Mixing
Monotonic equation
$begingroup$
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
- If it always fulfills
min(q3,q1) <= q2 <= max(q3,q1)andmin(h3,h1) <= h2 <= max(h3,h1), it DOES NOT always satisfymin(u3,u1) <= u2 <= max(u3,u1). Right? Why? Any proof? - If it always fulfills
min(u3,u1) <= u2 <= max(u3,u1)andmin(h3,h1) <= h2 <= max(h3,h1), it ALWAYS satisfiesmin(q3,q1) <= q2 <= max(q3,q1). Right?
Many thanks.
PS: Please don't forget to consider the functions min and max.
calculus linear-algebra monotone-functions
asked Jan 23 at 21:53
bob.bob.bobbob.bob.bob
11
$endgroup$
add a comment |
$begingroup$
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
- If it always fulfills
min(q3,q1) <= q2 <= max(q3,q1)andmin(h3,h1) <= h2 <= max(h3,h1), it DOES NOT always satisfymin(u3,u1) <= u2 <= max(u3,u1). Right? Why? Any proof? - If it always fulfills
min(u3,u1) <= u2 <= max(u3,u1)andmin(h3,h1) <= h2 <= max(h3,h1), it ALWAYS satisfiesmin(q3,q1) <= q2 <= max(q3,q1). Right?
Many thanks.
PS: Please don't forget to consider the functions min and max.
calculus linear-algebra monotone-functions
asked Jan 23 at 21:53
bob.bob.bobbob.bob.bob
11
$endgroup$
add a comment |
$begingroup$
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
- If it always fulfills
min(q3,q1) <= q2 <= max(q3,q1)andmin(h3,h1) <= h2 <= max(h3,h1), it DOES NOT always satisfymin(u3,u1) <= u2 <= max(u3,u1). Right? Why? Any proof? - If it always fulfills
min(u3,u1) <= u2 <= max(u3,u1)andmin(h3,h1) <= h2 <= max(h3,h1), it ALWAYS satisfiesmin(q3,q1) <= q2 <= max(q3,q1). Right?
Many thanks.
PS: Please don't forget to consider the functions min and max.
calculus linear-algebra monotone-functions
asked Jan 23 at 21:53
bob.bob.bobbob.bob.bob
11
$endgroup$
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
- If it always fulfills
min(q3,q1) <= q2 <= max(q3,q1)andmin(h3,h1) <= h2 <= max(h3,h1), it DOES NOT always satisfymin(u3,u1) <= u2 <= max(u3,u1). Right? Why? Any proof? - If it always fulfills
min(u3,u1) <= u2 <= max(u3,u1)andmin(h3,h1) <= h2 <= max(h3,h1), it ALWAYS satisfiesmin(q3,q1) <= q2 <= max(q3,q1). Right?
Many thanks.
PS: Please don't forget to consider the functions min and max.
calculus linear-algebra monotone-functions
calculus linear-algebra monotone-functions
asked Jan 23 at 21:53
bob.bob.bobbob.bob.bob
11
asked Jan 23 at 21:53
bob.bob.bobbob.bob.bob
11
edited Jan 25 at 19:25
bob.bob.bob
asked Jan 23 at 21:53
bob.bob.bobbob.bob.bob
11
asked Jan 23 at 21:53
bob.bob.bobbob.bob.bob
11
asked Jan 23 at 21:53
bob.bob.bobbob.bob.bob
11
11
add a comment |
add a comment |
1 Answer
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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
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.
answered Jan 23 at 22:26
SergGrSergGr
1162
$endgroup$
$begingroup$
Apologize I forgot to mention, see the revision. Can you please edit your answer, so that I can vote it? Thanks!
$endgroup$
– bob.bob.bob
Jan 25 at 12:09
$begingroup$
@bob.bob.bob I still insist that 2 inequalities do not add any new information. If I break one inequality, I break both. I've extended my examples to show that.
