Transcript LESSON PLAN CLASS 10th SUBJECT MATHS TIME:35min.

LESSON PLAN
CLASS 10th
SUBJECT MATHS
TIME:
35min.
Name of the topic
REAL NUMBERS
Sub topic
EUCLID’SDIVISION ALOGRITHIM
Time Management
1.
2.
3.
4.
5.
P K Testing
Motivation
Presentation
Student Activity
Evaluation & Conclusion
3 minutes.
2 minutes.
12 minutes.
10 minutes.
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:
•
•
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•
•
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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 as
6q+1 or 6q+3 or 6q+5
Since 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
4095
2
1365
3
3
455
91
5
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=2×3×5×17
And 92=2×2×23 so HCF(510,92)=2
And 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)