If x < y, is square(x) < square(y)?
(1) y > 0
(2) x > 0
Could someone solve it algebricaly please.
[/list]
(1) y > 0
(2) x > 0
Could someone solve it algebricaly please.
[/list]
my humble try
if x<y is x^2<y^2, after little transforming
is x^2-y^2<0
is (x-y)(x+y)<0. two cases are possible
1) x-y>0 x>y but accrding to the problem x<y so
2) x-y<0, x<y and x+y>0 and now the question: is x+y>0
(1) y>0 if y>0 x can be also >0 or <0 not suffiicent
x=2,y=3 x+y>0
y=3, x=-4 x+y<0
(2) is x>0 then y must be>0 as x<y and x+y will be always +ve
so my pick for B
