Show whether √3 is rational or irrational.
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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 and b r co-prime numbers.
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
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 and b r co-prime numbers.
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
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Solution :-
☆
suppose √3 is rational, then √3=a/b for some (a,b) suppose we have a/b in simplest form.
√3=ab
a*2 = 3b*2
if b is even, then a is also even in which case a/b is not in simplest form.
if b is odd then a is also odd. Therefore:
ab(2n+1)24n2+4n+12n2+2n2(n2+n)=2n+1=2m+1=3(2m+1)2=12m2+12m+3=6m2+6m+1=2(3m2+3m)+1a=2n+1b=2m+1(2n+1)2=3(2m+1)24n2+4n+1=12m2+12m+32n2+2n=6m2+6m+12(n2+n)=2(3m2+3m)+1
Since (n^2+n) is an integer, the left hand side is even. Since (3m^2+3m) is an integer, the right hand side is odd and we have found a contradiction, therefore our hypothesis is false
Thus we get a contradiction that root 3 is irrational no
☆ ☆ ☆ Hop its helpful ☆ ☆ ☆
☆ Regards :- ♡♡《 Nitish kr singh 》♡♡
Solution :-
☆
suppose √3 is rational, then √3=a/b for some (a,b) suppose we have a/b in simplest form.
√3=ab
a*2 = 3b*2
if b is even, then a is also even in which case a/b is not in simplest form.
if b is odd then a is also odd. Therefore:
ab(2n+1)24n2+4n+12n2+2n2(n2+n)=2n+1=2m+1=3(2m+1)2=12m2+12m+3=6m2+6m+1=2(3m2+3m)+1a=2n+1b=2m+1(2n+1)2=3(2m+1)24n2+4n+1=12m2+12m+32n2+2n=6m2+6m+12(n2+n)=2(3m2+3m)+1
Since (n^2+n) is an integer, the left hand side is even. Since (3m^2+3m) is an integer, the right hand side is odd and we have found a contradiction, therefore our hypothesis is false
Thus we get a contradiction that root 3 is irrational no
☆ ☆ ☆ Hop its helpful ☆ ☆ ☆
☆ Regards :- ♡♡《 Nitish kr singh 》♡♡
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