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real number concept map

Real Number

Irrational number

Rational number

Terminating

X = p/q and q=0 (q is in the form of 2 and 5 Wherem n are non negative integers , then it is terminating decimal)
3/22*5 = 0.15

Non terminating

x = p/qand q*0(a is not in theform of2*5*where m, n arenon-negativeintegers, thenit isnon- terminatingdecimal)e 910 - 3.3

Euclid divisionLemma

Given positive integers aand b there exist uniqueintegers q and r satisfyinga = ba = r Oar

Euclid divisionalgorithm

If 'a' and "b'are positiveintegers such that a = bq + P,then every common divisorof 'a and "b' is a commondivisor of "b' and 'r', andvice-versa.e.g. HCF of 420 & 48420 = 48 × 8 + 3648 = 36 x1 + 1236 = 12 × 3 + 0- H.C.F, of 420 & 48 is 12

Fundamental theormof arithmetic

Every composite numbercan be expressed orfactorized as a product ofprimeseg. 48 = 2 x 2* 2 * 2.420 = 2 * 2 y 3 * 5*7

Application

e.g. Check whether 6* can end with the digit O for any naturalnumber 'n'I Sol. Any positive integer ending with the digit zero is divisibleby 5 and so its prime factorization must contain the prime 5.6 = (2 x 3) =2 x 3The prime in the factorization of G' is 2 and 35 does not occur in the prime factorization of 6* for any n.6 does not end with the digit zero for any natural number n

HCF

48 = 2 x 2 x 2 x 2 x 3420 = 2*2 x 3 x5 x7HCF 2*2*3

HCF. (48, 420) × L.C.M. (48,420)= 48> 420(This is true for 2 numbers only)

LCM

eg, 48 = 2 x 2 x 2 x 2 x 3420 = 2 x2 x3 x 5 x7LCM = 2 y 2x 2x 2 x 3 x 5 x 7LCM 1680

Some Important Results1. Let p' be a prime number and a be a positiveinteger. If 'p' divides a*, then 'p' divides a2. HCF × LCM = Prosuct of two numbers.3. LCM is always divisible by HCF

Real number concept map

95

real number concept map

Real Number

Irrational number

Rational number

Terminating

Non terminating

Euclid division

Lemma

Euclid division

algorithm

Fundamental theorm

of arithmetic

X = p/q and q=0 (q is

in the form of 2 and 5 Where

m n are non negative integers ,

then it is terminating decimal)

3/22*5 = 0.15

x = p/q

and q*0

(a is not in the

form of2*5*

where m, n are

non-negative

integers, then

it is

non- terminating

decimal)

e 9

10 - 3.3

Given positive integers a

and b there exist unique

integers q and r satisfying

a = ba = r Oar

If 'a' and "b'are positive

integers such that a = bq + P,

then every common divisor

of 'a and "b' is a common

divisor of "b' and 'r', and

vice-versa.

e.g. HCF of 420 & 48

420 = 48 × 8 + 36

48 = 36 x1 + 12

36 = 12 × 3 + 0

- H.C.F, of 420 & 48 is 12

Every composite number

can be expressed or

factorized as a product of

primes

eg. 48 = 2 x 2* 2 * 2.

420 = 2 * 2 y 3 * 5*7

Application

e.g. Check whether 6* can end with the digit O for any natural

number 'n'

I Sol. Any positive integer ending with the digit zero is divisible

by 5 and so its prime factorization must contain the prime 5.

6 = (2 x 3) =2 x 3

The prime in the factorization of G' is 2 and 3

5 does not occur in the prime factorization of 6* for any n.

6 does not end with the digit zero for any natural number n

HCF

HCF. (48, 420) × L.C.M. (48,420)

= 48

> 420

(This is true for 2 numbers only)

LCM

48 = 2 x 2 x 2 x 2 x 3

420 = 2*2 x 3 x5 x7

HCF 2*2*3

eg, 48 = 2 x 2 x 2 x 2 x 3

420 = 2 x2 x3 x 5 x7

LCM = 2 y 2x 2

x 2 x 3 x 5 x 7

LCM 1680

Some Important Results

1. Let p' be a prime number and a be a positive

integer. If 'p' divides a*, then 'p' divides a

2. HCF × LCM = Prosuct of two numbers.

3. LCM is always divisible by HCF

real number concept map

Real Number

Irrational number

Rational number

Terminating

X = p/q and q=0 (q is in the form of 2 and 5 Wherem n are non negative integers , then it is terminating decimal)
3/22*5 = 0.15

Non terminating

x = p/qand q*0(a is not in theform of2*5*where m, n arenon-negativeintegers, thenit isnon- terminatingdecimal)e 910 - 3.3

Euclid divisionLemma

Given positive integers aand b there exist uniqueintegers q and r satisfyinga = ba = r Oar

Euclid divisionalgorithm

If 'a' and "b'are positiveintegers such that a = bq + P,then every common divisorof 'a and "b' is a commondivisor of "b' and 'r', andvice-versa.e.g. HCF of 420 & 48420 = 48 × 8 + 3648 = 36 x1 + 1236 = 12 × 3 + 0- H.C.F, of 420 & 48 is 12

Fundamental theormof arithmetic

Every composite numbercan be expressed orfactorized as a product ofprimeseg. 48 = 2 x 2* 2 * 2.420 = 2 * 2 y 3 * 5*7

Application

e.g. Check whether 6* can end with the digit O for any naturalnumber 'n'I Sol. Any positive integer ending with the digit zero is divisibleby 5 and so its prime factorization must contain the prime 5.6 = (2 x 3) =2 x 3The prime in the factorization of G' is 2 and 35 does not occur in the prime factorization of 6* for any n.6 does not end with the digit zero for any natural number n

HCF

48 = 2 x 2 x 2 x 2 x 3420 = 2*2 x 3 x5 x7HCF 2*2*3

HCF. (48, 420) × L.C.M. (48,420)= 48> 420(This is true for 2 numbers only)

LCM

eg, 48 = 2 x 2 x 2 x 2 x 3420 = 2 x2 x3 x 5 x7LCM = 2 y 2x 2x 2 x 3 x 5 x 7LCM 1680

Some Important Results1. Let p' be a prime number and a be a positiveinteger. If 'p' divides a*, then 'p' divides a2. HCF × LCM = Prosuct of two numbers.3. LCM is always divisible by HCF

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