Presentation on theme: "LESSON PLAN CLASS 10th SUBJECT MATHS TIME: 35min."— Presentation transcript:
1 LESSON PLAN CLASS 10th SUBJECT MATHS TIME: 35min. Name of the topicREAL NUMBERSSub topicEUCLID’SDIVISION ALOGRITHIM
2 Time Management 1. P K Testing 3 minutes. 2. Motivation 2 minutes. 3. Presentation minutes.4. Student Activity 10 minutes.5. Evaluation & Conclusion 10 minutes.
3 GENERAL OBJECTIVES At the end of the lesson students will enable to: 1.Define Real numbers2.Define Division Alogrithm
4 SPECIFIC OBEJECTIVES 1.Define Rational numbers and Irrational numbers. 2.Use Euclid’s Division Algorithm to find the H.C.F. of two positive integers.3.Define Fundamental Theorem of Arithmetic's.
5 Previous knowledge testing Dear students ,what is a rational number?Expected answerIf 2 a rational number ?Of courseWhat will be the L. C. M. of 4 and 5What would be the H.C.F.of 455 and 42Students will be unable to answer.
6 INTRODUCTION OF THE TOPIC Dear students , as you know that every composite number can be expressed as the product of primes in a unique way , is known as fundamental principal of Arithmatics.
7 Consider the folk puzzle A trader was moving along a road selling eggsAn idler who did not have much work to do , started to get the trader into a wordy duel . This grew into a fight , he pulled his bucket 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 Panchyat 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 bucket cannot accommodate more than 150 eggs .
8 So, how many eggs were there. Let us try and solve the puzzle So, how many eggs were there? Let us try and solve the puzzle. Let the numberof 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 = 6q+ 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 = 2 u + 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.
9 The moment we write down such equations we are using Euclid’s division lemma, Getting back to our puzzle, do you have any idea how you will solve it? Yes!You must look for the multiples of 7 which satisfy all the conditions. By trial and error method. (using the concept of LCM), you will find he had 119 eggs.
10 So as you have seen a=bq+r : 0≤r<b. Now we shall use this method to find HCF of 455 and 42 .455=42x10+3542=35x1+735=7x5+0Here the remainder becomes 0 so we can say 7 is a divisor of 42 and 455.
12 Similarly we can show that for any odd positive integer x , can be expressed as 6q+1 or 6q+3 or 6q+5Since a=bq+r :0 ≤ r<bTaking b=6 we get the remainders 0,1,2,3,4,5.so we can write a=6q, or 6q+1, or 6q+2, or 6q+3, or 6q+4, or 6q+5.Since a is odd therefore remainders may not be even i.e. r ≠ 0,2,4Hence a can be written in the form of 6q+1, or 6q+3, or 6q+5.
13 Now we shall proceed further Take the collection of primes say 2,3,5,7,11 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.Let us list a few 7x11x23=1771, 3x7x11x23=5313, 2³ x3x7³ =8232 and so on
14 We are going to factor tree with which you are all familiar We are going to factor tree with which you are all familiar . Let us take some large number , say , 8190.
16 So we have factorised 8190 as 2×3×3×5×7×13 as product of primes and we have seen every composite number can be written as the product of powers of primes once we have decided that the order will be ascending ,then the way the number is factorised , is unique.The fundamental theorem of artihmatics has many applications ,both with in mathematics and in other fields .
17 Let us take some examples Example; consider whether 6n for any n, can be end with the digit zero.Solution ; if the number 6n ,for any n, will end with the digit 0.then it would be divisible by 5.i. e. , the prime factorisation of 6n would contain the prime 5.this is not possible because 6n =(2×3)n So the prime factorisation of 6n are 2and3. So the fundamental theorem of Arithmatics guarantees that there are no other primes in the factorisation of 6n So ,there is no natural number n for which 6n ends with the digit 0.Now Dear students using prime factorisation method we shall find LCM and HCF of 510 and 92.
18 And LCM(510,92)=22 ×5×17×23 as you have done in your earlier classes . Since 510=2×3×5×17And 92=2×2×23 so HCF(510,92)=2And LCM(510,92)=22 ×5×17×23 as you have done in your earlier classes .Using HCF and LCM we seeHCF(510,92)×LCM(510,92)=2×3220=510×92
19 STUDENT ACTIVITY Using prime factorisation find: Group1 : HCF(12,15,21)Group2 : LCM(12,15,21)Group3: Prove HCF(306,657)=9