i think i see what ben is getting at here. in a sense, you guys are both correct -- you're just talking past each other.

YOU are using the transitive property:
if a > b, and b > c, then a > c.
this is of course true.

BEN, however, is not talking so much about the inequalities themselves as he's talking about their solution sets.
for instance:
let's say that x > y and y > 10.
in this case,
* we do know that x > 10 (in other words, it's impossible that x > 10 is false); in that sense, you are correct.
* where you are going wrong here, though -- and what ben is trying to point out -- is that you are assuming that, given the above two inequalities, that x can automatically assume ANY value > 10. in other words, it seems that you are taking the two inequalities x > y and y > 10, together, as a guarantee that x can actually equal 10, or any greater number.
this isn't necessarily true, depending upon the value of y. for instance, if y is 15, then x is now restricted to numbers that are 15 and higher.

hope this makes more sense

2. |x-3| <= - y
|x-3| <= -(0) or |x-3| >= 0
x <=3 or x >= 3

these steps are invalid here, because (-y) is a negative number.
as a simpler analogy, consider the following inequalities:
|x - 3| < 4
|x - 3| < -4
the first of these can be solved with normal techniques (i.e., change it into -4 < x - 3 < 4).
the second can't -- it has no solutions at all, since an absolute value can't be negative.

if you still aren't seeing how this applies to the situation at hand, try numbers until you see what is happening.
i.e., go ahead and actually try to find specific values of "x" and "y" (with "y" positive, as required) that satisfy the equation |x - 3| < -y.
after you try enough values and observe what happens, (a) you'll realize that the equation can only work if x = 3 and y = 0, and (b) you'll see why.

Before I look at the values of Y, I will write the equations

Step 1 :|x - 3| <= -y or |x - 3| >=y

Case 2: |x - 3| >= 0
so x >= 3.

this is a big mistake -- this kind of equation (in which an absolute value is less than a number) NEVER splits into "or".

if you are going to approach this algebraically, then all such equations become "and":
|blah| < A
becomes
blah < A and blah > -A, or
-A < blah < A.

so, here, that would be |x - 3| < -y and |x - 3| > y, or
y < |x - 3| < -y

since y is 0 or greater, the only possible way for this to work is y = 0. (if y is positive, then you get an impossible condition.) so, same conclusion.


Thanks ron. Now, I realised the mistake. I was using "OR" instead of AND.Can I say, when I have something like |x - 3| <= -y the equation transfers to AND instead of OR?




i don't know where this is coming from.

case 1 (stuff in abs value is negative)
...
case 2 (stuff in abs value is positive)

dave, the problem can be seen in your labels here:



these labels aren't 100% accurate. specifically, your "case 1" covers all the possibilities in which the "stuff" is negative or zero, and your "case 2" covers all the possibilities in which the "stuff" is positive or zero. that explains why the equality is turning up in both cases: because it's true in both cases.

if you don't have a good grip on this, try some simpler equations: |x| = x
|x| = -x
the first of these is true when x > 0 (not just when x > 0), and the second is true when x < 0 (not just when x < 0). BOTH of them are true when x is zero.

wow thanks Ron. if you didn't list that rudimentary example it would have taken me a while to see that!

i think it is a mistake then to write |x| ,for example, into two cases (which i notice is a knee-jerk reaction in how most people do this algebraically)

x < 0
x > 0

it's actually a negative of the entire expression for any value of x. not the value of x being negative or positive.