1st Chapter - Real Numbers

Exercise 1.1

Question 1. Use Euclid's division algorithm to find the HCF of :
(i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255
Answer: (i)
Here 225 > 135 we always divide greater number with smaller one.
Divide 225 by 135 we get 1 quotient and 90 as remainder so that
225= 135*1 + 90
Divide 135 by 90 we get 1 quotient and 45 as remainder so that
135= 90*1 + 45
Divide 90 by 45 we get 2 quotient and no remainder so we can write it as
90 = 2*45+ 0
As there are no remainder so divisor 45 is our HCF

(ii)
38220>196 we always divide greater number with smaller one.
Divide 38220 by 196 then we get quotient 195 and no remainder so we can write it as
38220 = 196 * 195 + 0
As there is no remainder so divisor 196 is our HCF

(iii)
867>255 we always divide greater number with smaller one.
divide 867 by 255 then we get quotient 3 and remainder is 102
so we can write it as
867 = 255 * 3 + 102
Divide 255 by 102 then we get quotient 2 and remainder is 51
So we can write it as
255 = 102 * 2 + 51
Divide 102 by 51 we get quotient 2 and no remainder
So we can write it as
102 = 51 * 2+ 0
As there is no remainder so divisor 51 is our answer

Question 2: Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Answer:
Let take a as any 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, 5 because 0 <= r < b and the value of b is 6
So total possible forms will 6q+0 , 6q+1 , 6q+2,6q+3,6q+4,6q+5
6q+0=6 is divisible by 2 so it is a even number
6q+1=6 is divisible by 2 but 1 is not divisible by 2 so it is a odd number
6q+2=6 is divisible by 2 and 2 is also divisible by 2 so it is a even number
6q+3=6 is divisible by 2 but 3 is not divisible by 2 so it is a odd number
6q+4=6 is divisible by 2 and 4 is also divisible by 2 it is a even number
6q+5=6 is divisible by 2 but 5 is not divisible by 2 so it is a odd number
So odd numbers will in form of 6q + 1, or 6q + 3, or 6q + 5

Question 3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Answer:
To get the maximum number column here we always find HCF and for minimum number we find LCM
So can use Euclid's algorithm to find the HCF.
Here 616> 32 so always divide greater number with smaller one
When we divide 616 by 32 we get quotient 19 and remainder 8
So we can write it as
616 = 32 x 19 + 8
Now divide 32 by 8 we get quotient 4 and no remainder
So we can write it as
32 = 8 x 4 + 0
As there are no remainder so our HCF will 8
So that maximum number of columns in which they can march is 8.

4. Use Euclid'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.
Answer:
Let take a as any positive integer and b = 3.
Then using Euclid's algorithm we get a = 3q + r here r is remainder and value of q is more than or equal to 0 and r = 0, 1, 2 because 0 < r < b and the value of b is 3 So our possible values will 3q+0 , 3q+1 and 3q+2

Now find the square of values
Use the formula (a+b)² = a² + 2ab +b² to open the square bracket
(3q)² = 9q² if we divide by 3 we get no remainder

we can write it as 3*(3q²) so it is in form of 3m here m = 3q²
(3q+1)²  = (3q)² + 2*3q*1 + 1²
=9q² + 6q +1 now divide by 3 we get 1 remainder

so we can write it as 3(3q² + 2q) +1 so we can write it in form of 3m+1 and value of m is 3q² + 2q here
(3q+2)²  = (3q)² + 2*3q*2 + 2²
=9q² + 12q +4 now divide by 3 we get 1 remainder

so we can write it as 3(3q² + 4q +1) +1 so we can write it in form of 3m +1 and value of m will 3q² + 4q +1
Square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Exercise 1.2

Question 1. Express each number as a product of its prime factors:
(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429
Answer:
(i) 140 = 2 x 2 x 5 x 7 =2² x 5 x 7
(ii)156 =2 x 2 x 3 x 13 =2² x 3 x 13
(iii)3825 =3 x 3 x 5 x 5 x 17 =3² x 5² x 17
(iv)5005 =5 x 7 x 11 x 13 =5 x 7 x 11 x 13
(v)7429 =17 x 19 x 23 =17 x 19 x 23
Question 2. Find the LCM and HCF of the following pairs of integers and verify that LCM*HCF = product of the two numbers.
(i)26 and 91 (ii) 510 and 92 (iii) 336 and 54
Answer:
(i)26 and 91
26 = 2 x 13
91 =7 x 13
HCF = 13
LCM  =2 x 7 x 13 =182
Product of two value  26 x 91= 2366
Product of HCF and LCM 13 x 182  = 2366
Hence, product of two numbers = product of HCF X LCM

