- Flashback of IX real Number's
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- Euclid's Division Lemma
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- Proof of Euclid's Division Lemma
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- HCF (Highest common factor)
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- What is Prime Numbers
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- What is Composite Numbers
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- Fundamental Theorom of Arithmetic
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- HCF and LCM by prime factorization method
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- Irrational Numbers
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- How to prove the irrational numbers or Rational numbers

- Real number problem and Solutions
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- Real number Worksheet
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- Real number problems
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- Real number Important questions

Given below are the

a) Concepts questions

b) Calculation problems

c) Multiple choice questions

d) Long answer questions

e) Fill in the blank's

- 7/25
- 3/7
- 29/343
- 6/15
- 77/210
- 11/67
- 15/27
- 11/6
- 343445/140

Those rational number which can be expressed in form x/2

Terminating decimal: (a), (d)

Non terminating repeating decimal: (b), (c), (e), (f), (g).(h) ,(i)

a) 867 and 225

b) 616 and 32

a)

Using Euclid theorem

867=225X3 +192

225=192X1 +33

192=33X5+ 27

33=27X1+6

27=6X4+3

6=3X2+0

So solution is 3

b) 8

a) 4 and 5

b) 1/2 and 1/3

a) 1.232323….

b) 1.25

c) 3.67777777

a) Prove that 2+√3 is a irrational number

b) Prove that 3√3 a irrational number

a) Let’s take this as rational number

$\frac{a}{b}=2+\sqrt{3}$

Or

$\frac{a-2b}{b}=\sqrt{3}$

Since a rational number can’t be equal to irrational number, our assumption is wrong

b) Let’s take this as rational number

q=3√3

q/3=√3

Since a rational number can’t be equal to irrational number, our assumption is wrong

a) Number of the form 2n +1 where n is any positive integer are always odd number

b) Product of two prime number is always equal to their LCM

c) √3X √12 is a irrational number

d) Every integer is a rational number

e) The HCF of two prime number is always 1

f) There are infinite integers between two integers

g) There are finite rational number between 2 and 3

h) √3 Can be expressed in the form √3/1,so it is a rational number

i) The number 6

j) Any positive odd integer is of the form 6m+1 or 6m+3 or 6m +5 where q is some integer

- True
- True
- False, as it is written as 6
- True ,as any integer can be expressed in the form p/q
- True
- False,There are finite integer between two integers
- False
- False
- False
- True

a) 25

b) 9

c) 27

d) 54

LCM X HCF=aXb

a) Non terminating repeating

b) Terminating

c) Non terminating non repeating

d) None of these

a) LCM (a, b) =aXb

b)HCF (a, b)= aXb

c) a=br

d) None of these

a) 2 or 5 only

b) 2 or 3 only

c)3 or 5 only

d) 3 or 7 only

a) x is a irrational number

b) y is a rational number

c) z is rational number

d) All of the above

a) 27,81

b) 120 ,144

c) 29029 ,580

27= 3X3X3

81=3X3X3X3

HCF=27

LCM=81

b)

120=2X2X3X2X5

144=2X2X3X2x2X3

HCF=2

LCM=720

c)

29029=29X13X11X7

580=29X5X4

HCF=29

LCM=29X13X11x7X4X5=580580

p

(q-p)(q+p)=16

So we have

Case 1

q-p=8 and q+p=2 which gives p=3

Case 2

q-p=4 and q+p=4 which gives p=0

Case 3

q-p=2 and q+p=8 which gives p=3

So the answer is 3 only

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