REAL NUMBERS

Sub. Topics:- 1. Euclid’s Division Lemma.

2. Fundamental Theorem of Arithmetic.

3. Revisiting Irrational Numbers .

4. Decimal Expansion of Rational numbers.

Euclid’s Division Lemma

If ‘a’ and ‘b’ be two given positive integers, then there exist unique integers q and r such that a=bq+r, o≤ 𝑟 < 𝑏. It is known as called, Euclid’s Division lemma

FIND THE HCF OF THE FOLLOWING BY USING EUCLID’SDIVISION LEMMA

Home work ( Date:- 03/04/20)

1)135 and 225 6) 650 and 1170

2)24,36 and 40 7) 196 and 38220

3)120,224,and 256 8) 867 and 255

4)274170 and 17017 9) 321 and 396

5)225,1309,1326 10) 455 and 42

1) Find the greatest number which divides 230, 1314 and

1331 and leaving remainder 5 in each case.

2) Find the greatest number which divides 121,226 and

259 and leaves remainder 1,2 and 3 respectively.

3) Find the greatest number which divides 90,115 and

140 and leaves remainder 1,2 and 3 respectively.

APPLICATION BASED ON E.D.L1) Show that any positive odd integer is of the from 6q+1 or 6q+3 or 6q+5, where q is some integer.

Sol:- Let us assume that ‘a’ be any positive odd integer. We apply the division algorithm with ‘a’ and b=6.i.e a = 6q+r , where 0≤ 𝑟 < 6, i.e r=0,1,2,3,4,5Put r=0 than a = 6q+0 ⇒ a = 6q

r=1 than a = 6q+1 r=2 than a = 6q+2 r=3 than a = 6q+3 r=4 than a = 6q+4 r=5 than a = 6q+5

∴ a = 6q, 6q+1, 6q+2, 6q+3, 6q+4, 6q+5 are positive integers. ∵ a is positive odd integer, a cant not be 6q,6q+2 and 6q+4 since these are divisible by 2 ∴ any odd integer is of the form 6q+1, 6q+3 or 6q+5

Home Work ( Date:- 04/04/20)

1) Show that any positive integer is of the form

3q or 3q+1 or 3q+2, for some integer q.

2) Show that any positive odd integer is of the

form 4q+1 or 4q+3, where q is some integer.

Use Eculid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m+1 for

some integer m• Sol:- Let us assume that ‘a’ be any

positive integer. • We apply the division algorithm with

‘a’ and b=3.• i.e a = 3q+r , where 0≤ 𝑟 < 3, i.e

r=0,1,2• Put r=0 than a = 3q+0

⇒ a = 3q

• r=1 than a = 3q+1 • r=2 than a = 3q+2 1) a=3qSquaring on both side • 𝑎2 = 3𝑞 2 ⇒ 9𝑞2=3(3𝑞2)=3m

• (∵ 3𝑞2= m)

2) a=3q+1 S.O.B.S

𝑎2 = 3𝑞 + 1 2, 𝑎 + 𝑏 2=𝑎2+2ab+𝑏2

𝑎2= 3𝑞 2+2(3q)(1)+ 1 2

𝑎2=9 𝑞2+6q+1𝑎2=3(3 𝑞2+2q)+1 𝑎2=3m+1 ( ∵ 3 𝑞2+2q=m) 3) a=3q+2

S.O.B.S𝑎2 = 3𝑞 + 2 2, 𝑎 + 𝑏 2=𝑎2+2ab+𝑏2

𝑎2= 3𝑞 2+2(3q)(2)+ 2 2

𝑎2=9 𝑞2+12q+4𝑎2= 9 𝑞2+12q+3+1𝑎2=3(3 𝑞2+4q+1)+1 𝑎2=3m+1 ( ∵ 3 𝑞2+4q+1=m)

Use Eculid’s division lemma to show that the cube of any positive integer is either of the form 9m ,9m+1 or 9m+8 for

some integer m • Sol:- Let us assume that ‘a’ be any

positive odd integer. • We apply the division algorithm

with ‘a’ and b=3.• i.e a = 3q+r , where 0≤ 𝑟 < 3, i.e

r=0,1,2• Put r=0 than a = 3q+0

⇒ a = 3q

• r=1 than a = 3q+1 • r=2 than a = 3q+2 1) a=3qCubing on both side • 𝑎3 = 3𝑞 3 ⇒ 27𝑞3=9(3𝑞3)=9m

• (∵ 3𝑞3= m)

2) a=3q+1 C.O.B.S

𝑎3 = 3𝑞 + 1 3, ሺሻ

𝑎 +𝑏 3=𝑎3+3𝑎2b+3a𝑏2+𝑏3

𝑎3= 3𝑞 3+3 3𝑞 2 (1)+3(3q) 1 2+ 1 3

𝑎3=27 𝑞3+27𝑞2+9q+1𝑎3=9(3 𝑞3+3q+q)+1 𝑎3=9m+1 ( ∵ 3 𝑞3+3q+q=m) 3) a=3q+2

C.O.B.S𝑎3 = 3𝑞 + 2 3,• Home work

Express the numbers as product of prime factors and representing on tree diagram

1) 140 2 140

2 70

5 35

7 7

1

Tree diagram

140

2 70

2 35

5 7

Express the numbers as product of prime factors and representing on tree

diagram (07/04/20)

