try question 8
try question 7plz
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the answer of the Q no.7 is in the picture
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can u try 8 also
actually i can't
ok thanks for 7th question
no no it's ok
my pleasure
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Hello dear,
suppose root 3 is a rational no.
such that root 3 = a/b , where a and b both r integers and b = nonzero integer 'a ' and 'b ' have no common factor other than 1
therefore root 3 = a/b .......where a nad b r coprime nos.
therefore , b root 3 = a
therefore, 3b2 = a2 (squaring both sides)..... (1)
therefore , b2 = a2/3
therefore, 3 divides a2 ,so 3 divides a
so we write, a = 3c..........where 'c ' is an integer.....(2)
a2 = (3c)2 .....squaring both sides
therefore, 3b2 = 9c2 .........substuting (2) in (1)
therefore, b2 = 3 c2
therefore, c2 = b2 / 3
3 divides b2 , means 3 divides b
therefore 'a ' and 'b ' have at least 3 as common factor but it was stated before that it was stated that a and b had no common factors other than 1.
this contradiction arises because we have assumed that root 3 is rational. therefore, root 3 is irritional no.
hope its help you dear
suppose root 3 is a rational no.
such that root 3 = a/b , where a and b both r integers and b = nonzero integer 'a ' and 'b ' have no common factor other than 1
therefore root 3 = a/b .......where a nad b r coprime nos.
therefore , b root 3 = a
therefore, 3b2 = a2 (squaring both sides)..... (1)
therefore , b2 = a2/3
therefore, 3 divides a2 ,so 3 divides a
so we write, a = 3c..........where 'c ' is an integer.....(2)
a2 = (3c)2 .....squaring both sides
therefore, 3b2 = 9c2 .........substuting (2) in (1)
therefore, b2 = 3 c2
therefore, c2 = b2 / 3
3 divides b2 , means 3 divides b
therefore 'a ' and 'b ' have at least 3 as common factor but it was stated before that it was stated that a and b had no common factors other than 1.
this contradiction arises because we have assumed that root 3 is rational. therefore, root 3 is irritional no.
hope its help you dear
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