Question

Prove that the following numbers are irrational root 7​

Answer 1

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.

Answer 2

Answer↷

•To Prove: √7 is an irrational number.

•Proof: Let us assume that √7 is a rational number.

So...

$\sqrt{7} = \frac{a}{b}$

•where a and b are Coprime integers.

•Squaring Both sides:

${ \sqrt{7} }^{2} = \frac{ {(a)}^{2} }{ {(b)}^{2} } \\ 7 = \frac{ {a}^{2} }{ {b}^{2} } \\ {b}^{2} = \frac{ {a}^{2} }{7}$

•This implies that 7 divides a², so it will also divide a.

•Let us replace a by 7 c.

${b}^{2} = \frac{ {(7c)}^{2} }{7} \\ {b}^{2} = \frac{49 {c}^{2} }{7} \\ {b}^{2} = {7c}^{2} \\ {c}^{2} = \frac{ {b}^{2} }{7}$

• This implies that 7 divides b², so it will also divide b.

• This results that our assumption was incorrect that a and b were Coprime integers as 7 divides both of them.

Hence √7 is an irrational number.

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