$endgroup$
– SergGr
Jan 25 at 18:32
$begingroup$
Many thanks. So for the point 1: Is it correct to say: IT DOES NOT ALWAYS GUARANTEE, right? Have you also considered the MIN and MAX functions for the boundary conditions? Thanks.
$endgroup$
– bob.bob.bob
Jan 25 at 19:21
$begingroup$
@bob.bob.bob, do you see that my counter-examples break theminandmaxboundaries as well (just because everything with index1is theminand with index3is themaxexcept for the "conclusion" which I break)? If no, then please clarify what do you mean? Please check all your statements against those 2 examples. If one of those examples breaks the statement, don't ask about it again - it is already false and you know it.
$endgroup$
– SergGr
Jan 25 at 19:27
add a comment |
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1 Answer
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1 Answer
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$begingroup$
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
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.
answered Jan 23 at 22:26
SergGrSergGr
1162
$endgroup$
$begingroup$
Apologize I forgot to mention, see the revision. Can you please edit your answer, so that I can vote it? Thanks!
$endgroup$
– bob.bob.bob
Jan 25 at 12:09
$begingroup$
@bob.bob.bob I still insist that 2 inequalities do not add any new information. If I break one inequality, I break both. I've extended my examples to show that.
$endgroup$
– SergGr
Jan 25 at 18:32
$begingroup$
Many thanks. So for the point 1: Is it correct to say: IT DOES NOT ALWAYS GUARANTEE, right? Have you also considered the MIN and MAX functions for the boundary conditions? Thanks.
$endgroup$
– bob.bob.bob
Jan 25 at 19:21
$begingroup$
@bob.bob.bob, do you see that my counter-examples break theminandmaxboundaries as well (just because everything with index1is theminand with index3is themaxexcept for the "conclusion" which I break)? If no, then please clarify what do you mean? Please check all your statements against those 2 examples. If one of those examples breaks the statement, don't ask about it again - it is already false and you know it.
$endgroup$
– SergGr
Jan 25 at 19:27
add a comment |
$begingroup$
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
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.
answered Jan 23 at 22:26
SergGrSergGr
1162
$endgroup$
$begingroup$
Apologize I forgot to mention, see the revision. Can you please edit your answer, so that I can vote it? Thanks!
$endgroup$
– bob.bob.bob
Jan 25 at 12:09
$begingroup$
@bob.bob.bob I still insist that 2 inequalities do not add any new information. If I break one inequality, I break both. I've extended my examples to show that.
$endgroup$
– SergGr
Jan 25 at 18:32
$begingroup$
Many thanks. So for the point 1: Is it correct to say: IT DOES NOT ALWAYS GUARANTEE, right? Have you also considered the MIN and MAX functions for the boundary conditions? Thanks.
$endgroup$
– bob.bob.bob
Jan 25 at 19:21
$begingroup$
@bob.bob.bob, do you see that my counter-examples break theminandmaxboundaries as well (just because everything with index1is theminand with index3is themaxexcept for the "conclusion" which I break)? If no, then please clarify what do you mean? Please check all your statements against those 2 examples. If one of those examples breaks the statement, don't ask about it again - it is already false and you know it.
$endgroup$
– SergGr
Jan 25 at 19:27
add a comment |
$begingroup$
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
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.
answered Jan 23 at 22:26
SergGrSergGr
1162
$endgroup$
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
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.
answered Jan 23 at 22:26
SergGrSergGr
1162
edited Jan 25 at 18:31
answered Jan 23 at 22:26
SergGrSergGr
1162
answered Jan 23 at 22:26
SergGrSergGr
1162
answered Jan 23 at 22:26
SergGrSergGr
1162
1162
$begingroup$
Apologize I forgot to mention, see the revision. Can you please edit your answer, so that I can vote it? Thanks!
$endgroup$
– bob.bob.bob
Jan 25 at 12:09
$begingroup$
@bob.bob.bob I still insist that 2 inequalities do not add any new information. If I break one inequality, I break both. I've extended my examples to show that.