1) 156 2) 3825 3) 5005

4) 7429 5) 176 6)256

7) 4825 8)12673

Let L be the L.C.M and H be the H.C.F of any two given number

than L X H = product of two given

number i.e L x H = a x b

Fundamental theorem of Arithmetic

Every composite number can be expressed as a product of primes and this expression is unique, apart from the order in which they appear

Find the L.C.M and H.C.F of the following pairs of integers by applying the fundamental theorem of

arithmetic method

1) 275,225 and 175 5 275 5 225 5 1755 55 5 45 5 35

11 11 3 9 7 71 3 3 1

1275= 5 x 5 x 11 225 = 5 x 5 x 3 x 3 175 = 5 x 5 x 7

H.C.F = 5 X 5 = 25 L.C.M = 25 X 11 X 3 X 3 X 7= 17325

Find the LCM and HCF of 455 and 78 and verify it.

5 455 2 787 91 3 3913 13 13 13

1 1

455 = 5 x 7 x 13 78 = 2 x 3 x 13 HCF = 13LCM = 13 x 2 x 3 x 5 x 7 = 2730

Verification:-a = 455, b = 78 , L= 2730 and H = 13

L X H = a x b13 X 2730 = 455 X 78

35490 = 35490

Home work

a) 455 and 78 b) 408 and 170

c) 275,225 and 175 d) 765,510 and 408

e) 32 and 54 f) 18 and 24

g) 56 and 88 h) 475 and 495

i) 75 and 243 j) 240 and 6552

k) 155 and 1385 l) 105 and 120

Revisiting Irrational Numbers

Let p be a prime number. If p divides 𝑎2, then p divides a, where a is a positive integer.

Prove that 2 is an irrational numberLet as assume that 2 is a rational number

i.e 2 = 𝑝

𝑞where p and q are co-prime and q≠ 0

Squaring on both sides

22

= 𝑝

𝑞

2

2 = 𝑝2

𝑞2

𝑞2 = 𝑝2

2……………………………………..(1)

i.e 2 divides 𝑝2 then 2 divides p Put p = 2r in equ. (1)

𝑞2 = 2𝑟 2

2

𝑞2 = 4𝑟2

2

𝑞2 = 2𝑟2

𝑞2

2= 𝑟2

i.e 2 divides 𝑞2, than 2 divides q

∴ p and q have at least 2 as a common factor .But this contradicts , the fact that p and q have no common factor So, our assumption is wrong

Hence 2 is irrational

Prove that 3+5 2 is an irrational number

Let as assume that 3+5 2 is a rational number

i.e 3+5 2 = 𝑝

𝑞where p and q are co-prime and q≠ 0

5 2 = 𝑝

𝑞- 3

5 2 = 𝑝−3𝑞

𝑞

2 = 𝑝−3𝑞

5𝑞

Since p and q are integer than 𝑝−3𝑞

5𝑞is rational number

But 2 is irrational number

∴ 2 ≠𝑝−3𝑞

5𝑞

but this contradicts the fact that 3+5 2 is rational is wrong

Hence 3+5 2 is an irrational

Home work Prove that the following are irrational numbers

a) 2 b) 3 c) 5 d) 7

e) 3 2 f) 2

7g)

3

2 5h) 2- 3

i) 3+ 2 j) 5-2 3 k) 2 3-1 l) 4-5 2

m) 2+ 3 n) 3+2 5 o) 3+ 4

p) 5+ 3.

2+ 3

Let as assume that 2 + 3 is a rational number

i.e 2 + 3= 𝑝

𝑞where p and q are co-prime and q≠ 0

Squaring on both sides

2 + 32

= 𝑝

𝑞

2

𝑎 + 𝑏 2 = 𝑎2 + 2𝑎𝑏 + 𝑏2

22+ 2 2 3 + 3

2= 𝑝2

𝑞2

2 + 2 6 + 3 = 𝑝2

𝑞2

5 + 2 6 = 𝑝2

𝑞2

2 6 = 𝑝2

𝑞2- 5

2 6 = 𝑝2−5𝑞2

𝑞2

6 = 𝑝2−5𝑞2

2𝑞2

Since p and q are integer than𝑝2−5𝑞2

2𝑞2is rational number

But 6 is irrational number

∴ 6 ≠𝑝2−5𝑞2

2𝑞2

but this contradicts the fact

that 2+ 3is rational is wrong

Hence 2+ 3is an irrational

Decimal expansion of rational numbers

A rational number in the form of 𝑝

𝑞with p and

q are co-prime q ≠ 0 can be expressed in the form of terminating decimal expansion if prime factors of denominator i.e ‘q’ has either only 2𝑚, only 5𝑛 or both 2𝑚 and 5𝑛.

1)12

75

Sol.:-12

75=

4

25

Science prime factor of denominator 25 is of the form 20 X 52

Hence 12

75is a

terminating decimal .4

20𝑋52X22

22

4𝑋4

10 2 = 16

10 2 = 0.16

2) 11

53𝑋23𝑋74

Science the factor denominator is 53𝑋23𝑋74 so, it is not in the form of 2𝑚𝑋5𝑛

Hence 11

53𝑋23𝑋74is a

non- terminating decimal

Without actually performing the long division, state the following rational

numbers will have a terminating decimal expansion or non-terminating decimal

expansion. And also write down the decimal expansions of those terminating

decimal expansion