(ii)
510  = 2 x 3 x 5 x 17
92 =2 x 2 x 23
HCF =2
LCM =2 x 2 x3 x 5 x 17 x 23 = 23460
Product of both values 510x92 = 46920
Product of HCF and LCM 2x23460 =46920
Hence, product of two numbers = product of HCF X LCM

(iii)
336 = 2 x 2 x 2 x 2 x 3 x 7
54 = 2 x 3 x 3 x 3
HCF = 2 x 3  = 6
LCM = 2 x 2 x 2 x 2 x 3 x 3 x 3 x 7 =3024
Product of number 336 x 54 =18144
Product of HCF and LCM 6 x 3024 = 18144
Hence, product of two numbers = product of HCF X LCM

Question 3. Find the LCM and HCF of the following integers by applying the prime factorization method.
(i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25
Answer:
(i)
12 = 2 x 2 x 3
15 =3 x 5
21=3 x 7
HCF = 3
LCM = 2 x 2 x 3 x 5 x 7 = 420

(ii)

17= 1 x 17
23 = 1 x 23
29 = 1 x 29
HCF = 1
LCM= 1 x 17 x 19 x 23 = 11339

(iii)
8 =1 x 2 x 2 x 2
9 =1 x 3 x 3
25 =1 x 5 x 5
HCF =1
LCM = 1 x 2 x 2 x 2 x 3 x 3 x 5 x 5 = 1800

Question 4. Given that HCF (306, 657) = 9, find LCM (306, 657).
Answer:

We have the formula that
Product of LCM and HCF = product of number
LCM x 9  =  306 x 657
Divide both side by 9 we get
LCM = (306 x 657)/9 = 22338

Question 5. Check whether 6n can end with the digit 0 for any natural number n.
Answer:
If any digit has last digit 10 that means It is divisible by 10 And the factors of
10 = 2x5 So value 6n should be divisible by 2 and 5 both 6n is divisible by 2 but not divisible by 5 So it can not end with 0.

Question 6. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?
Answer:
They will be meet again after LCM of both values at the starting point.
To get the LCM we have to factorize the number.
18 = 2 X 3 X 3
12 = 2 X 2 X 3
LCM = 2 x 2 x 3 x 3 = 36
Therefore, they will meet together at the starting point after 36 minutes.

Exercise 1.3

Question 1. Prove that √5 is irrational.
Answer:
Let take √5 as rational number
If a and b are two co prime number and b is not equal to 0.
We can write √5 = a/b
Multiply by b both side we get
b√5 = a
To remove root, Squaring on both sides, we get
5b² = a² (1)
Therefore, 5 divides a² and according to theorem of rational number, for any prime number p which is divides a² then it will divide a also.
That means 5 will divide a. So we can write
a = 5c
and plug the value of a in equation (1) we get
5b² = (5c)²
5b² = 25c²
Divide by 25 we get
b²/5 = c²
again using same theorem we get that b will divide by 5
and we have already get that a is divide by 5
but a and b are co prime number. so it is contradicting .
Hence √5 is a non rational number

Question 2. Prove that 3 + 2√5 is irrational.

Answer:
Let take that 3 + 2√5 is a rational number.
So we can write this number as
3 + 2√5 = a/b
Here a and b are two co prime number and b is not equal to 0
Subtract 3 both sides we get
2√5 = a/b  3
2√5 = (a-3b)/b
Now divide by 2 we get √5 = (a-3b)/2b
Here a and b are integer so (a-3b)/2b is a rational number so √5 should be a rational number But √5 is a irrational number so it contradict the fact
Hence result is 3 + 2√5 is a irrational number


