Is zero a rational number? Can you write it in the form qp, where p and q are integers and q ≠ 0?
Answer
Yes 0 (zero) is a rational number. We can write it in the form of qp like 10, 20, 30, 40, ........
where p = 0 , q ≠ 0
Question 2
Find six rational numbers between 3 and 4.
Answer
Since we want six rational numbers, we write 3 and 4 as rational numbers with denominator 3 + 4 = 7,
i.e.,
3=73×7=721
and
4=74×7=728
∴ Rational numbers between 3 and 4 are, 722, 723, 724, 725, 726, 727
Question 3
Find five rational numbers between 53 and 54.
Answer
Since we want five rational numbers, we write 53 and 54 as rational numbers with denominator 5 + 1 = 6, i.e.,
53=5×63×6=3018
and
54=5×64×6=3024
∴ Rational numbers between 53 and 54 are, 3019, 3020, 3021, 3022, 3023
Question 4
State whether the following statements are true or false. Give reasons for your answers.
(i) Every natural number is a whole number.
(ii) Every integer is a whole number.
(iii) Every rational number is a whole number.
Answer
(i) True Reason — Whole numbers are = 0, 1, 2, 3, 4, 5,.......∞ Natural numbers are = 1, 2, 3, 4, 5,......∞ From above we see all natural numbers are whole numbers.
(ii) False Reason — Integers can be positive as well as negative i.e., integers are = 0, +(-1), +(-2), +(-3), +(-4)......∞ whereas whole number are 0 or positive only i.e., whole numbers are = 0, 1, 2, 3, 4,.......∞
State whether the following statements are true or false. Justify your answers.
(i) Every irrational number is a real number.
(ii) Every point on the number line is of the form √m , where m is a natural number.
(iii) Every real number is an irrational number.
Answer
(i) True Reason — Every irrational number is a real number, because rational number, natural number, whole number all the numbers comes into real number.
(ii) False Reason — Every point on the number line represents a unique real number.
(iii) False Reason — For example 2 is real but not irrational.
Question 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, square root of all positive integers are not always irrational.
Example = 4=(2×2) = 2, and 2 is a rational number
Question 3
Show how 5 can be represented on the number line.
Answer
Representing 5 as the sum of squares of two natural numbers:
5 = 4 + 1 = (2)2 + (1)2
Let l be the number line. If point O represents number 0 and point A represents number 2 then draw a line segment OA = 2 units.
At A, draw AC ⊥ OA. From AC, cut off AB = 1 unit.
We observe that OAB is a right angled triangle at A. By Pythagoras theorem, we get:
(OB)2 = (OA)2 + (AB)2
(OB)2 = (2)2 + (1)2
(OB)2 = 4 + 1
(OB)2 = 5
OB = 5 units
With O as centre and radius = OB, we draw an arc of a circle to meet the number line l at point P.
As, OP = OB = 5 units, the point P will represent the number 5 on the number line as shown in the figure below:
Exercise 1.3
Question 1
Write the following in decimal form and say what kind of decimal expansion each has :
Hence, 400329 = 0.8225. This is a terminating decimal expansion.
Question 2
You know that 71 = 0.142857. Can you predict what the decimal expansions of 72, 73, 74, 75, 76 are, without actually doing the long division? If so, how?
[Hint : Study the remainders while finding the value of 71 carefully.]
Answer
Given,
71 = 0.142857
(i) 72 = 2×71
= 2×0.142857
= 0.285714
(ii) 73 = 3×71
= 3×0.142857
= 0.428571
(iii) 74 = 4×71
= 4×0.142857
= 0.571428
(iv) 75 = 5×71
= 5×0.142857
= 0.714285
(v)76 = 6×71
= 6×0.142857
= 0.857142
Question 3
Express the following in the form qp, where p and q are integers and q ≠ 0.
(i) 0.6
(ii) 0.47
(iii) 0.001
Answer
(i) 0.6
Let x = 0.6
x = 0.666666........... (1)
Here one digit 6 is repeated so we multiply both side by 10 in equation (1)
10x = 6.6666.......
we can write
10x = 6.66666.......
10x = 6 + 0.666666...........
From equation (1)
10x = 6 + x
10x - x = 6
9x = 6
x = 96 = 32
Hence, 0.6 = 32
(ii) 0.47
Let x = 0.47777777........ (1)
Here one digit 7 is repeated so we multiply both side by 10 in equation (1)
We can write
10x = 4.77777..... (2)
On subtracting equation (2) from equation (1)
10x - x = 4.777777........ - 0.477777.........
9x = 4.300000......
x = 94.3
Multiplying numerator and denominator by 10 we get,
x = 9×104.3×10 = 9043
Hence, 0.47 = 9043
(iii) 0.001
Let x = 0.001001001...... (1)
Here three digit 001 are repeating so we multiply both side by 1000 in equation (1)
1000x = 0001.001001........ (2)
On subtracting equation (2) from equation (1)
1000x - x = 0001.001001...... - 0.001001......
999x = 1.0000000.....
x = 9991
Hence, 0.001 = 9991
Question 4
Express 0.99999 .... in the form qp. Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.
