EXERCISE
1.1
1. Is zero a
rational number? Can you write it in the form p/q, where p and q are integers
and q ≠ 0?
Solution:
Yes. 0 is a rational number. We can write 0 as 0/5, 0/6, 0/7 and so on.
2. Find six
rational numbers between 3 and 4.
There are infinite rationals numbers that lies between 3 and 4. One easy way to find them is to write the
decimals that lies between these values and then convert them to the form p/q.
Six rational numbers between 3 and 4 are
3.1 = 31/10
3.2 = 32/10
3.3 = 33/10
3.4 = 34/10
3.5 = 35/10
3.6 = 36/10 ...
3. Find five
rational numbers between 3 /5 and 4/ 5
To find the equivalent fractions of the given numbers and then go on increasing the numerator keeping the
denominator the same.
. Fractions equivalent to 3/5 and 4/5 are 30/50 and 40/50. And rational number between these numbers
are 31/50, 32/50,33/50,34/50,35/50.
4. State whether
the following statements are true or false. Give reasons for your answers.
(i) Every natural
number is a whole number.
TRUE
Natural numbers
are 1, 2, 3,4,5,6…
Whereas whole numbers
are 0, 1, 2, 3, 4, 5, 6….
It can be
inferred that all natural numbers belongs to whole numbers.
(ii) Every integer
is a whole number.
FALSE.
All positive
integers can be whole numbers but the negative integers are not whole numbers and
hence
not all integers are whole numbers.
(iii) Every rational number is a whole
number.
FALSE.
Rational numbers are of the form p/q and could be negative
or positive whereas Whole numbers cannot
be negative or partial.
EXERCISE 1.2
1.
State whether the following statements are true or false.
Justify your answers.
(i) Every
irrational number is a real number.
Answer:
True. By definitions real number includes all
rational and irrational numbers.
(ii) Every point on
the number line is of the form √m ,
where m is a natural number.
Answer:
False. Negative
numbers cannot be expressed as the square root of any other number.
(iii) Every real
number is an irrational number.
Answer:
False. Every irrational number is a real number but
vice versa is not true. Not every real number is irrational.
2.
Are the square roots of all positive integers irrational? If
not, give an example of the square root of a number that is a rational number.
Answer:
No. Not all positive integers have irrational roots.
For example the roots of the positive integers 4, 9, 16, 25, 36… are rational.
3.
Show how √5 can be represented on the number line.
EXERCISE
1.3
1.
Write the following in decimal form and say what kind of decimal
expansion each has:
(i) 36/100
Answer: 36/100=0.36
Terminating decimal
(ii) 1/11
Answer: 0.09090909090909090909090909090909
Non-terminating recurring decimal
(iii) 4(1/8)
= 33/8
Answer: 4.125
Terminating decimal
(iv) 3/13
Answer:
3/13= 0.23076923076923076923076923076923
Non-terminating recurring decimal
(v) 2/11
Answer:
2/11=0.18181818181818181818181818181818
Non-terminating recurring decimal
(vi)329/400
Answer:
329/400 = 0.8225
Terminating decimal
2.
You know that
1/7= 0.142857.
Can you predict what the decimal expansions of?
2 /7, 3/7, 4/7, 5/7, 6/7
are, without actually doing the long division? If so, how?
[Hint : Study
the remainders while finding the value of 1/7carefully.]
Answer:
Since 1/7 = 0.142857
2/7 =
2(1/7) = 2(0.142857) = 0.285714
3/7 =
3(1/7) = 3(0.142857) = 0.428571
4/7 = 4(1/7) = 4(0.142857) = 0.571428
5/7
= 5(1/7) = 5(0.142857) = 0.714285
6/7 = 6(1/7) = 6(0.142857) = 0.857142
3.
Express the following in the form p/q ,
where p and q are integers and q ≠ 0.
(i) 0.6
Answer:
Let x = 0.6666...
Then, 10x = 6.6666...
