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Question

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Prove that √7 is an irrational number.​

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Iris 10 months 2021-07-07T15:51:52+00:00 2 Answers 0 views 0

Answers ( )

  1. Everleigh
    0
    2021-07-07T15:53:02+00:00
    Reply

    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.

  3. Lydia
    0
    2021-07-07T15:53:25+00:00
    Reply

    To prove : \sf{\sqrt{7}} is irrational.

    Proof :

    Assume that \sf{\sqrt{7}} is rational.

    \sf{ \sqrt{7} = \dfrac{a}{b} \: - - - \gray{(a \: and \: b \: are \: coprime \: numbers)}}

    \sf{\longrightarrow} \: \sqrt{7} = \dfrac{a}{b}

    \sf{\longrightarrow} \: \sqrt{7}b = a

    \sf{\longrightarrow} \: (\sqrt{7}b)^{2} = {a}^{2} - - \gray{(Squaring \: both \: the \: sides)}

    \sf{\longrightarrow} \: {7b}^{2} = {a}^{2} - - -\gray{ (1)}

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

    \sf{\longrightarrow} \:a = 7c - - -\gray{(2)}

    Substitute Eq. 2 in Eq. 1

    \sf{\longrightarrow} \: {7b}^{2} = {(7c)}^{2}

    \sf{\longrightarrow} \:7 {b}^{2} = 49 {c}^{2}

    \sf{\longrightarrow} \: {b}^{2} = {7c}^{2}

    7 and c² are factors of b². Which means 7 is a factor of b too.

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

    Hence proved, \sf{\sqrt{7}} is irrational.

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