Real Numbers

Euclid’s Division is a method for finding the HCF (highest common factor) of two given integers. According to Euclid’s Division Algorithm, For any two positive integers, ‘a’ and ‘b’, there exists a unique pair of integers ‘q’ and ‘r’ which satisfy the relation:

a = bq + r , 0 ≤ r ≤ b

Given integers 900 and 270. Clearly 900>270.

By applying division lemma

⇒ 900 = 270×3 + 90

Since remainder 0, applying division lemma on 270 and 90

⇒ 270 = 90×3 + 0

∵ remainder = 0,

∴ the HCF of 900 and 270 is 90.

Euclid’s Division is a method for finding the HCF (highest common factor) of two given integers. According to Euclid’s Division Algorithm, For any two positive integers, ‘a’ and ‘b’, there exists a unique pair of integers ‘q’ and ‘r’ which satisfy the relation:

a = bq + r , 0 ≤ r ≤ b

Given integers 196 and 38220. Clearly 38220>196.

By applying division lemma

⇒ 38220 = 196×195 + 0

Since remainder 0

∴ the HCF of 196 and 38220 is 195.

Euclid’s Division is a method for finding the HCF (highest common factor) of two given integers. According to Euclid’s Division Algorithm, For any two positive integers, ‘a’ and ‘b’, there exists a unique pair of integers ‘q’ and ‘r’ which satisfy the relation:

a = bq + r , 0 ≤ r ≤ b

Given integers 1651 and 2032. Clearly 2032>1651.

By applying division lemma

⇒ 2032 = 1651×1 + 381

Since remainder 0, applying division lemma on 1651 and 381

⇒ 1651 = 381×4 + 127

Since remainder 0, applying division lemma on 381 and 127

⇒ 381 = 127×3 + 0

Since remainder = 0,

∴ the HCF of 1651 and 2032 is 127.

Let a be any odd positive integer and b = 6

Then using Euclid’s algorithm, we get a = 6q + r here r is remainder and value of q is more than or equal to 0 and r = 0,1,2,3,4,5because 0≤r<band the value of b is 6

So total form available will be 6q, 6q + 1, 6q + 2, 6q + 3, 6q + 4, 6q + 5 ,6q + 6 is divisible by 2, so it is an even number.

6q + 1, 6 is divisible by 2 but 1 is not divisible by 2, so it is an odd number.

6q + 2 , 6 is divisible by 2 but 2 is also divisible by 2, so it is an even number.

6q + 3, 6 is divisible by 2 but 3 is not divisible by 2, so it is an odd number.

6q + 4, 6 is divisible by 2 but 4 is also divisible by 2, so it is an even number.

6q + 5, 6 is divisible by 2 but 5 is not divisible by 2, so it is an odd number.

∴ so odd numbers will in form of 6q + 1 or 6q + 3 or 6q + 5

Let be any positive integer. Then, it is form 3q or, 3q + 1 or, 3q + 2

So, we have the following cases:

Case I When a = 3q

In this case, we have

a² = (3q)² = 9q² = 3q(3q) = 3p, where p = 3q²

Case II When a = 3q + 1

In this case, we have

a² = (3q + 1)² = 9q² + 6q + 1 = 3q(3q + 2) + 1 = 3p + 1,

where p = q(3q + 2)

Case III When a = 3q + 2

In this case, we have

a² = (3q + 2)² = 9q² + 12q + 4 = 9q² + 12q + 3 + 1

= 3(3q² + 4q + 1) + 1 = 3p + 1

where p = 3q² + 4² + 1

Hence, a is the form of 3p or 3p + 1 or 3p + 2

Let a be any positive integer. Then, it is of the form 3q or, 3q + 1 or, 3q + 2.

We know that according to Euclid's division lemma:
a = bq + r
So, we have the following cases:

Case I When a = 3q

In this case, we have

a³ = (3q)³ = 27q³ = 9(3q³ ) = 9m, where m = 3q³

Case II When a = 3q + 1

In this case, we have

a³ = (3q + 1)³

⇒ 27q³ + 27q² + 9q + 1

⇒9q(3q² + 3q + 1) + 1

⇒ a³ = 9m + 1, where m = q(3q² + 3q + 1)

Case III When a = 3q + 2

In this case, we have

a³ = (3q + 1)³

⇒ 27q³ + 54q² + 36q + 8

⇒9q(3q² + 6q + 4) + 8

⇒ a³ = 9m + 8, where m = q(3q² + 6q + 4)

Hence, a³ is the form of 9m or, 9m + 1 or, 9m + 8

We know that any positive integer is of the form 3q or, 3q + 1 or,

3q + 2 for some integer and one and only one of these possibilities can occur.