Now,
10x - x = 6.6666... - .66666= 6
=> 9x = 6
=> x = 6/9
Therefore, 0 .66666... = 6/9
(ii) 0.47
Answer:
Let x = 0.477777....
Then, 100x = 47.7777..
10x = 4.777777
Now,
100x -10 x =47.777..... - 4.777....
=> 90x = 47-4
=> 90x = 43
=> x = 43/90
Therefore, 0.477.... = 43/90
(iii) 0.001
Let x = 0.001111....
Then, 1000x = 1.1111...
And 100x = 0.11111....
now,
1000x - 100x = 1.1111.... - 0.11111
=> 900x = 1
=>x = 1/900
Therefore, 0.0011111....= 1/900
4.
Express 0.99999.... In the form p/q .
Are you surprised by your answer? With your teacher and classmates discuss why
the answer makes sense.
Answer:
Let x = 0.9999...
then, 10 x = 9.999...
and 10x - x = 9.999... -0 .9999
=> 9x = 9
=> x = 1
5.
What can the maximum number of digits be in the repeating block
of digits in the decimal expansion of 1/17?
Perform the division to check your answer.
Answer:
The maximum number of digits could be 16 ( 17 - 1) as the maximum value of remainder could only be 16.
Check:
1/17 = 0.058823529411764705882352941176470588235294117647....
The block that is repeating is 0588235294117647 and have 16 digits..
6.
Look at several examples of rational numbers in the form p/q (q
≠ 0), where p and q are integers with no common factors other
than 1 and having terminating decimal representations (expansions). Can you
guess what property q must satisfy?
Answer:
For this to happen the factors of the denominator could only be powers of 2 and 5.
7.
Write three numbers whose decimal expansions are non-terminating
non-recurring.
Answer:
1.4142135623730950488016887242097...
1.4142135623730950488016887242097...
1.7320508075688772935274463415059...
2.2360679774997896964091736687313....
8.
Find three different irrational numbers between the rational
numbers
5/7 and
9/11.
Answer:
5/7 = 0.71428571428571428571428571428571...
9/11 = 0.81818181818181818181818181818182...
Thus, the required numbers also lie between the values .07 and 0.8
Now, write three numbers that are non-terminating non-recurring to get the required irrational numbers.
3 such examples are
1. 0.756945988652479876544565....
2. 0.8054115545778574524475745887....
3. 0.79851452687455682368745....
9.
Classify the following numbers as rational or irrational :
(i) √23
√23= 4.7958315233127195415974380641627....
It is non-terminating and non-recurring. Thus it is irrational.
(ii) √225
√225 = 15. Therefore it is a rational number.
(iii) 0.3796
Rational number.
(iv) 7.478478...
478 is the repeating block, Thus it non-terminating but recurring decimal .Thus it is a rational number.
(v) 1.101001000100001...
It is non-terminating and non-recurring. Thus it is irrational.
EXERCISE
1.5
1.
Classify the following numbers as rational or irrational:
(i) 2-√5
= 2 - 2.2360679774997896964091736687313...
= - 0.23606797749978969640917366873128...
Thus, it is an irrational number.
(ii) (3 + √23) - √23
= 3 + √23 - √23
= 3
3 is a rational number.
(iii) 2√7/7√7
= 2/7 = 0.28571428571428571428571428571429...
Which is a rational number.
(iv) 1/√2 = 1/1.4142135623730950488016887242097... = 0.7071067811865475244008443621052...
It is non terminating and non repeating. Thus, it is an irrational number.
(v) 2π
= 2 x 3.1415926535897932384626433832795..
= 6.283185307179586476925286766559....
Thus, 2π is also irrational.
2.
