1.

Prove that √3 is an irrational number and then prove that 7 – 2√3 is an irrational number.

Answer»

Answer:

If possible , let

3

be a rational number and its simplest form be

b

a

then, a and b are INTEGERS having no common factor

other than 1 and b

=0.

Now,

3

=

b

a

⟹3=

b

2

a

2

(On SQUARING both sides )

or, 3B

2

=a

2

.......(i)

⟹3 divides a

2

(∵3 divides 3b

2

)

⟹3 divides a

Let a=3c for some integer c

Putting a=3c in (i), we get

or, 3b

2

=9c

2

⟹b

2

=3c

2

⟹3 divides b

2

(∵3 divides 3c

2

)

⟹3 divides a

Thus 3 is a common factor of a and b

This contradicts the fact that a and b have no common factor other than 1.

The contradiction ARISES by assuming

3

is a rational.

Hence,

3

is irrational.

2

nd

part

If possible, Let (7+2

3

) be a rational number.

⟹7−(7+2

3

) is a rational

∴ −2

3

is a rational.

This contradicts the fact that −2

3

is an irrational number.

Since, the contradiction arises by assuming 7+2

3

is a rational.

Hence, 7+2

3

is irrational.

Proved.


Discussion

No Comment Found

Related InterviewSolutions