Question

1) prove
that
√3
is irrational​

Answer 1

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 ) be a rational number.

√3

⟹7−(7+2 ) is a rational

√3

∴ −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.

Answer 2

Answer:

let us assume that √3 is a rational number(contradiction). ... 3 will be a factor of both p and q which contradicts are statement. that they bioth have a hcf of 1. so we make a conclusion that although 2 is rational but √3 is irrational.