Simplify each of the following expressions:
(i) (2 + √3)(2 + √2)
Answer:
(2 + √3)(2 + √2)
= 4 + 2√2 + 2√3 + √6
= 4 + 2√2 + 2√3 + √6
(ii) (3 + √3)(3 - √3)
(3 + √3)(3 - √3)
= 9 - 3√3 + 3√3 - 3
= 9 - 3
= 6
(iii) (5 + √2)²
(5 + √2)²
= 25 + 2 + 10√2
= 27 + 10√2
(iv) (√5 - √2)(√5-√2)
(√5 - √2)(√5-√2)
= 5 - 2
= 3
3.
Recall, π is defined as the ratio of the circumference (say c)
of a circle to its diameter (say d). That is, π =c/ d . This
seems to contradict the fact that π is irrational. How will you resolve this
contradiction?
Answer:
π could be irrational if either c or d is irrational.
4.
Represent 9.3 on the number line.
5. Rationalize the denominators of the following:
(i) 1/√7
1/√7
= (1/√7)x(√7/√7)
= (√7/7)
(ii) 1/(√7 - √6)
1/(√7 - √6)
= (√7 + √6)/(7 - 6)
= (√7 + √6)
(iii) 1/(√5 + √2)
1/(√5 + √2)
= (√5 - √2)/(5 - 2)
= (√5 - √2)/3
(iv) 1/(√7 - 2)
1/(√7 - 2)
= (√7 + 2)/(7 - 4)
=(√7 + 2)/3
1/√7
= (1/√7)x(√7/√7)
= (√7/7)
(ii) 1/(√7 - √6)
1/(√7 - √6)
= (√7 + √6)/(7 - 6)
= (√7 + √6)
(iii) 1/(√5 + √2)
1/(√5 + √2)
= (√5 - √2)/(5 - 2)
= (√5 - √2)/3
(iv) 1/(√7 - 2)
1/(√7 - 2)
= (√7 + 2)/(7 - 4)
=(√7 + 2)/3
EXERCISE
1.6
1.
Find :
\ (i) (64)1/2
(64)1/2
= (82)1/2
=(8)2x1/2
= 8
(ii) (32)1/5
(32)1/5
= (25)1/5
= (2)5x1/5
= 2
(iii) (125)1/3
(125)1/3
=(53)1/3
=(5)3x1/3
= 5
\ (i) (64)1/2
(64)1/2
= (82)1/2
=(8)2x1/2
= 8
(ii) (32)1/5
(32)1/5
= (25)1/5
= (2)5x1/5
= 2
(iii) (125)1/3
(125)1/3
=(53)1/3
=(5)3x1/3
= 5
2.
2.
Find : (i)
(9)3/2
=(32)3/2
=(3)2x3/2
= (3)3
= 27
(ii) (32)2/5
= (25)2/5
= (2)5x2/5
= (2)2
(iii) (16)3/4
= (24)3/4
= (2)4x3/4
= (2)3
= 8
(iv) (125)-1/3
= (53)-1/3
= (5)3x(-1/3)
= (5)-1
= 1/5
(9)3/2
=(32)3/2
=(3)2x3/2
= (3)3
= 27
(ii) (32)2/5
= (25)2/5
= (2)5x2/5
= (2)2
= 4
(iii) (16)3/4
= (24)3/4
= (2)4x3/4
= (2)3
= 8
(iv) (125)-1/3
= (53)-1/3
= (5)3x(-1/3)
= (5)-1
= 1/5
3.
Simplify :
(i) (2)2/3.(2)1/5
=(2)2/3+1/5
=(2)13/15
(ii) (1/33)7
= (3-3)7
=(3)-27
(i) (2)2/3.(2)1/5
=(2)2/3+1/5
=(2)13/15
(ii) (1/33)7
= (3-3)7
=(3)-27
(iii) (11)1/2/(11)1/4
= (11)1/2-1/4
= (11)1/4
(iv) (7)1/2.(8)1/2
= (7)1/2(23)1/2
= (7)1/2.(2)3/2
= (11)1/2-1/4
= (11)1/4
(iv) (7)1/2.(8)1/2
= (7)1/2(23)1/2
= (7)1/2.(2)3/2
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