prove that 2√3 is irrational number
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irrational number multiplied by rational number gives the product as irrational. 2 is rational multipled by
![\sqrt{ 3 } \sqrt{ 3 }](https://tex.z-dn.net/?f=+%5Csqrt%7B+3+%7D+)
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let us assume that 2√3 is a rational no.
2√3=a/b(where a Nd b are co prime no.)
squaring both side
12=a^2/b^2
12b^2=a^2............k
12 is a factor of a^2
12 will be factor of a.....................1
let a=12c
squaring both side
a^2=144c^2...................ll
putting ll in k
12b^2=144c^2
b^2=12c^2
12 is a factor of b^2
12 will be factor of b..............lll
from 1&lll
12 is the factor of both a&b
so, they are not co prime no.
hence our assumption is wrong
2√3 is an irrational no.
2√3=a/b(where a Nd b are co prime no.)
squaring both side
12=a^2/b^2
12b^2=a^2............k
12 is a factor of a^2
12 will be factor of a.....................1
let a=12c
squaring both side
a^2=144c^2...................ll
putting ll in k
12b^2=144c^2
b^2=12c^2
12 is a factor of b^2
12 will be factor of b..............lll
from 1&lll
12 is the factor of both a&b
so, they are not co prime no.
hence our assumption is wrong
2√3 is an irrational no.
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