So, we have following cases:

Case I When n = 3q

In this case, we have

n = 3q, which is divisible by 3

Now,

⇒ n + 2 = 3q + 2,

⇒ n + 2 leaves remainder 2 when divided by 3

⇒ n + 2 is not divisible by 3

Again, n = 3q

⇒ n + 4 = 3q + 4 = 3(q + 1) + 1

⇒ n + 4leaves remainder 1 when divided by 3

⇒ n + 4 is not divisible by 3

Thus, n is divisible by 3 but n + 2 and n + 4 are not divisible by 3.

Case II When n = 3q + 1

In this case, we have

n = 3q + 1

⇒ n leaves remainder 1 when divided by 3

⇒ n is not divisible by 3

Now, n = 3q + 1

⇒ n + 2 = (3q + 1) + 2 = 3(q + 1),

⇒ n + 2 is divisible by 3

Again, n = 3q + 1

⇒ n + 4 = (3q + 1) + 4 = 3q + 5 = 3(q + 1) + 2

⇒ n + 4 leaves remainder 2 when divided by 3

⇒ n + 4 is not divisible by 3

Thus, n + 2 is divisible by 3 but n and n + 4 are not divisible by 3.

Case III When n = 3q + 2

In this case, we have

n = 3q + 2

⇒ n leaves remainder 2 when divided by 3

⇒ n is not divisible by 3

Now, n = 3q + 2

⇒ n + 2 = 3q + 2 + 2 = 3(q + 1) + 1,

⇒ n + 2 leaves remainder 1 when divided by 3

⇒ n + 2 is not divisible by 3

Again, n = 3q + 2

⇒ n + 4 = 3q + 2 + 4 = 3(q + 2)

⇒ n + 4 is divisible by 3

Thus, n + 4 is divisible by 3 but n and n + 2 are not divisible by 3.

I. 140 = 2×2×5×7 = 2²×5×7

II. 156 = 2×2×3×13 = 2²×3×13

III. 3825 = 3×3×5×5×17 = 3²×5²×17

IV. 5005 = 5×7×11×13

V. 7429 = 17×19×23

I. 12,15 and 21

12 = 2²×3

15 = 3×5

21 = 3×7

LCM = 2²×3×5×7 = 420

HCF = 3

II. 17,23 and 29

17 = 1×17

23 = 1×23

29 = 1×29

LCM = 1×17×23×29 = 11339

HCF = 1

III. 8, 9 and 5

8 = 2³

9 = 3²

5 = 1×5

LCM = 2³×3²×5 = 360

HCF = 1

IV. 72 and 108

72 = 2³×3²

108 = 2²×3³

LCM = 2⁵×3⁵ = 7776

HCF = 2²×3² = 4×9 = 36

V. 306 and 657

306 = 2×3²×17

657 = 3²×73

LCM = 2×3²×17×73 = 22338

HCF = 3² = 9

If any number end with digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as

Prime factorization of 6ⁿ = (2×3)ⁿ

It can be observed that 5 is not in the prime factorization of 6ⁿ.

Hence, for an value of n, 6ⁿ will not visible by 5.

∴ 6ⁿ cannot end with the digit 0 for any natural number n.

Numbers are of two types – composite and prime. Prime numbers can be divided by 1 and only itself, whereas composite numbers have factors other than 1 and itself.

It can be observed that

7×11×13 + 13 = 13×(7×11 + 1) = 13×(77 + 1) = 13×78

= 13×13×6

The given expression has 6 and 13 as its factors.

∴ , it is a composite factor.

7×6×5×4×3×2×1 + 5 = 5×(7×6×4×3×2×1 + 1)

= 5×(1008 + 1) = 5×1009

1009 cannot be factorized further. Therefore, the given expression has 5 and 1009 as its factors. Hence, it is a composite number.

