prove that root 3 + root 5 is irrational no.
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To prove: √3+√5 is irrational
To prove it let us assume it to be a rational number
Rational numbers are the ones which can be expressed in p/q form where p,q are integers and q isn't equal to 0.
√3+√5=p/q
√3=(p/q)-√5
Squatting on both sides.
3=p²/q²-(2√5p)/q+5
(2√5p)/q=5-3-p²/q²
2√5p/q=(2q²-p²)/q²
√5=(2q²-p²)/q²*q/2p
√5=(2q²-p²)/2pq
As p and q are integers RHS is rational
As RHS is rational LHS is also rational i.e √5 is rational
But this contradicts the fact that √5 is irrational.
This contradiction arosebecause of our false assumptionSo
√3+√5 is irrational.
To prove it let us assume it to be a rational number
Rational numbers are the ones which can be expressed in p/q form where p,q are integers and q isn't equal to 0.
√3+√5=p/q
√3=(p/q)-√5
Squatting on both sides.
3=p²/q²-(2√5p)/q+5
(2√5p)/q=5-3-p²/q²
2√5p/q=(2q²-p²)/q²
√5=(2q²-p²)/q²*q/2p
√5=(2q²-p²)/2pq
As p and q are integers RHS is rational
As RHS is rational LHS is also rational i.e √5 is rational
But this contradicts the fact that √5 is irrational.
This contradiction arosebecause of our false assumptionSo
√3+√5 is irrational.
Answered by
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hey!
here is the answer
this regards by utsav
√3+√5 is irrational
prove that ....
solution :
let √3+√5 be a rational number, then 3+√5 = a/b where a and b are integers and b is not equal to zero..
squaring both side , we get
( √3+√5)^2 =(a/b)^2
°-° let's see what happens
here we use the formula (a+b)^2
then (√3)^2 +2×√3×√5+(√5)^2 =a^2/b^2
3+2√15+5=a^2/b^2
8+2√15 = a^2/b^2
2√15= a^2/b^2 -8
2√15= a^2-8b^2/b^2
√15= a^2-8b^2 /2b^2
here a^2-8b^2/2b^2 is a rational number
but √15 is irrational number
so,the contradiction we supposed is wrong
hence , √3+√5 is an irrational number
hope it helps
thanx
here is the answer
this regards by utsav
√3+√5 is irrational
prove that ....
solution :
let √3+√5 be a rational number, then 3+√5 = a/b where a and b are integers and b is not equal to zero..
squaring both side , we get
( √3+√5)^2 =(a/b)^2
°-° let's see what happens
here we use the formula (a+b)^2
then (√3)^2 +2×√3×√5+(√5)^2 =a^2/b^2
3+2√15+5=a^2/b^2
8+2√15 = a^2/b^2
2√15= a^2/b^2 -8
2√15= a^2-8b^2/b^2
√15= a^2-8b^2 /2b^2
here a^2-8b^2/2b^2 is a rational number
but √15 is irrational number
so,the contradiction we supposed is wrong
hence , √3+√5 is an irrational number
hope it helps
thanx
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