Question

prove that √3 is an irrational number​

Answer 1

Answer:

Yes, √3 is an irrational number

Step-by-step explanation:

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

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Answer 2

Answer:

Let us consider √3 as a rational number

  therefore √3 can be written as in the form of p/q

           √3 = p/q

On squaring both sides we get

                 3 = $p^{2} / q^{2}$

                 3$p^{2}$ = $q^{2}$

that is 3 divides  $q^{2}$ and q

            therefore we can consider that

                      q = 3c [where value of 3 be any integer]

                     $q^{2}$  =9$c^{2}$

 we know that    3$p^{2}$ = $q^{2}$

                         3$p^{2}$ = 9$c^{2}$

                           $p^{2}$ = 3$c^{2}$

so 3 divides $p^{2}$ and p

therefore we can observe that p and q has a common factor 3.

this contradicts the fact that a and b are co-prime. This means that our assumption is not correct.

Hence √3  is irrational.