STD - X �CHAPTER- 1

REAL NUMBERS

PREVIOUS KNOWLEDGE

NEW TERMS USED

• Algorithm - A series of well defined steps which gives a procedure for solving a type of problem
• Lemma : A proven statement used for proving another statement

CONTENT

EUCLID ‘S DIVISION LEMMA

• Given positive integers a and b , there exist unique integers q and r satisfying
• a = b q + r where 0 ≤ r < b.

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E UCLID’S DIVISION ALGORITHM

• This is based on Euclid’s division lemma. According to this, the HCF of any two positive integers c and d, with c > d is obtained as follows:
• Step 1: Apply the division lemma to c and d. we can find whole numbers q and r such that c = d q + r , 0 ≤ r < d.
• Step 2 : If r = 0, the HCF is d. If r ≠ 0, apply Euclid’s lemma to d and r.
• Step 3 : Continue the process till the remainder is zero. The divisor at this stage will be HCF ( c,d).

FUNDAMENTAL THEOREM OF ARITHMETIC

• Every composite number can be expressed as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.

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HOW TO FIND HCF AND LCM OF NUMBERS

HCF

• 1)Write the prime factorisation of the numbers
• 2)List the common prime factors of the numbers
• 3)Write the smallest power of each common prime factor
• 4)Find the product of these

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LCM

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• 1)Write the prime factorisation of the numbers
• 2)List all the factors involved in the numbers
• 3)Write the greatest power of the prime factors
• 4)Find the product of these

If a and b are two positive integers then�HCF(a,b)X LCM(a,b) =a X b

• If a and b are two co- prime positive integers then
• HCF(a,b) = 1

SHOW THAT √ 2 IS IRRATIONAL.

Let us assume the contrary that√2 is rational.

• Then, √2 = a/b
• where a and b are co-prime integers and b not zero. ……………………… (1)
• Or √2b = a or 2b² = a² It means 2 divides a²
• and hence 2 divides a. ……………………….(2)
• Let, a= 2c for some integer c.
• then, 2b² = (2c)² = 4c² or b² = 2c² It means 2 divides b²
• and hence 2 divides b. ……………………………(3)
• From statements (2) and (3), it is clear that 2 is a common factor of a and b both.

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• But, from statement (1), p and q are co-prime which means that they cannot have any common
• factor other than 1.
• This indicates that our supposition that √2 is rational, is wrong.
• Hence, √2 is irrational.

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CONDITION FOR A RATIONAL NUMBER TO HAVE TERMINATING OR NON-TERMINATING REPEATING DECIMAL EXPANSION.

• Terminating: A rational number p/q, where p and q are co- prime integers will have a terminating decimal expansion if the denominator is of the form 2ⁿ × 5^m, where n and m are non- negative integers.
• Non- Terminating: A rational number p/q, where p and q are co- prime integers will have a non- terminating repeating decimal expansion if the denominator is not of the form 2ⁿ × 5^m, where n and m are non- negative integers.

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