Question

Prove that √3 a an number. Irrational number​

Answer 1

Answer:

Let us assume on the contrary that

3

is a rational number.

Then, there exist positive integers a and b such that

3

=

b

a

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

Now,

3

=

b

a

⇒3=

b

2

a

2

⇒3b

2

=a

2

⇒3 divides a

2

[∵3 divides 3b

2

]

⇒3 divides a...(i)

⇒a=3c for some integer c

⇒a

2

=9c

2

⇒3b

2

=9c

2

[∵a

2

=3b

2

]

⇒b

2

=3c

2

⇒3 divides b

2

[∵3 divides 3c

2

]

⇒3 divides b...(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.

Answer 2

Question :-

• Prove that √3 is an Irrational number.

Answer :-

• Let us assume that √3 is rational

• Hence, √3 = ${ \sf{ \frac{p}{q} }}$

where p and q are co - prime and q ≠ 0

$\dashrightarrow{ \sf{ \sqrt{3} = \frac{p}{q} }}$

$\dashrightarrow{ \sf{ \ \sqrt{3}q = p}}$

${ \dashrightarrow{ \sf{Now, \: squaring \: both \: sides}}}$

$\dashrightarrow{ \sf{ \ (\sqrt{3}q)^{2} = p ^{2} }}$

$\dashrightarrow{ \sf{ \ {3}q^{2} = p ^{2} }}$

$\dashrightarrow{ \sf{ \ \frac{ \: p ^{2} }{3} = q ^{2} }}$

• Here, 3 divides p² and 3 also divide p ------ (1 Eq.)

• P = 3 K where k, is any positive integer.

• Putting P = 3k

$\dashrightarrow{ \sf{3q ^{2} = (3k)^{2}}}$

$\dashrightarrow{ \sf{3q ^{2} = 9k^{2}}}$

$\dashrightarrow{ \sf{q^{2} = \frac{1}{3} \times 9k^{2} }}$

$\dashrightarrow{ \sf{q^{2} = \frac{1}{ \cancel3} \times { \cancel9}k^{2} }}$

$\dashrightarrow{ \sf{q^{2} = 3k^{2} }}$

$\dashrightarrow{ \sf{ \ \frac{q ^{2} }{3} = k^{2} }}$

• Here, 3 divides q² and 3 also divide q ------ (2 Eq.)

From (i) and (ii) Eq., we observe that p and q have at least 3 as a common factor. But, this contradicts the fact that p and q are co-prime. So, this contradiction is arisen because of our wrong assumption.

Hence, we conclude that √3 is an Irrational number.