Which of the following lists consists of all RATIONAL numbers?
A. √81, 9√3, 3.716, 2.444
B. √14, v38, √2/4, 7.235
C. 6, √64, 7.25, 2√25
D. 4.132, √13, 1/3, √49
Which of the following lists consists of all IRRATIONAL numbers?
A. √81, 9√7, 3.716, 2.444
B. √14, √38, √2/4, 7.235
C. 6, √64, 7.25, 2√25
D. 4.132, √13, 1/3, √49
Answers
Answer:
option C
Step-by-step explanation:
all of the above numbers can be expressed as p/q
Prove that the following are irrational:
(i)
2
1
(ii) 7
5
(iii) 6+
2
Medium
Solution
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(i)
2
1
Let us assume
2
1
is rational.
So we can write this number as
2
1
=
b
a
---- (1)
Here, a and b are two co-prime numbers and b is not equal to zero.
Simplify the equation (1) multiply by
2
both sides, we get
1=
b
a
2
Now, divide by b, we get
b=a
2
or
a
b
=
2
Here, a and b are integers so,
a
b
is a rational number,
so
2
should be a rational number.
But
2
is a irrational number, so it is contradictory.
Therefore,
2
1
is irrational number.
(ii) 7
5
Let us assume 7
5
is rational.
So, we can write this number as
7
5
=
b
a
---- (1)
Here, a and b are two co-prime numbers and b is not equal to zero.
Simplify the equation (1) divide by 7 both sides, we get
5
=
7b
a
Here, a and b are integers, so
7b
a
is a rational
number, so
5
should be a rational number.
But
5
is a irrational number, so it is contradictory.
Therefore, 7
5
is irrational number.
(iii) 6+
2
Let us assume 6+
2
is rational.
So we can write this number as
6+
2
=
b
a
---- (1)
Here, a and b are two co-prime number and b is not equal to zero.
Simplify the equation (1) subtract 6 on both sides, we get
2
=
b
a
−6
2
=
b
a−6b
Here, a and b are integers so,
b
a−6b
is a rational
number, so
2
should be a rational number.
But
2
is a irrational number, so it is contradictory.
Therefore, 6+
2
is irrational number.
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textbook
Mathematics
NCERT
Exercise 1.3
SIMILAR QUESTIONS
star-struck
State whether True or False :
All the following numbers are irrationals.
(i)
7
2
(ii)
2
5
3
(iii) 4+
2
(iv) 5
2
Medium
View solution
>
Each of the following numbers is irrational
i) (5+3
2
)
ii) 3
7
iii)
5
3
iv) (2−3
5
)
v) (
3
+
5
)