Question

proove that root 5 is irrational

Answer 1

let us assume on the contrary that root 5 is a rational number there exist a coprime integer so A and B root 5 equals to a upon B

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

squaring on both side

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

5 = a^2/b^2

5b^2 = a^2

a is divisible by 5................(1)

when a = 5c

we have a^2 = 5b^2

put the value

(5c) ^2 = 5b^2

25c^2 = 5b^2

b^2 = 25/5 c^2

b^2 = 5c^2

b is divisible by 5

so, our supposition is wrong 5 is a common factor between both of them thus,

$\sqrt{5}$

is irrational number........

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Answer 2

Let us assume that

√5 is not irrational

√5 becomes rational

√5 = p/q where p,q belongs to integers and q ≠ 0 and p , q are co-primes .

Squaring on both sides

( √5 ) 2 = ( p/q ) 2

5 = p2 / q2

p2 = 5q2 ===> equation 1

p2 is divisible by 5

p is also divisible by 5

let p = 5k

p2 = 5q2

( 5k )2 = 5 q2

25k2 = 5 q2

5k2 = q2 or 5p2 = q2

q2 = 5p2 ===> equation 2

q2 is divisible by 5

q is also divisible by 5

=> ' 5 ' is the common factor for both p and q .

But p , q are co-primes

It is contradiction to our assumption

Therefore our assumption is wrong

√5 is an irrational .