Numbers are of two types – composite and prime. Prime numbers can be divided by 1 and only itself, whereas composite numbers have factors other than 1 and itself.

It can be observed that

(17×11×2) + (17×11×5) = 17{(11×2) + (11×5)}

= 17×{11×{(2) + (5)}} = 17×11×7

The given expression has 17, 11and 7 as its factors.

∴ it is a composite factor.

This is related to concept of numbers in the unit digits place of the powers of natural number. The power of 6 any index repetition 6 i.e. (6)ⁿ the last digit is 6 only.

Example:

i: 6¹ = 6

ii: 6² = 36

iii: 6³ = 216

The last digit in the expansion of 6¹⁰⁰ is 6.

Since, this decimal has finite number of digits

∴ it is terminating

Since, this decimal has finite number of digits

∴ it is terminating.

Since, this decimal has finite number of digits

∴ it is terminating.

Since the decimal continues endlessly, it is non-terminating and repeating.

∴ it is non-terminating and repeating.

Since, this decimal has finite number of digits

∴ it is terminating.

⇒ 13 and 3125 are co-prime.

Now, We have to write the denominator 3125 in the form of 2ⁿ5^m

where, n and m are the non-negative numbers.

3125 = 5 × 5 × 5 × 5 × 5 = 5⁵

3125 ⇒ 1325 = 1 × 5⁵ = 2⁰ × 5⁵

∴ denominator is of the form 2ⁿ5^m where, n = 0 and m = 5

Thus, is a Terminating decimal

⇒ 11 and 12 are co-prime.

Now, We have to write the denominator 12 in the form of 2ⁿ5^m where, n and m are the non-negative numbers.

12 = 2×2×3

12 = 2²×3

∴ denominator is not of the form 2ⁿ5^m where, n = 2 and m = 0

Thus, is Non-terminating and repeating decimal

⇒ 64 and 455 are co-prime.

Now, We have to write the denominator 455 in the form of 2ⁿ5^m where, n and m are the non-negative numbers.

455 = 5×7×13

∴ denominator is not of the form 2ⁿ5^m where, n = 0 and m = 1

Thus, is Non-terminating and repeating decimal

⇒ 3 and 320 are co-prime.

Now, We have to write the denominator 320 in the form of 2ⁿ5^m where, n and m are the non-negative numbers.

320 = 2×2×2×2×2×2×5 = 2⁶×5

∴ denominator is in the form 2ⁿ5^m where, n = 6 and m = 1

Thus, is terminating decimal.

⇒ 29 and 343 are co-prime.

Now, We have to write the denominator 343 in the form of 2ⁿ5^m where, n and m are the non-negative numbers.

343 = 7×7×7 = 7³

∴ denominator is not of the form 2ⁿ5^m where, n = 0 and m = 0

Thus, is Non-terminating and repeating decimal

Thus, is Non-terminating decimal.

⇒ 23 and 2³ 5² are co-prime.

Now, we have to write the denominator 2² 5⁷ 7⁵ in the form of 2ⁿ5^m where, n and m are the non-negative numbers.

2³ 5²

∴ denominator is in the form 2ⁿ5^m where, n = 3 and m = 2

Thus, is terminating decimal

⇒ 23 and 2² 5⁷ 7⁵ are co-prime.

Now, we have to write the denominator 2² 5⁷ 7⁵ in the form of 2ⁿ5^m where, n and m are the non-negative numbers.

2³ 5²

2²×5⁷×7⁵

∴ denominator is not of the form 2ⁿ5^m where, n = 2 and m = 7. Due to one more factor it is not in the form

Thus, is Non-terminating decimal.

⇒ 2 and 5 are co-prime.

Now, we have to write the denominator 5 in the form of 2ⁿ5^m where, n and m are the non-negative numbers.

2⁰ 5

∴ denominator is in the form 2ⁿ5^m where, n = 0 and m = 1

Thus, is Terminating decimal.

⇒ 7 and 10 are co-prime.

Now, we have to write the denominator 10 in the form of 2ⁿ5^m where, n and m are the non-negative numbers.

10 = 2×5

∴ denominator is in the form 2ⁿ5^m where, n = 1 and m = 1

Thus, is terminating decimal.

