Math, asked by 7302314472, 10 months ago



prove that
prove that  \sqrt{3}  is irrational

Answers

Answered by JaiJT
0

Step-by-step explanation:

The contradiction arises by assuming 3 is a rational. Hence, 3 is irrational. If possible, Let (7+23 ) be a rational number. ∴ −23 is a rational.

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Answered by rohith7718
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.

Step-by-step explanation:

put √3 in place of 3

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