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Can someone please help me understand how to change the sign

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Joined: 10 Apr 2008
Posts: 53
Can someone please help me understand how to change the sign [#permalink]

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New post 19 Apr 2009, 21:33
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

I am very confused about how sign changes work in inequalities.
I've studied the MGMAT guides but it hasn't helped.

For example:\(x/y > 1 => x > y\)

y could be negative and I want to multiply both sides by y.
Is it enough to do:
if y < 0
then x < y ??

Or should I be adding a negative sign somewhere?

What if there is something like \(2xy < x^2\)
I can't do 2y < x right?
Also, here is safe to assume that x is not equal to 0?
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Joined: 21 Apr 2009
Posts: 10
Location: DC
Re: Can someone please help me understand how to change the sign [#permalink]

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New post 21 Apr 2009, 15:48
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Hey there, Thinkblue!

What's funny about these questions is that they touch upon not only inequality rules, but also division by zero and absolute value.

First, some general rules, some of which you know, some of which may be new:
R1) You may add or subtract to both sides of an inequality as normal; no flipping involved.
R2) When you multiply or divide both sides of an inequality by a positive number, then nothing fancy happens.
3) When you multiply or divide both sides of an inequality by a negative, then you must flip the sign.

Here are the fancier ones:
R4) If you multiply both sides of an inequality by zero, you can end up breaking things.
R5) If you multiply both sides of an inequality by a variable with an unknown sign, you no longer know whether to flip or not. However, this can be handled, carefully, with casewise statements. The same holds for division.
R6) Obviously, you won't ever divide both sides by zero itself. However, there's a potential issue where you can LOSE SOLUTIONS if you divide both sides by a variable that CAN equal zero. This actually leads to proof by contradiction - if you try dividing both sides by a variable, and you get a contradiction, then you know the variable CAN'T be zero.

Starting with your first example: Begin with the statement x/y>1.

At face value, this already says to me a couple things:
O1) X and Y must have the same sign, such that their quotient is positive.
O2) X must have a larger absolute value than Y.
For example, x= -5, y= -4.
O3) Y is clearly not zero; otherwise x/y wouldn't be defined, and thus couldn't be greater than 1. Therefore, Y is definitely either positive or negative.
O4) Remember how we said X and Y must have the same sign? Furthermore, zero divided by anything will never yield something bigger than one. Therefore, X is also nonzero.

Now let's talk about some algebra we can do:
If indeed Y is positive, then you can multiply both sides by it and nothing fancy happens. If, however, as you point out, it is negative, then multiplying both sides by Y forces you to flip signs.

Therefore, we can sum this up with this casewise statement:
If Y is positive, then x > y.
If Y is negative, then x < y.

Now here's the wonky thing; this is actually less information than we had at the beginning! Compare this set of statements with the observations (O1 through O4) made above. These statements actually lose the nonzero restrictions. The only PROPER way to simulate the original statement is with THIS set of casewise statements:

If Y positive, then X > Y > 0.
If Y negative, then X < Y < 0.
Y cannot be zero.
(As a result, X can never be zero.)

Isn't that wild? It seems that it might have just been easier to analyze the original statement.

Everyone knows the rule "flip signs when you multiply by a negative". However, this has an interesting corollary: You must also flip signs when you multiply by a VARIABLE that has a negative value. Furthermore, one doesn't have to SEE negative signs in order to have to flip!

The big thing is this: when you see an x with a negative sign in front of it, the best thing to do is think of it as "negative one times x", the result of which can be either positive or negative, depending on the VALUE of x ITSELF. In other words, a variable can contain a negative value. If x has the value of -4 (negative), then -x is actually positive! It equals (-1) * (-4) = (+4).

For example, let's suppose x > y, and z < 0. Z has to be negative. Therefore, it absolutely follows that xz < yz. See? I had to flip, and there were no visible negative signs anywhere; z itself contained a negative.
(check it with, say, x = 4, y = 3, and z = -1)

This has been a great source of confusion for many. Hope this clarifies the issue.

Let's also talk about dividing by zero. The issue comes into play when you have a statement like \(2x=x^2\). There are two correct ways (that I can think of...) to obtain the value(s) of x here:
1) Subtract 2x from both sides, factor as x(x-2)=0, then realize x = {0, 2}.
2) By inspection, realize that plugging in x = 2 works, and hey, so does x = 0. (Hopefully, you just do this, in the interest of time!)

But see, what a lot of students will do is try to divide both sides by x, which leads to the misleading statement x=2. I say misleading because many folks will miss that x can also equal zero, and the equation will still hold.

Here's a famous math trick:
1) Suppose 1x = 2x. (For example, 1*0 = 0 = 2*0, n'est-ce pas?)
2) Now, divide out the X.
3) 1 = 2. WTF??

Not only does dividing by zero give you problems, but dividing by variables that CAN equal zero will also give you problems. Just handle these things very carefully.

By the way, that trick reveals that contradiction idea I presented earlier, in Rule R6: Since dividing both sides by X yields a contradiction, that means we shouldn't divide both sides by X. And when should you not do that? When X = 0. Therefore, we KNOW that X = 0. (Indeed, that's the only time 1x=2x.)

Your second question speaks to the above issue a little. The second question asked "If \(2xy < x^2\), is it not right to say \(2y < x\) ? Also, can I assume \(x \neq 0\) ?"

Let me get to the second part of the question first, and the answer is that you are correct. If you were to actually plug in x = 0, then you would get the statement "0 > 0", which, while it looks like a cute cartoon bird, is FALSE. Therefore, we cannot allow x to equal zero.

Now, actually, since we now know that x can't be zero, we KNOW that we avoid the loss of solution issue, which means we CAN divide by x, casewise!

If X positive, then 2y < x. (Note that Y can be positive, negative, or zero).
If X negative, then 2y > x. (Again, Y can be whatever.)

And then from there, you do your thing.

Hope this answers your question in exquisite and excruciating detail! And I hope you are less confused as a result. Good luck!
Joined: 02 Jul 2007
Posts: 112
Re: Can someone please help me understand how to change the sign [#permalink]

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New post 01 May 2009, 13:49
I couldn't hope to be as thorough as Liquidhypnotic, but I wanted to give my two cents in case it was helpful (and just in case I'm way off base myself someone can correct me :shock: )

With sign problems, I always find it helpful to draw quick charts. I always screw up signs when doing these problems and charts help force me to keep things in order. In the first example, (x/y)>1 ----> x>y

We know off the bat that both x and y have to have the same sign ((+/+)= + and (-/-)= -.

x y ---> (x/y)
+ + +
- - +

When both x and y are positive:

(x/y)>1 -----> x>y

When both x and y are negative:
(-x/-y)>1 [multiply both sides by –y and switch the inequality] –x<-y [multiply by -1 and switch the inequality again] x>y

For the second example, 2xy<x^2

We can never divide by a variable unless the question explicitly says it does not equal 0, so:

2xy - x^2 < 0
x(2y – x)<0
x<0 or 2y – x<0

For what it’s worth, this is how I do problems like these. But it may be completely off base and have a ton of flaws. If anyone out there sees anything glaringly wrong with what I wrote, please let me know.
Re: Can someone please help me understand how to change the sign   [#permalink] 01 May 2009, 13:49
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Can someone please help me understand how to change the sign

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