Can any one prove that root 2 is a irrational number....
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Hi!
here is your answer
Let,
√2 be a rational number.
.°. √2 = a/b , where a and b ( b ≠ 0 ) is a integer .
now ,
√2 = a/b
=> 2b² = a²
=> 2 | a² [ °.° 2 | 2b² ]
=> 2 | a ( from law ) -----------( 1 )
=> a = 2c , any integer for c
=> a² = 4c²
2b² = 4c² [ °.° a² = 2b² ]
=> b² = 2c²
=> 2 / b² [ °.° 5 / 5c² ]
=> 2 / b ( from law )--------------( 2 )
now from 1 and 2 we get that a and b have only one general factor like" 5 " .
Which is against our thinks
,°. √2 is an irrational number
here is your answer
Let,
√2 be a rational number.
.°. √2 = a/b , where a and b ( b ≠ 0 ) is a integer .
now ,
√2 = a/b
=> 2b² = a²
=> 2 | a² [ °.° 2 | 2b² ]
=> 2 | a ( from law ) -----------( 1 )
=> a = 2c , any integer for c
=> a² = 4c²
2b² = 4c² [ °.° a² = 2b² ]
=> b² = 2c²
=> 2 / b² [ °.° 5 / 5c² ]
=> 2 / b ( from law )--------------( 2 )
now from 1 and 2 we get that a and b have only one general factor like" 5 " .
Which is against our thinks
,°. √2 is an irrational number
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