Prove that 3/2root5 is irrational
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let suppose root5 is rational
therefore. root5=a/b(a and b has only 1 as it common factor)
thus,a=5b
sq.both sides,we get
(a)
![a^{2} = {5b}^{2} \: \: \: \: \: \: (i) a^{2} = {5b}^{2} \: \: \: \: \: \: (i)](https://tex.z-dn.net/?f=+a%5E%7B2%7D++%3D+%7B5b%7D%5E%7B2%7D+++%5C%3A++%5C%3A++%5C%3A++%5C%3A+%5C%3A++%5C%3A+%28i%29)
so a=5c (c is an another integer)
sq.both side,we get
![{a}^{2} = 25c {a}^{2} = 25c](https://tex.z-dn.net/?f=+%7Ba%7D%5E%7B2%7D++%3D+25c)
also a=5b (from i)
thus,5b=25 c
b=5c which is not possible as b has only 1 as its common factor
so our supposition is wrong
therefore,
3/2root5 is irrational
therefore. root5=a/b(a and b has only 1 as it common factor)
thus,a=5b
sq.both sides,we get
(a)
so a=5c (c is an another integer)
sq.both side,we get
also a=5b (from i)
thus,5b=25 c
b=5c which is not possible as b has only 1 as its common factor
so our supposition is wrong
therefore,
3/2root5 is irrational
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