Question

Question 14. Show that the number 7-2√3 is an irrational number
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Answer 1

Let 3 be a rational number and its simplest form be 

a/b then, a and b are integers having no common factor

other than 1 and b is not equsl to 0.

Now, 3=ba⟹3=b²a²    (On squaring both sides )

or, 3b²=a²         .......(i)

⟹3 divides a²  (∵3 divides 3b²)

⟹3 divides a

Let a=3c for some integer c

Putting a=3c in (i), we get

or, 3b²=9c²⟹b²=3c²

⟹3 divides b2   (∵3 divides 3c²)

⟹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.

Let (7-2√3) be a rational number.

⟹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.

Hence Proved.

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