Download Algorithm is a series of well defined steps which gives a procedure

NOTES FOR REAL NUMBERS
Basic Concepts
• Algorithm : Algorithm is a series of well defined steps which gives a procedure for solving a
type of problem.
• Lemma : A lemma is a proven statement used for proving another statement.
• Euclid’s Division Lemma : Given positive integers a and b there exists unique integers q
and r statisfying a = bq + r, 0  r  b.
• Euclid’s Division Algorithm : It is a technique to compute the Highest Common Factor
(HCF) of two given positive integers.
Euclid’s Division algorithm to obtain the HCF of two positive integers say m and n with m  n
is :
• Step I : We find whole numbers, q and r such that m = nq + r, 0  r  n
• Step II : If r = 0 then n is the HCF of m and n. If r  0 apply the division lemma to n and r.
• Step III : Continue this process till the remainder is zero. The divisor at this stage will be the
required HCF.
Note : Let a and b be positive integers. If a = bq + r, 0  r  b, then HCF (a, b) = HCF (b, r)
1.
Sol.
Use Euclid’s Division algorithm to find the HCF of 504 and 980.
Start with the larger integer, that is 980. Using Euclid’s division algorithm, we get
980 = 504 × 1 + 476
Now considering the divisor 504 and the remainder 476, and applying the division algorithm, we
get
504 = 476 × 1 + 28
Now considering the divisor 476 and remainder 28, and applying the division algorithm, we get
476 = 28 × 17 + 0
Notice that the remainder has become zero and we cannot proceed further. Hence the HCF of
504 and 980 is 28.
Note : 28 = HCF (476, 28) = HCF (504, 476)
= HCF (980, 504)
2.
Show that any odd positive integer is of the form 6q + 1 or 6q + 3 or 6q + 5, where q is some
integer.
Let a is a odd positive integer and apply division algorithm with a,
a = bq + r
Here b = 6, \ 0  r  6
The possible remainders are 0, 1, 2, 3, 4, 5
i.e., a can be 6q, 6q + 1, 6q + 2, 6q + 3,
6q + 4, 6q + 5, where q is the quotient.
Since a is odd, neglect 6q, 6q + 2, 6q + 4 as they are divisible by 2.
So any odd positive integer is of the form 6q + 1, 6q + 3 or 6q + 5.
Sol.
Fundamental Theorem of Arithmetic
Every composite number can be expressed as product of primes and this decomposition is unique,
apart from the order in which prime factors occur.
Let p be a prime number and a be a positive integer. If p divides a2, then p divides a.
For two positive numbers a and b.
HCF × LCM = a × b.
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3.
Sol.
Decompose 32760 into prime factors.
32760 = 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13
= 23 × 32 × 5 × 7 × 13
Irrational Number
A number is said to be irrational if it cannot be written in the form
p
, where p and q are integers
q
and q  0.
5.
Prove that
Sol.
Let
2
2 is irrational.
p
where p and q are coprime integers.
q
q 2 =p
 q2.2 = p2
2 divides p2 as p2 is even. If p2 is even the p should also be even.
Let p = 2c for some integer, then equation (i) becomes
 q2.2 = 4c2
 q2 = 2c2
 q2 is even
 q is also even.
This contradicts our assumption that p and q are coprime as they both have a common factor 2.
Hence 2 is irrational.
•
For any rational number
p
with terminating decimal representation, the prime factorisation of q
q
is of the form 2n 5m where n and m are non-negative integers.
•
For any rational number
p
where prime factorisation of q is of the form 2n 5m, where n and m
q
are non-negative, the decimal representation is terminating.
•
For any rational number
p
where prime factorisation of q is not of the form 2n 5m, where n and
q
m are non-negative integers, the decimal representation is non-terminating and repeating.
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