⇒ 11 and 30 are co-prime.

Now, We have to write the denominator 30 in the form of 2ⁿ5^m where, n and m are the non-negative numbers.

30 = 2×3×5

∴ denominator is not of the form 2ⁿ5^m where, n = 1 and m = 1. Due to one more factor it is not in the form

Thus, is Non-terminating and repeating decimal.

According to Euclid’s Division Algorithm,

For any two positive integers, ‘a’ and ‘b’, there exists a unique pair of integers ‘q’ and ‘r’ which satisfy the relation:

a = bq + r , 0 ≤ r ≤ b

(i)

(ii)

(iii)

(iv)

(v)

(i) 43.123456789

43.123456789 is terminating.

So, it would be a rational number

Hence, 43.123456789 is now in the form of .

And the prime factors of q are in terms of 2 and 5.

(ii) 0.120120012000120000….

0.120120012000120000…. is non-terminating and non-repeating.

So, it is not a rational number.

(iii) 43.

43. is non- terminating but terminating.

So, it would be a rational number.

In a non-terminating, repeating expansion of

q will have factors other than 2 or 5.

Let be rational. Then, there exists positive co-primes a and b such that

⇒

⇒

⇒

is rational as a and b are integers

∴√2 is rational which contradicts to the fact that √2 is irrational.

Hence, our assumption is false and is irrational.

Let us suppose that is rational.

Let be rational equal to , where a and b are integers and a≠0 and b≠0 Then,

⇒

⇒ [squaring both sides]

⇒

⇒

⇒

⇒

Since a and b are integers, is rational. So, √3 is rational

Now, this contradicts the fact that √3 is rational.

Hence √3 + √5 is irrational.

6 + √2

Let 6 + √2 be a rational number equal to , where a,b are positive co-primes. Then,

6 + √2 =

⇒ √2

⇒

Since a and b are integers, is also rational and hence, √2should be rational. This is contracdicts the fact that√2is irrational. Therefore , our assumption is false and hence, 6 + √2 is irrational.

√5

Let take √5as rational number equal to , where a, b are positive co-primes. Then,

⇒ √5 b = a

⇒ 5b² = a²[squaring both sides] …I

Therefore, 5 divides a² and according to theorem of rational number, for any prime number p divides a² then it will divide a also.

∴ a = 5c

Put value of a in Eq. I, we get

5b² = (5c)²

⇒ 5b² = 25c²

⇒ [divide by 25 both sides]

Using same theorem we get that b will divide by 5 and we have already get that a is divided by 5. This contradicts our assumption.

Hence, √5 is irrational.

3 + 2√5

Let us assume on the contrary that 3 + 2√5 is rational. Then, there exist co-prime positive integers a and b such that

3 + 2√5

⇒

⇒

⇒ is rational [∵ a,b are integers ∴ is a rational]

This contradicts the fact that is irrational. So, our supposition is incorrect.

Hence, is an irrational number.

Let √p + √qbe rational

⇒ √p + √q [where a, b are co-primes and integers]

Squaring both sides

⇒

⇒

⇒ 2√pq = a^2/b^2 -p-q

⇒

⇒ ….I

∴ from Eq. I our assumption contradicts here, because p is rational

Hence, √p + √q is irrational number.

The logarithmic form of log₂₅5

Using property of logarithmic, log_b x = and log_a a^x = x

=

The logarithmic form of log₈₁3

Using property of logarithmic, log_b x = and log_a a^x = x

=

The logarithmic form of log₂

Using property of logarithmic,

= log₂ 1 – log₂16

= 0 – log₂2⁴

= 0 – 4 = -4

The logarithmic form of log₇ 1

= 0

The logarithmic form of log_x √x

Using the property of logarithmic, log_a a^x = x

= log_x x^1/2

=

The Logarithmic form of log₂512

Using the property of logarithmic, log_a a^x = x

= log₂2⁹

= 9

The Logarithmic form of log₁₀0.01

= log₁₀

Using property of logarithmic, and log_a a^x = x

= log₁₀1 – log₁₀100

= 0 – log₁₀10²

= 0 – 2 = -2

The Logarithmic form of

Using the property of logarithmic, log_a a^x = x

=

=

= 3

The Logarithmic form of 2^{2 + log}₂³

Using property of logarithmic, a^{m + n} = a^m × aⁿ and a^log_a^x = x

= 2^{2 + log}₂³

= 2² × 2^log₂³

= 4 × 3

= 12

Some basic logarithmic formulas are

1. a^log_ab = b
2. log_a 1 = 0
3. log_a a = 1
4. log_a(x · y) = log_ax + log_ay
5. log_a xy = log_ax - log_ay
6. log_a 1x = -log_ax
7. log_a x^p = p log_ax
8. log_a^k x = 1k log_a x, for k ≠ 0