Question 3. Prove that the following are irrationals:
(i) 1/√2 (ii) 7√5 (iii) 6 + √2
Answer:
(i) Let take that 1/√2 is a rational number.
So we can write this number as
1/√2 = a/b
Here a and b are two co prime number and b is not equal to 0
Multiply by ?2 both sides we get
1 = (a√2)/b
Now multiply by b
b = a√2
divide by a we get
b/a = √2
Here a and b are integer so b/a is a rational number so √2 should be a rational number But √2 is a irrational number so it is contradict
Hence result is 1/√2 is a irrational number


(ii) Let take that 7√5 is a rational number.
So we can write this number as
7√5 = a/b
Here a and b are two co prime number and b is not equal to 0
Divide by 7 we get √5 =a/(7b)
Here a and b are integer so a/7b is a rational number so √5 should be a rational number but √5 is a irrational number so it is contradict
Hence result is 7√5 is a irrational number.

(iii) Let take that 6 + √2 is a rational number.
So we can write this number as
6 + √2 = a/b
Here a and b are two co prime number and b is not equal to 0
Subtract 6 both side we get √2 = a/b  6√2 = (a-6b)/b
Here a and b are integer so (a-6b)/b is a rational number so √2 should be a rational number But √2 is a irrational number so it is contradict
Hence result is 6 + √2 is a irrational number

Exercise 1.4

Question 1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:
(i)13/3125 (ii)17/8 (iii)64/455
(iv)15/1600
(v)29/343
(vi)23/2³*5²
(vii)129/2²* 57* 75
(viii)6/15
(ix)35/50
(x)77/210

Answer:

(i) 13/3125
Factorize the denominator we get
3125 =5 x 5 x 5 x 5 x 5 = 55
So denominator is in form of 5m so 13/3125 is terminating .
(ii) 17/8
Factorize the denominator we get
8 =2 x 2 x 2 = 2³
So denominator is in form of 2n so 17/8 is terminating .
(iii) 64/455
Factorize the denominator we get
455 =5 x 7 x 13
There are 7 and 13 also in denominator so denominator is not in form of 2n*5m . hence 64/455 is not terminating.
(iv) 15/1600
Factorize the denominator we get
1600 =2 x 2 x 2 x2 x 2 x 2 x 5 x 5 = 26 x 5²
so denominator is in form of 2n x 5m
Hence 15/1600 is terminating.
(v) 29/343
Factorize the denominator we get
343 = 7 x 7 x 7 = 7³
There are 7 also in denominator so denominator is not in form of 2n x 5m
Hence it is none - terminating.
(vi) 23/(2³ x 5²)
Denominator is in form of 2n x 5m
Hence it is terminating.
(vii) 129/(2² x 57 x 75 )
Denominator has 7 in denominator so denominator is not in form of 2n x 5n
Hence it is none terminating.
(viii) 6/15
divide nominator and denominator both by 3 we get 2/5
Denominator is in form of 5m so it is terminating.

(ix) 35/50 divide denominator and nominator both by 5 we get 7/10
Factorize the denominator we get
10=2 x 5
So denominator is in form of 2n x5m so it is terminating
(x) 77/210.
simplify it by dividing nominator and denominator both by 7 we get 11/30
Factorize the denominator we get
30=2 x 3 x 5
Denominator has 3 also in denominator so denominator is not in form of 2n x 5n
Hence it is none terminating.

Question 2. Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.
(i)13/3125 = 0.009375 it is terminating

(ii)17/8 it is terminating

(iii)64/455 none- terminating

(iv)15/1600 it is terminating

(v)29/343 it is none terminating

(vi)23/23*52 = 23/200 it is terminating

(vii)129/22* 57*75 it is none terminating

(viii) 6/15 = 2/5 = 0.4 it is terminating

(ix)35/50 it is terminating

(x)77/210 it is none terminating


Question 3. The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form p , q you say about the prime factors of q?
Answer:
(i) 43.123456789
it has certain number of digits so they can be represented in form of p/q . Hence they are rational number
As they have certain number of digit and the number which has certain number of digits is always terminating number and for terminating number denominator has prime factor 2 and 5 only .

(ii) 0.120120012000120000. . .
In this problem repetitions number are not same so it is not a irrational number
so prime factor of denominator Q will has a value which is not equal to 2 or 5. And irrational number is always none terminating
(iii) 43.123456789
In this number 0.123456789 repeating again and again so it is a rational number and it is none terminating so that the prime factor has a value which is not equal to 2 or 5