Question

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Prove that √7 is an irrational number.​

Answer 1

TO PROVE

• √7 is irrational number

PROOF

Lets assume √7 as a rational number

√7 = a/b

• a & b are co-prime numbers

⟶ √7 = a/b

⟶ √7 b = a

• Squaring both the sides

⟶ (√7 b)² =a²

⟶ 7b² = a² _(i)

7 and b² is a factor of a², which means 7 is also a factor of a.

⟶ a =7c _(ii)

• Substitute Eq. (ii) in Eq. (i)

⟶ 7b² = (7c)²

⟶ 7b² = 49c²

⟶ b² = 7c²

7 and c² are factors of b² i.e 7 is a factor of b also.

CONCLUSiON

• 7 is a common factor of both a and b.
• But a & b were co-prime numbers.
• This is a contradiction.
• The contradiction was arisen due to wrong assumption.

★ Hence proved, √7 is irrational.

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Answer 2

Answer:

Given √7

To prove: √7 is an irrational number.

Proof:

Let us assume that √7 is a rational number.

So it t can be expressed in the form p/q where p,q are co-prime integers and q≠0

√7 = p/q

Here p and q are coprime numbers and q ≠ 0

Solving

√7 = p/q

On squaring both the side we get,

=> 7 = (p/q)2

=> 7q2 = p2……………………………..(1)

p2/7 = q2

So 7 divides p and p and p and q are multiple of 7.

⇒ p = 7m

⇒ p² = 49m² ………………………………..(2)

From equations (1) and (2), we get,

7q² = 49m²

⇒ q² = 7m²

⇒ q² is a multiple of 7

⇒ q is a multiple of 7

Hence, p,q have a common factor 7. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number

√7 is an irrational number.