LESSON PLANCLASS 10th

SUBJECT MATHS

TIME: 35min.

Name of the topic

REAL NUMBERSSub topic

EUCLID’SDIVISION ALOGRITHIM

Time Management

1. P K Testing 3 minutes.2. Motivation 2 minutes.3. Presentation 12 minutes.4. Student Activity 10 minutes.5. Evaluation & Conclusion 10 minutes.

GENERAL OBJECTIVES

• At the end of the lesson students will enable to:

• 1.Define Real numbers• 2.Define Division Alogrithm

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.

Previous knowledge testing

• Dear students ,what is a rational number?• Expected answer • If 2 a rational number ?

Of course• What will be the L. C. M. of 4 and 5• Expected answer• What would be the H.C.F.of 455 and 42• Students will be unable to answer.•

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.

• Consider the folk puzzle• A trader was moving along a road selling eggs • An 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 .

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 = 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.

• 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.

• 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+35• 42=35x1+7• 35=7x5+0• Here the remainder becomes 0 so we can say

7 is a divisor of 42 and 455.

STUDENTS ACTIVITY

• Group 1. Find HCF(135,225).• Group 2. Find HCF (26,91).• Group 3. Find HCF (336,54)

• Similarly we can show that for any odd positive integer x , can be expressed as6q+1 or 6q+3 or 6q+5Since a=bq+r :0 ≤ r<b

• Taking 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 ≠ 0,2,4

• Hence a can be written in the form of 6q+1, or 6q+3, or 6q+5.

• 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

• We are going to factor tree with which you are all familiar . Let us take some large number , say , 8190.

8190

2 4095

3 1365

3 455

591

7

13

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 .

• 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.

• 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 see • HCF(510,92)×LCM(510,92)=2×3220=510×92

STUDENT ACTIVITY

• Using prime factorisation find:

• Group1 : HCF(12,15,21)

• Group2 : LCM(12,15,21)• Group3: Prove HCF(306,657)=9

Prepared by • 1. Rajesh Kumar Sharma(lect. in Maths.) GSSS Bharari Distt. Bilaspur( H.P.)• 2. Kamal Kishore Sharma (Lect.in Maths.) GSSS Kasauli Distt. Solan (H.P.)• 3. Bal Krishan Saini (Lect.in Maths.) GSSS Heeran Distt. Una (H.P.)• 4. Mukesh Kumar Sharma (TGT N/M) GMS Tialu Distt. Hamirpur (H.P.)• 5.Desh Raj Negi (Lect.in Maths) GSSS Banjar Distt. Kullu (H.P)• 6. Roshan Lal Sharma (Lect.in Maths) GSSS (B) Kullu Distt. Kullu (H.P)

Thanks