Math, asked by singh077, 11 months ago

prove that underroot5 is a irrational

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Answered by Anonymous

Step-by-step explanation:

Lets assume that √5 is rational. Then it must in the form of p/q [q is not equal to 0][p and q are co-prime].

√5=p/q

=> √5 * q = p squaring on both sides

=> 5*p^2 = p^2 ------> 1 p^2 is divisible by 5 p is divisible by 5 p = 5c [c is a positive integer] [squaring on both sides ] p^2 = 25c^2 --------- > 2 sub p*p in 1 5*q^2 = 25*c^2 q^2 = 5*c^2

=> q is divisble by 5

thus q and p have a common factor 5 there is a contradiction

as our assumsion p &q are co prime but it has a common factor so

√5 is rational

√5 is an irrational

Answered by Anonymous

Hiii Friend

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Prove That $\sqrt{5}$

Is Irrational

we have to prove

$\sqrt{5}$

is Irrational

Let Use Assume The opposite ,

i.e.,

$\sqrt{5}$

is rational

Hence

$\sqrt{5}$

can be written in the form

$\frac{a}{b}$

where a and b (b not equal 0)are co-prime

(no common factor other than 1)

Hence,

$\sqrt{5 } = \frac{a}{b}$

$\sqrt{5} b= a$

Squaring both sides

$( \sqrt{5} ) {}^{2} = {a}^{2}$

$5 {b}^{2} = {a}^{2}$

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prove that underroot5 is a irrational