Answer:
let√5 be a rational number
√5=p\q
(squaring both the side)
√5 =p2q2 = p2= 5q2
:: p2 is an even number
:: p is an even number
p2= 2r
p2= 4r2
2r2
:: q2 is an even number,so q is an even number
this contradice the fact that √5 is an rational number
√5 ia an irrational number
Step-by-step explanation:
Let us assume for contrary that root 5 is a rational. Then it can be written in a/b form where a and b are co prime no.
root5= a/b
Squaring both sides
5=a^2/b^2
a^2=5b^2
Here, a^2 is divisible by 5 then a is also divisible by 5.
Put a=5c
(5c)^2=5b^2
25c^2=5b^2
25c^2/5=b^2
5c^2=b^2
Here,b^2 is divisible by 5 then b is also divisible by 5.
Therefore, a and b have common factor 5. There is a contradiction has arisen that a and b are co-prime no.
This contradiction has arisen because of our wrong assumption that root 5 is rational. So we conclude that root 5 is irrational.
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Answers & Comments
Answer:
let√5 be a rational number
√5=p\q
(squaring both the side)
√5 =p2q2 = p2= 5q2
:: p2 is an even number
:: p is an even number
p2= 2r
p2= 4r2
2r2
:: q2 is an even number,so q is an even number
this contradice the fact that √5 is an rational number
√5 ia an irrational number
Step-by-step explanation:
Let us assume for contrary that root 5 is a rational. Then it can be written in a/b form where a and b are co prime no.
root5= a/b
Squaring both sides
5=a^2/b^2
a^2=5b^2
Here, a^2 is divisible by 5 then a is also divisible by 5.
Put a=5c
(5c)^2=5b^2
25c^2=5b^2
25c^2/5=b^2
5c^2=b^2
Here,b^2 is divisible by 5 then b is also divisible by 5.
Therefore, a and b have common factor 5. There is a contradiction has arisen that a and b are co-prime no.
This contradiction has arisen because of our wrong assumption that root 5 is rational. So we conclude that root 5 is irrational.