$endgroup$
– SergGr
Jan 25 at 18:32
$begingroup$
Many thanks. So for the point 1: Is it correct to say: IT DOES NOT ALWAYS GUARANTEE, right? Have you also considered the MIN and MAX functions for the boundary conditions? Thanks.
$endgroup$
– bob.bob.bob
Jan 25 at 19:21
$begingroup$
@bob.bob.bob, do you see that my counter-examples break theminandmaxboundaries as well (just because everything with index1is theminand with index3is themaxexcept for the "conclusion" which I break)? If no, then please clarify what do you mean? Please check all your statements against those 2 examples. If one of those examples breaks the statement, don't ask about it again - it is already false and you know it.
$endgroup$
– SergGr
Jan 25 at 19:27
add a comment |
$begingroup$
Apologize I forgot to mention, see the revision. Can you please edit your answer, so that I can vote it? Thanks!
$endgroup$
– bob.bob.bob
Jan 25 at 12:09
$begingroup$
@bob.bob.bob I still insist that 2 inequalities do not add any new information. If I break one inequality, I break both. I've extended my examples to show that.
$endgroup$
– SergGr
Jan 25 at 18:32
$begingroup$
Many thanks. So for the point 1: Is it correct to say: IT DOES NOT ALWAYS GUARANTEE, right? Have you also considered the MIN and MAX functions for the boundary conditions? Thanks.
$endgroup$
– bob.bob.bob
Jan 25 at 19:21
$begingroup$
@bob.bob.bob, do you see that my counter-examples break theminandmaxboundaries as well (just because everything with index1is theminand with index3is themaxexcept for the "conclusion" which I break)? If no, then please clarify what do you mean? Please check all your statements against those 2 examples. If one of those examples breaks the statement, don't ask about it again - it is already false and you know it.
$endgroup$
– SergGr
Jan 25 at 19:27
$begingroup$
Apologize I forgot to mention, see the revision. Can you please edit your answer, so that I can vote it? Thanks!
$endgroup$
– bob.bob.bob
Jan 25 at 12:09
$begingroup$
Apologize I forgot to mention, see the revision. Can you please edit your answer, so that I can vote it? Thanks!
$endgroup$
– bob.bob.bob
Jan 25 at 12:09
$begingroup$
@bob.bob.bob I still insist that 2 inequalities do not add any new information. If I break one inequality, I break both. I've extended my examples to show that.
$endgroup$
– SergGr
Jan 25 at 18:32
$begingroup$
@bob.bob.bob I still insist that 2 inequalities do not add any new information. If I break one inequality, I break both. I've extended my examples to show that.
$endgroup$
– SergGr
Jan 25 at 18:32
$begingroup$
Many thanks. So for the point 1: Is it correct to say: IT DOES NOT ALWAYS GUARANTEE, right? Have you also considered the MIN and MAX functions for the boundary conditions? Thanks.
$endgroup$
– bob.bob.bob
Jan 25 at 19:21
$begingroup$
Many thanks. So for the point 1: Is it correct to say: IT DOES NOT ALWAYS GUARANTEE, right? Have you also considered the MIN and MAX functions for the boundary conditions? Thanks.
$endgroup$
– bob.bob.bob
Jan 25 at 19:21
$begingroup$
@bob.bob.bob, do you see that my counter-examples break the
min and max boundaries as well (just because everything with index 1 is the min and with index 3 is the max except for the "conclusion" which I break)? If no, then please clarify what do you mean? Please check all your statements against those 2 examples. If one of those examples breaks the statement, don't ask about it again - it is already false and you know it.$endgroup$
– SergGr
Jan 25 at 19:27
$begingroup$
@bob.bob.bob, do you see that my counter-examples break the
min and max boundaries as well (just because everything with index 1 is the min and with index 3 is the max except for the "conclusion" which I break)? If no, then please clarify what do you mean? Please check all your statements against those 2 examples. If one of those examples breaks the statement, don't ask about it again - it is already false and you know it.$endgroup$
– SergGr
Jan 25 at 19:27
add a comment |
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