Prove this one with proper steps pls
Answers
let us assume that root 5 be rational then it will be of the form a by b where a, b are integers and b is not equal to zero again let A and B have no common factor other than one therefore root 5 is equal to a by b where A and B are coprime integers on swelling both the sides we get 5 is equal to a square by b square is equal to 5 a b is equal to b square equation when then I can be run as 5 and where n is an integer on putting a is equal to 5 and in equation when we get 5 b square is equal to 5 and whole square is equal to 5 b square is equal to 10 m square which is equal to b square is equal to 5 and Square 45 / b square and V B from theorem 1-5 is a common factor of a and b by this contradicts that the fact A and B have no common factor other than one the contradiction arises by assuming that root 5 is irrational hence root 5 is irrational
hence proved ...
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√5=a\b
(ab belong to integers and the hcf of ab=1)
squaring both side we get
(√5)²=a²\b²
5b²=a² (eq 1)
This show that a² is divisible by 5
it follo2s that a is divisible by 5. (eq2)
a=5m for some integer m
substituting a=5m in eq 1
5b²=(5m)²=25m²
or
b²=5m²
b² is divisible by 5
and hence b is divisible by 5
ab are common integer
It contradicts our assumption that √5 is
irrational number
Hope it will help u
Thanks u