Answer
Let x = 0.99999 ........ (1)
Here one digit 9 is repeating so we multiply both side by 10 in equation (1)
10 x = 9.99999.......... (2)
By subtracting equation (2) - equation (1)
10 x -x = 9.99999........ - 0.99999........
9x = 9
x = 99
x = 1
Therefore, (0.99999...) is equivalent to (1), which might seem surprising at first glance, but it makes sense mathematically. This result can be demonstrated by the fact that any number infinitesimally close to 1 but less than 1, when infinitely added to itself, equals 1. For example, if we take (0.9) and add (0.09), we get (0.99), and if we add (0.009) to that, we get (0.999), and so on. As we keep adding these infinitesimal increments, we approach (1). So, in a way, (0.99999...) is just another way to represent the number (1).
Question 5
What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 171 ? Perform the division to check your answer.
Look at several examples of rational numbers in the form qp (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
Suppose, qp = 52 = 0.4
or
109 = 0.9
or
23 = 1.5
q must contain factors of 2 or 5 or both.
For example,
In 87, 8 = 2 x 2 x 2
and
in 107, 10 = 2 x 5
Question 7
Write three numbers whose decimal expansions are non-terminating non-recurring.
Answer
π = 3.1415......
2 = 1.414213......
3 = 1.732050......
Question 8
Find three different irrational numbers between the rational numbers 75 and 119 .
Irrational numbers between the rational numbers 75 and 119
0.75075007500075000075. . .
0.767076700767000767. . .
0.808008000800008. . .
Question 9
Classify the following numbers as rational or irrational :
(i) 23
(ii) 225
(iii) 0.3796
(iv) 7.478478...
(v) 1.101001000100001...
Answer
(i) 23 is a prime number. It is irrational because 23 does not have a complete root.
(ii) 225 is not a prime number. 15 is root of 225 so it is rational.
(iii) 0.3796 is terminating decimal expansion. Hence, it is rational.
(iv) 7.478478... = 7.478, it is non-terminating but repeating decimal expansion. Hence, it is rational.
(v) 1.101001000100001... is non-terminating and non-repeating decimal expansion. Hence, it is irrational.
Exercise 1.4
Question 1
Classify the following numbers as rational or irrational:
(i) 2 − 5
(ii) 3 + 23 − 23
(iii) 7727
(iv) 21
(v) 2π
Answer
(i) 2 − 5
If in any rational number we add, subtract, multiply or divide any irrational number it becomes irrational number.
Hence, 2 − (5) is an irrational number.
(ii) 3 + 23 − 23
= 3 + 23 − 23
= 3
Hence, (3 + 23) − 23 is a rational number
(iii) 7727
7727 = 72
It is in qp where q ≠ 0.
Hence, is a rational number
(iv) 21
21=irrationalrational = irrational number
If in any rational number we add, subtract, multiply or divide any irrational number it becomes irrational
Hence, 21 is an irrational number
(v) 2π
2 x π = irrational number [∵ rational x irrational number = irrational number]
Hence, 2π is an irrational number
Question 2(i)
Simplify the following expression:
(3+3)(2+2)
Answer
(3+3)(2+2)
= 3 x 2 + 3 x 2 + 2 x 3 + 3 x 2
= 6 + 3 2 + 2 3 + 6
Hence, (3+3)(2+2)=6+32+23+6
Question 2(ii)
Simplify the following expression:
(3+3)(3−3)
Answer
(3+3)(3−3)
= (3)2 - (3)2 [∵ (a + b)(a - b) = (a)2 - (b)2]
= 9 - 3
= 6
Hence, (3+3)(3−3) = 6
Question 2(iii)
Simplify the following expression:
(5+2)2
Answer
(5+2)2
= (a + b)2
= a2 + 2ab + b2
= (5)2 + 2 x (5) x (2) + (2)2 [∵ (a + b)2 = a2 + 2ab + b2]
= 5 + 2 (10) + 2
= 7 + 2 (10)
Hence,(5+2)2 = 7 + 2 (10)
Question 2(iv)
Simplify the following expression:
(5 - 2)(5 + 2)
Answer
(5 - 2)(5 + 2)
= a2 - b2
= (5)2−(2)2 [∵ (a + b)(a - b) = (a)2 - (b)2]
= 5 - 2
= 3
Hence, (5 - 2)(5 + 2) = 3
Question 3
Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, π = dc. This seems to contradict the fact that π is irrational. How will you resolve this contradiction?
Answer
In the question given π is irrational, π = dc There is no contradiction. When we measure a length with a scale or any other device we only get an approximate rational value. So, we may not realize that either c or d is irrational.
Question 4
Represent 9.3 on the number line.
Answer
Steps:
Draw a line and take AB = 9.3 units on it.
From B, measure a distance of 1 unit and mark C on the number line. Mark the midpoint of AC as O.
With 'O' as center and OC as radius, draw a semicircle.
At B, draw a perpendicular to cut the semicircle at D.
With B as center and BD as radius draw an arc to cut the number line at E. Thus, taking B as origin the distance BE = 9.3
Hence, point E represents 9.3 on the number line.