real numbers(class X)

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Real Numbers Presented by- Kaushlesh Sinha

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The numbers along a continuous line that can be located (on the number line) are real numbers. Real numbers consist of all the rational and irrational numbers. The real number system has many subsets: Natural Numbers Whole Numbers Integers Rational Numbers Irrational numbers Real Numbers

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Natural Numbers Natural numbers are the set of counting numbers which starts from 1 . They are denoted by N Example : {1, 2, 3,…} Whole Numbers Whole numbers are the set of numbers that include 0 plus the set of natural numbers. Example : {0, 1, 2, 3, 4, 5,…}

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An integer is a whole number (not a fractional number) that can be positive, negative, or zero. It is denoted by Z . Example : Z = {..., -3, -2, -1, 0, 1, 2, 3, ...} Integers

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Rational Numbers Rational numbers are any numbers that can be expressed in the form of p /q , where p and q are integers, and q ≠ 0. They can always be expressed by using terminating decimals or repeating decimals. They are denoted by Q. Example : 2/3, 6/7,1

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Terminating Decimals Terminating decimals are decimals that contain a finite number of digits. Examples: 36.8 0.125 4.5 Repeating Decimals Repeating decimals are decimals that contain a infinite number of digits. Examples: 0.333… 7.689689… Non Terminating Decimals While expressing a fraction into a decimal by the division method, if the division process continues indefinitely, and zero remainder is never obtained then such a decimal is called Non-Terminating non-repeating Decimal . A non-terminating non-recurring. decimal is a decimal never repeats and never terminates. Example : 0.076923....,  0.05882352..... ,  0.01001000100001.....

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Euclid's Division Lemma Euclid's division lemma states that : “ For any two positive integers a and b, there exist integers q and r such that a= bq+r , 0≤ r< b ”. Example : For a= 15,b=3 it is observed that 15=3(5)+0 where q=5 and r=0 Lemma : A lemma is a proven statement used for proving another statement

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Euclid Division Algorithm Algorithm : An algorithm is a series of well defined steps which gives a procedure for solving a type of problem. Euclid division algorithm can be used to find the HCF of two numbers. It can also be used to find some common properties of numbers . To obtain the HCF of two positive integers,say c and d, with c>d , we have to follow the steps below: STEP 1 : Apply Euclid division lemma, to c and d. So, we find whole numbers, q and r such that c= dq+r STEP 2 : If r=0,d is the HCF of c and d. If r does not equal to 0 , apply the division lemma to d and r. STEP 3 : Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.

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Question : Use Euclid’s algorithm to find the HCF of 455 and 42 ? Solution : Since 455>42, we apply division lemma to 455 and 42 , to get 455= 42 × 10+35 Remainder is not zero therefore we apply lemma to 42 and 35, 42=35 ×1+7 Again remainder is not zero therefore we apply lemma to 35 and 7 35=7 ×5+0 The remainder has become zero , and we cannot proceed any further ,therefore the HCF of 455 and 42 is the divisor at this stage , i.e. 7

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The Fundamental Theorem of Arithmetic Every composite number can be expressed as a product of primes . This representation is called prime factorization of the number. This factorization is unique, apart from the order in which the prime factors occur.

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The HCF of two numbers is equal to the product of the terms containing the least powers of common prime factors of the two numbers. Highest Common Factor(HCF) : The LCM of two numbers is equal to the product of the terms containing the greatest powers of all prime factors of the two numbers. Lowest Common Multiple(LCM) :

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Relationship between HCF and LCM Example : if a=3 and b=6 HCF(3,6) × LCM(3,6) = 3× 6 3 × 6 = 18 18 = 18 Hence verified…….. For any two integers a and b HCF( a,b ) × LCM( a,b ) = a ×b

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Question : Given that HCF(306,657)=9, find LCM(306,657) S olution : We know that the product of the HCF and the LCM of two numbers is equal to the product of the given numbers . HCF(306,657) × LCM(306,657) = 306 × 657 9 × LCM(306,657) = 306 × 657 LCM(306,657) = (306 × 657)/9 LCM =22338

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Irrational Numbers Irrational numbers are any numbers that cannot be expressed as a/b . They are expressed as non-terminating, non-repeating decimals ; decimals that go on forever without repeating a pattern. Examples of irrational numbers: 0.34334333433334… 45.86745893… Pi

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Things to Remember . Let p be a prime number. If p divides a 2 , then p also divides a , where a is a positive integer. . When prime factorisation of q is of the form 2 m 5 n . Then x( any rational number with denominator q) has a decimal expansion which terminates and when q is not of the form 2 m 5 n , then x has a decimal expansion which is non-terminating repeating or recurring.

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Proof: √2 is irrational. Proof- Assume that √2 is rational. ⇒ √2 = p/q ( p,q ∈ Z, q≠0) ⇒ √2= a/b [ HCF( a,b )=1] Squaring and cross multiplying ⇒ 2b²=a² ⇒ 2|a² ⇒2|a ( i ) ⇒ a²=2c ( c ∈ Z) ⇒ 2b²=(2c)² ⇒ 2b²=4c² ⇒ b²=2c² ⇒ 2|b² ⇒2|b (ii) From ( i ) & (ii) 2|a & 2|b ⇒ HCF( a,b )=2 This a contradiction to our assumption that HCF( a,b )=1. This is due to our assumption that √2 is rational. Hence, √2 is irrational . Similarly it can be proved that √3, √5, etc. are irrationals.

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. Does 17/8 have a terminating decimal expansion? Sol : We have 17/8 =17/2 3 ×5 0 So , the denominator 8 of 17/8 is of the form 2 m ×5 n therefore it has a terminating decimal expansion. . Does 29/343 have a terminating decimal expansion? Sol : We have 29/243 =29/3 5 Since, 243 is not of the form 2 m ×5 n therefore it has a non terminating decimal expansion.

BIBLIOGRAPHY:

BIBLIOGRAPHY NCERT text book for Mathematics class X en.wikipedia.org www.mathisfun.com

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Presented by- Kaushlesh K Sinha Class- X Roll No.- 1026 Subject- Mathematics Guided by- Chandrashekhar Sir THANK YOU

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