That took my a while but I think I got it. Kudos to you! Thanks!
The most important passage is the first one: x^2 < x - |y| from this I divide the function into 2 cases:
y>0 => x^2<x-y
y<0 => x^2<x+y
I want to make this clear, so now:
1)It's an useful technique! When you have 2 inequalities, you know what they say taken on their own, but what do they say together?
You can sum(not subtract) inequalities as long as they have the same operator (> or <)
2)I divided the first equation into two cases : y>0 and y<0
Now I take case y>0:
x^2<+x-y, sum (literally sum) the equation of statement 1
-y<-x, and I obtain this:
x^2+(-y)<(-x)+x-y, now I semplify its form into x^2<0. So in this scenario the equation is never true, 0 cannot be greater than a squared number.
Now I repeat the same passages for the other case y<0:
x^2<+x+y, sum the equation of statement 1
-y<-x, and I obtain this:
x^2+(-y)<(-x)+x+y, and I get x^2<2y. But since in this scenario I am considering y<0 the right term 2y will be negative.
So once more, can a negative number be greater than a squared number? x^2<-ve? No, so also in this scenario the answer to the main question is NO.
To summarize what we did here: we have divided the main question into two cases, we have studied each case on its own, and since the answer to the main question is the same (NO in both) we can say that statement 1 is sufficient
Zarrolou wrote:
WholeLottaLove wrote:
Hi Zarrolou!
Could you run me through this:
(i) y >x
or -y<-x (we need to have the same operator to sum inequalities). Sum this to each case:
I)x^2+(-y)<(-x)+x-y or x^2<0 Never true
II)x^2+(-y)<(-x)+x+y or x^2<2y, but since in this case we are considering y<0 that equation is never true
x^2<-ve Never true.
Sufficient
Firstly, why do we plug in -y<-x? I get where that comes from, but why do we use it here?
Secondly, where do we get x^2+(-y)<(-x)+x-y from? In other words, how does x^2 < x - |y| translate to x^2+(-y)<(-x)+x-y?
Thanks!
Could you run me through this:
(i) y >x
or -y<-x (we need to have the same operator to sum inequalities). Sum this to each case:
I)x^2+(-y)<(-x)+x-y or x^2<0 Never true
II)x^2+(-y)<(-x)+x+y or x^2<2y, but since in this case we are considering y<0 that equation is never true
x^2<-ve Never true.
Sufficient
Firstly, why do we plug in -y<-x? I get where that comes from, but why do we use it here?
Secondly, where do we get x^2+(-y)<(-x)+x-y from? In other words, how does x^2 < x - |y| translate to x^2+(-y)<(-x)+x-y?
Thanks!
The most important passage is the first one: x^2 < x - |y| from this I divide the function into 2 cases:
y>0 => x^2<x-y
y<0 => x^2<x+y
I want to make this clear, so now:
1)It's an useful technique! When you have 2 inequalities, you know what they say taken on their own, but what do they say together?
You can sum(not subtract) inequalities as long as they have the same operator (> or <)
2)I divided the first equation into two cases : y>0 and y<0
Now I take case y>0:
x^2<+x-y, sum (literally sum) the equation of statement 1
-y<-x, and I obtain this:
x^2+(-y)<(-x)+x-y, now I semplify its form into x^2<0. So in this scenario the equation is never true, 0 cannot be greater than a squared number.
Now I repeat the same passages for the other case y<0:
x^2<+x+y, sum the equation of statement 1
-y<-x, and I obtain this:
x^2+(-y)<(-x)+x+y, and I get x^2<2y. But since in this scenario I am considering y<0 the right term 2y will be negative.
So once more, can a negative number be greater than a squared number? x^2<-ve? No, so also in this scenario the answer to the main question is NO.
To summarize what we did here: we have divided the main question into two cases, we have studied each case on its own, and since the answer to the main question is the same (NO in both) we can say that statement 1 is sufficient