Monotonic equation

Question

I have some equations like this:

q1 = u1 * h1 , q2 = u2 * h2 , q3 = u3 * h3 (note that u1, u2, and u3 can either be positive or negative, but h1, h2, and h3 are always positive).

REVISION

1. If it always fulfills min(q3,q1) <= q2 <= max(q3,q1) and min(h3,h1) <= h2 <= max(h3,h1), it DOES NOT always satisfy min(u3,u1) <= u2 <= max(u3,u1). Right? Why? Any proof?
2. If it always fulfills min(u3,u1) <= u2 <= max(u3,u1) and min(h3,h1) <= h2 <= max(h3,h1), it ALWAYS satisfies min(q3,q1) <= q2 <= max(q3,q1). Right?

Many thanks.

PS: Please don't forget to consider the functions min and max.

Answer 1

I think you missed the point of the original answer. The moment I break on side of the double inequality I break the whole inequality. I'm not sure why you want that much for a false fact to be a true one but it is still false nevertheless. You can't make any prediction in either of those two cases. Here are two examples (extension to me original ones) to show that the non-obvious cases are still possible:

Example #1:

u1 = 9, h1 = 10, q1 = 90
u2 = 2, h2 = 55, q2 = 110
u3 = 10, h3 = 60, q3 = 600

It is true that

h1 < h2 < h3
q1 < q2 < q3

but still u2 is the smallest:

u2 < u1 < u3

Example #2:

u1 = -2, h1 = 3, q1 = -6
u2 = -1, h2 = 8, q2 = -8
u3 = 3, h3 = 10, q3 = 30

It is true that

h1 < h2 < h3
u1 < u2 < u3

but still q2 is the smallest

q2 < q1 < q3

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Original Answer

I don't know why you use 3 triplets and 2 inequalities, it is enough to have 2 triplets and 1 inequality (the other one is exactly the same). So we know that:

q1 = u1 * h1
q2 = u2 * h2

and

0 <= h2 <= h1

Statement #1.

Given that you also know that q2 <= q1, can you say anything about whether u1 >= u2 or u1 <= u2? The answer is "no, you can't". To prove that something is not always true it is enough to provide one counter example.

Example #1.1

q1 = 110 = 2 * 55 = u1 * h1
q2 = 90 = 9 * 10 = u2 * h2

Clearly

q2 = 90 < 110 = q1
h2 = 10 < 55 = h1

but

u2 = 9 > 2 = u1

Example #1.2

q1 = 110 = 11 * 10 = u1 * h1
q2 = 90 = 9 * 10 = u2 * h2

Clearly

q2 = 90 < 110 = q1
h2 = 10 < 11 = h1

and

u2 = 9 < 10 = u1

So both cases are possible

Statement #2.

Given that you know that u2 <= u1 but both of us can be negative, can you say anything about whether u1 >= u2 or u1 <= u2? The answer is again "no, you can't".

Example #1.2 from the previous section says that the case when u2 <= u1, h2 <= h1 and q2 <= q1 is possible which is not a big surprise. An example for other possibility must use negative values for u. The intuition behind this example is that if you multiply a big positive number by something negative you get a big negative number which actually means it is a small number comparing to other negative numbers.

Example #2.1

q1 = -6 = -2 * 3 = u1 * h1
q2 = -8 = -1 * 8 = u2 * h2

Clearly

h1 = 3 < 8 = h2
u1 = -2 < -1 = u2

but

q1 = -6 >  -8 = q2

So again both cases are possible.

If we additionally knew that 0 <= u2 <= u1, then we could say that q2 <= q1. Consider q2 - q1:

q2 - q1 = u2*h2 - u1*h1 = (u1 + (u2-u1))*(h1 + (h2-h1)) - u1*h1 = 
  =  u1*h1 + (u2-u1)*h1 + (u2-u1)*(h2-h1) + u1*(h2-h1) - u1*h1 =
  =  (u2-u1)*h1 + (u2-u1)*(h2-h1) + u1*(h2-h1) >= 0

Obviously if h1 >= 0 and u1 >= 0, all three terms in the last line are non-negative, so q2 - q1 is non-negative or q2 >= q1.