Saturday, 5 July 2014

Exercise 1.3 Real Numbers | CBSE Class 10th Mathematics | Solutions

Question 1. Prove that √5 is irrational.

Solution : Let us assume, to the contrary, that √5 is rational.
That is, we can find integers a and b (≠ 0) such that :
a
√5=
b
Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.
So, b √5 = a⋅



Squaring on both sides, and rearranging, we get 5b2 = a2

[Note: As we know that if p is a prime number and If p divides a2, then p also divides a, where a is a positive integer.]
Therefore, a2 is divisible by 5
Then it a is also divisible by 5.
So, we can write a = 5c for some integer c.
Substituting for a, we get 5b2 = 25c2, that is, b2 = 5c2
This means that b2 is divisible by 5, and so b is also divisible by 5
Therefore, a and b have at least 5 as a common factor.
But this contradicts the fact that a and b are coprime.
This contradiction has arisen because of our incorrect assumption that √5 is rational.
So, we conclude that √5 is irrational.

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

Solution : Let us assume, to the contrary, that 3 + 2 √5 is rational.
That is, we can find coprime a and b (b ≠ 0) such that :
a
3 + 2 √5=
b
Therefore,
a
2 √5=
3
b

Or
a3
√5=

2b2

Since a and b are integers, we get :
a3


is rational
2b2
And therefore, √5 is rational.

⋅ But this contradicts the fact that √5 is irrational.
This contradiction has arisen because of our incorrect assumption that 3 + 2 √5 is rational.
So, 3 + 2 √5 is an irrational number.

Question 3. Prove that the following are irrationals :
1
(i)
(ii)7√ 5(iii)6 + √2
√ 2



Solution (i) Let us assume, to the contrary, that
1

is rational
√ 2
That is, we can find coprime a and b (b ≠ 0) such that
1a

=
√ 2b
1×√ 2a

=
√ 2×√ 2b

Or
√ 2a

=
2b

Rearranging, we get :
2a
√ 2=
b
Since a and b are integers,
2a

is rational, So √ 2 is rational
b
But this contradicts the fact that √ 2 is irrational.

So, we conclude that -
1

is irrational
√ 2



Solution (ii) Let us assume, to the contrary, that 7√5 is rational
That is, we can find coprime p and q (q ≠ 0) such that :
p
7√5=
q
Rearranging, we get:
p
√5=
7q
Since p and q are integers,
p

is rational, So √5 is rational
7q
But this contradicts the fact that √ 5 is irrational.

So, we conclude that 7√5 is rational


Solution (iii) Let us assume, to the contrary, that 6 + √2 is rational.
That is, we can find coprime p and q (q ≠ 0) such that :
p
6 + √2=
q
Therefore,
p
√2=
6
q


Since p and q are integers, we get :
p

− 6    is rational
q
And therefore, √2 is rational.

⋅ But this contradicts the fact that √2 is irrational.
Our  assumption that 6 + √2 is rational, is incorrect.
So, 6 + √2 is an irrational number.

Points to remember

  • A number ‘s’ is called irrational if it cannot be written in the form p/q , where p and q are integers and q ≠ 0. Examples : √2, √3, √15, π etc.
  • For p being a prime number , If p divides a2, then p also divides a, where a is a positive integer.
  • The sum or difference of a rational and an irrational number is irrational and
  • The product and quotient of a non-zero rational and irrational number is irrational.

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