how to prove  \sqrt{3} is irrational number ​

Question

how to prove
 \sqrt{3}
is irrational number

in progress 0
Sadie 3 weeks 2021-11-07T01:00:42+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-11-07T01:01:59+00:00

    ANSWER:

    • √3 is an irrational number.

    GIVEN:

    • Number = √3

    TO PROVE:

    • √3 is an Irrational number.

    SOLUTION:

    Let √3 be a rational number which can be expressed in the form of p/q where p and q have no common factor other than 1.

    => √3 = p/q

    => √3q = p

    Squaring both sides , We get;

    => (√3q)² = (p)²

    => 3q² = p². …(i)

    => 3 divides p²

    => 3 divides p ……(ii)

    Let p = 3m in eq(i) we get;

    => 3q² = (3m)²

    => 3q² = 9m²

    => q² = 3m²

    => 3 divides q²

    => 3 divides q. ….(iii)

    From (ii) and (iii)

    => 3 is the common factor of both p and q.

    => Thus our contradiction is wrong.

    => √3 is an irrational number.

    0
    2021-11-07T01:02:22+00:00

    To Prove :

    • √3 is irrational

    Theorem to be used :

    • If ‘p’ is a prime number and ‘p’ divides a² , then ‘p’ divides ‘a’ , where ‘a’ is a positive integer.

    Proof :

    Let us assume , to the contrary , that √3 is rational.

    Therefore , we can define √3 as :

    \sf \implies \sqrt{3} = \dfrac{a}{b} \: \: (where \: a \: and \: b \: are \: coprime ) \\\\ \sf \implies b\sqrt{3} = a

    Squaring both sides we have :

    \sf \implies 3b^{2} = a^{2}

    From above we get 3 divides a² , so 3 also divides ‘a’.

     \sf \implies a = 3m \: \: ( Let \: m \: be \: any \: positive \: integer)

    Again squaring both sides we have :

    \sf \implies a^{2} = (3m)^{2} \\\\ \sf \implies 3b^{2} = 9m^{2} \\\\ \sf \implies b^{2} = 3m^{2}

    From above we get 3 divides b² , so 3 also divides ‘b

    Thus ‘a’ is divisible by 3 and also ‘b’ is divisible by 3. It contradicts the assumption that ‘a’ and ‘b’ are coprime .

    This contradiction has raised due to the incorrect assumption that √3 is rational

    So , it can be concluded that √3 is irrational.

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