MATHS CLASS 10 CLICK ON THE PIS ABOVE ANSWER PLZ
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Let √3 be a rational number
√3 = a/b ( where a & b are integers & b is not equal to 0)
squaring both sides
3= a²/b²
=> 3b² = a²
=> a² is divisible by 3
=> a is divisible by 3
a = 3c (where c is an integer)
squaring both sides
a²= (53c)²
a²= 9c²
also,
a² = 3b²
=> 3b² = 9c²
=> b² = 3c²
=> b² is divisible by 3
=> b is divisible by 3
which is a contradiction as a & b are Co prime integers.
=> Our assumption is wrong.
=> √3 is an irrational number.
HOPE IT HELPS YOU ☺☺ !!!
rutujatayade98:
sry....bt i did't understand
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