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prove that (root of 2 + root of 5)2 is a irrational number.

let (root of 2 + root of 5 )2  be rational

2.root of 2 + 2. root of 5 = p by q ( where p & q are integers & not = to zero)

squaring on both sides

(2.root of 2 + 2. root of 5) 2   =  (p by q)2

28 + 8 . root of 10 = p2 by q2

8. root of 10 =p2 - 28q2 by q2

root of 10 = p2 - 28q2 by 8q2

p2,-28q2,8q2 are rational

therefore p2-28q2 by 8q2  is rational

this contradicts our assumption

therefore (root of 2 + root of 5 )2  is irrational

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(root 2 +root 5 )2 

(a+b)2 = a2 + b2 + 2ab    

:. , (root 2)2 + (root 5)2 + 2 (root 2)  ( root 5 )

 = 2 + 5 + 2 ( root 10 )

7 + 2 ( root 10 )

Let assume 7+ 2(root 10) is a rational no.

=> 7 + 2 (root 10) = a / b , where a and b are co - prime no.s , where b ( not equal to) 0

 => 2 (root 10) = a/b - 7

 => 2 (root 10) = a - 7b / b

 => root 10 = a - 7b / 2b

 Since,a and b are integers therefore, a-7b, 2b are also integers . We have b=0 therefore , 2b=0.

root 10 is rational no. which is a contradiction because root is irrational therefore , our assumption is wrong .

So, 7+2 root 5 irrational no. 

HOPE YOU UNDERSTOOD THIS

PLZzZz Thumbs up please

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 (root 2 +root 5 )2

(a+b)2 = a2 + b2 + 2ab

:. , (root 2)2 + (root 5)2 + 2 (root 2) ( root 5 )

= 2 + 5 + 2 ( root 10 )

7 + 2 ( root 10 )

Let assume 7+ 2(root 10) is a rational no.

=> 7 + 2 (root 10) = a / b , where a and b are co - prime no.s , where b ( not equal to) 0

=> 2 (root 10) = a/b - 7

=> 2 (root 10) = a - 7b / b

=> root 10 = a - 7b / 2b

Since,a and b are integers therefore, a-7b, 2b are also integers . We have b ( not equal to ) 0 therefore , 2b ( not equal to ) 0.

root 10 is rational no. which is a contradiction because root is irrational therefore , our assumption is wrong .

So, 7+2 (root 10) irrational no.

HOPE YOU UNDERSTOOD THIS

PLZzZz Thumbs up please

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 sorry there were two mistakes in 1st one

in second 1 all is right

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