Question

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

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

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.

Answer 2

Answer:

${\huge{\bf{\purple{\mathbb{\underline {Question:-}}}}}}$

Proof that √3 is irrational

${\huge{\bf{\purple{\mathbb{\underline {Proof:--}}}}}}$

Let us assume on the contrary that

√3 is a rational number.

Then, there exist positive integers a and b such that

√3=a/b

where, a and b, are co-prime i.e. their HCF is 1

Now,

on squaring both side,

3=a²/b²

3b²=a²

3 divides a². [∵3 divides 3b²]

3 divides a. ..(eq i)

a=3c for some integer c

a²=9c²

3b²=9c². [a²=3b²]

b²=3c²

3 divides b².

3 divides b. ..(eq ii)

From (i) and (ii), we observe that a and b have at least 3 as a common factor. But, this contradicts the fact that a and b are co-prime. This means that our assumption is not correct.

Hence,

√3 is an irrational number.