anybody ans me fast.....
prove that \/''''' 3 ( root 3 ) is an irrational number.......
lipi:
what do u want to prove, is it 33333333333333333333...*root 3
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Answered by
2
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
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
Answered by
1
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.
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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