I. log 2 + log 5

using the property of logarithm, log_a xy = log_a x + log_a y

= log(2 × 5)

= log 10

II. log 16 – log 2

using the property of logarithm,

and

= log (16/2)
= log 8
N = 8

III. 3 log 4

Using the property of logarithm,

log (4)³
= log 64
N = 64

IV. 2 log 3 - 3 log 2

using the property of logarithm,

and

= log 3²-log 2³

=

=

VI. log 10 + 2 log 3 - log 2

using the property of logarithm, log_a xy = log_a x + log_a y

= log 5 + log 2 + log 3² - log 2

= log 5 + log 9

= log (5×9)

= log 45

I. log₂15

⇒ log₂5 + log₂3 〗[∵]

⇒ x + y

II. log₂7.5

⇒ log₂

⇒ log₂15 - log₂2 [∵]

⇒ log₂5 + log₂3-log₂2[∵]

⇒ x + y-1

III.

⇒ log₂12 + log₂5 [∵]

⇒ log₂4 + log₂3 + log₂5 [∵]

⇒ 2log₂2 + log₂3 + log₂5

⇒ 2(1) + x + y = x + y + 2

IV. log₂6750

⇒ log₂54 + log₂125〗

⇒ log₂27 + log₂2 + log₂125 〗 〗

⇒ log₂3³ + log₂5³ + log₂2

⇒ 3 log₂3 + 3 log₂5 + log₂2

⇒ 3x + 3y + 1

i.

ii. =

⇒ log2⁷ - log5⁴

⇒ 7log2 - 4 log5

iii. logx² y³ z⁴

⇒ logx² + + logy³ + logz⁴

⇒ 2 logx + 3 logy + 4 logz

iv. logp² q³-logr

⇒ logp² + log q³-log r

⇒ 2 log p + 3 log q-log r

v. =

⇒

⇒ {3log x + 2 log y} = log x + log y

x² + y² = 25xy

Adding 2xy on both sides

x² + y² + 2xy = 27xy

(x + y)² = 27xy

Taking log both sides

log (x + y)² = log 27xy

2 log (x + y) = log 27 + log x + log y

2 log(x + y) = log 3³ + log x + log y

2 log (x + y) = 3 log 3 + log x + log y

Hence proved.

⇒

⇒ [∵] [∵]

Remove log from both sides

⇒

⇒ = xy

⇒ x² + y² + 2xy = 9xy

⇒ x² + y² = 7xy

Divide by both sides

⇒

2.3^x = 0.23^y = 1000

Consider 2.3^x = 1000

⇒ log 2.3^x = log10³

⇒ x log2.3 = 3 log 10

⇒ log2.3 = [∵ log10 = 1] …..I

Consider 0.23^y = 1000

⇒ log 0.23^y = log 1000

⇒ y log0.23 = log 10³

⇒ y log 0.23 = 3 log 10

⇒ log 0.23 = ……..II

Subtract eq. II from I

⇒

⇒

⇒

⇒

⇒

⇒

2^{(x + 1)} = 3^(1-x)

Taking log base both sides

x = (log 3 - log 2)/(log 2 + log 3)

Assume that log 2 is rational, that is,

…(1)

where p, q are integers.

Since, and therefore, p<q

From Eq. 1,

2^q = (2×5)^p

2^(q-p) = 5^p

Where p-q is an integer greater than 0.

Now, it can be seen that the L.H.S is even and the R.H.S is odd.

Hence, there is contradiction and log 2 is irrational.

Let us assume log 100 is rational

log 100 = log₁₀ 100

log₁₀ 10² = 2 log₁₀ 10 = 2

As 2 is rational number, ∴ log 100 is also rational.