the new divisor 272 and the new remainder 148, and apply the division lemma to get
272 = 148 x 1 + 124
We consider the new divisor 148 and the new remainder 124, and apply the division lemma to get
148 = 124 x 1 + 24
We consider the new divisor 124 and the new remainder 24, and apply the division lemma to get
124 = 24 x 5 + 4
We consider the new divisor 24 and the new remainder 4, and apply the division lemma to get
24 = 4 x6 +0
The remainder has now become zero,, so our procedure stops. Since the divisor at this stage is 4, the HCF of 12576 and 4052 is 4.
Notice that 4 = HCF (24,4) = HCF (124,24) = HCF (148, 124) = HCF (272 ,148) = HCF (420, 272) = HCF (4052, 420) = HCF (12576 , 4052).
Euclid’s division algorithem is not only useful for calculating the HCF of very large numbers, but also because it is one of the earliest examples of an algorithm that a computer has been programmed to carry out.
Remarks :
1. Euclid’s division lemma and algorithm are so closely interlinked that people often call former as the division algorithm also.
2. Although Euclid’s Division Algorithm is stated for only positive integers, it can be extended for all integers except zero i.e, b0. However , we shall not discuss this aspect here.
Solution :
Step 1 : Since 12576 > 4052, we apply the division lemma to 12576 and 4052, to get
12576 + 4052 x 3 + 420
Step 2 : Since the remainder 420 0, we apply the division lemma to 4052 and 420, to get
4052 = 420 x9 + 272
Step 3 : We consider the new divisor 420 and the new remainder 272, and apply the division lemma to get
420 = 272 x 1 + 148
We consider the new divisor 272 and the new remainder 148, and apply the division lemma to get
272 = 148 x 1 + 124
We consider the new divisor 148 and the new remainder 124, and apply the division lemma to get
148 = 124 x 1 + 24
We consider the new divisor 124 and the new remainder 24, and apply the division lemma to get
124 = 24 x 5 + 4
We consider the new divisor 24 and the new remainder 4, and apply the division lemma to get
24 = 4 x6 +0
The remainder has now become zero,, so our procedure stops. Since the divisor at this stage is 4, the HCF of 12576 and 4052 is 4.
Notice that 4 = HCF (24,4) = HCF (124,24) = HCF (148, 124) = HCF (272 ,148) = HCF (420, 272) = HCF (4052, 420) = HCF (12576 , 4052).
Euclid’s division algorithem is not only useful for calculating the HCF of very large numbers, but also because it is one of the earliest examples of an algorithm that a computer has been programmed to carry out.
Remarks :
1. Euclid’s division lemma and algorithm are so closely interlinked that people often call former as the division algorithm also.
2. Although Euclid’s Division Algorithm is stated for only positive integers, it can be extended for all integers except zero i.e, b0. However , we shall not discuss this aspect here.
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