$X+Y <0$ and $Y - X >0$. Can we determine which of $Y$ and $0$ is bigger? (GRE question)

Question

Given $X + Y < 0$, and $Y - X > 0$. Let quantity $A$ be $Y$, and quantity $B$ be $0$. Which of the following is true:

A. $A$ is always greater.

B. $B$ is always greater.

C. both are always the same.

D. can't be determined.

The given answer is D but I can't see the reason why? The following is my approach.

I first flip the second inequality so I get $X+Y<0$ and $X-Y<0$.

Then I subtract inequalities, giving $2y < 0$. Therefore I would get $y<0$.

So according to me the answer is B because $y$ is less than $0$.

Can anyone explain why I am wrong because the solution in book states the answer is D.

Answer 1

From the inequalities:

$$X + Y < 0$$ and $$X - Y < 0$$

you can add these two and thereby get:

$$2X < 0$$ from which it follows that $$X < 0.$$

Note that you cannot subtract, as follows: $$X + Y < 0$$ $$-(X - Y) > 0$$ because the inequality signs will be different. (Furthermore, note that subtracting reverses the sign of the inequality.)

Answer 2

As has been pointed out, you can't subtract one inequality from another, but you can add two inequalities.

To see why we can add inequalities, suppose $a < b$ and $c < d$. Adding $c$ to both sides of the first inequality, we have $a + c < b + c$. Now adding $b$ to both sides of the second inequality, we have $b + c < b + d$. Combining these two new inequalities, we have $a + c < b + d$, which is what some people call the sum of the two inequalities.

Why can't we do the same thing for subtracting inequalties? Well, let's try it and see what happens. Again, suppose $a < b$ and $c < d$. The subtraction inequality would compare $a-c$ and $b - d$, so let's try and get $a - c$ first. We can do this by subtracting $c$ from both sides of the first inequality, which gives $a - c < b - c$. As above, we will now try to show that the quantity of the right hand side of this new inequality is less than something else. To do this, multiply the second inequality by $-1$ to introduce the $-c$ term; this gives $-c > - d$. Now we add $b$ to both sides which results in $b - c > b -d$. Here's the problem. In the addition case we had the inequalities

$$a + c < b + c\qquad\text{and}\qquad b + c < b +d$$

which together implied $a + c < b + d$ by the transitivity property of $<$. However, in the subtraction case we have the two inequalities

$$a - c < b - c\qquad\text{and}\qquad b - c > b - d.$$

These two inequalities cannot be combined. We cannot compare $a - c$ and $b - d$, which is what the subtraction of inequalities would do if it worked. It could be the case that

• $a - c < b - d$ (i.e. $a = 0, b = 2, c = 0, d = 1$),
• $a - c > b - d$ (i.e. $a = 0, b = 1, c = 0, d = 2$), or
• $a - c = b - d$ (i.e. $a = 0, b = 1, c = 0, d = 1$).

This is why subtraction of inequalities fails.