Answers

2014-06-08T18:54:12+05:30
Let v!3 be rational number                (NOTE ! is denoted for root)
v!3=a%b                                                sq=square of the variable in bracket
!3a=vb

squarring on both sides
3(a)sq=(vb)sq
a&(a)sq factors  vq                     (let (a)sq =c)
 3c=(vb)sq   
c&(c)sq also factors vq
it contradicts our statement it has two factors
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2014-06-08T19:04:24+05:30
Let us assume to the contrary that √3 is a rational number.
Therefore √3 = p/q ,where p & q are co-prime & q≠0
So,         (√3)²= p²/q²
⇒              3   = p²/q²
⇒            3*q²= p²
So p² is divisible by 3,so p is also divisible by 3
⇒p² is divisible by 9.
                3    = p²/q²
⇒             q²   = p²/3
So,q² is also divisible by 3.
So p & q are not co-prime.
Therefore √3 is an  irrational number.
      

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