Question 201987: 1) when is the absolute value of difference the same with the difference of an absolute value?
2) when is the absolute value of difference greater than with the difference of an absolute value?
Found 2 solutions by RAY100, Theo:
Answer by RAY100(1637)   (Show Source):
You can put this solution on YOUR website! [a-b] = [a] - [b],,,,,,when is this true ???
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if both a and b are positive, this is true,,,+4,+3
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if both a and b are negative, this is true,,,-4, -3
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if a and b are not the same sign, this is an inequality
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[ (-4) - (+3)] = [-7] = 7,,,,[-4] -[+3] =4-3=1
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[(+4) - (-3)] = [+7] = 7,,,,,[+4] -[-3] = 4-3 =1
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have a great weekend
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! av(d) = absolute value of the difference = av(a-b).
d(av) = difference of the absolute value = av(a) - av(b).
av(a-b) = absolute value of (a - b).
av(a) - av(b) = avsolute value of (a) minus absolute value of (b).
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As best I can determine, the following rules apply:
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There are no cases where av(d) < d(av)
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The cases where av(d) > d(av) are:
b > 0 and a < b
b < 0 and a > b
a > 0 and b > a
a > 0 and b < 0
a < 0 and b > 0
a < 0 and b < a
a = 0 and b > 0
a = 0 and b < 0
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The cases where av(d) = d(av) are:
b > 0 and a >= b
b < 0 and a <= b
b = 0
a > 0 and b <= a and b >= 0
a < 0 and b >= a and b <= 0
a = 0 and b = 0
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you can spot check to see that they're accurate.
I basically crunched the numbers in excel and calculated d(av) and calculated av(d) and saw where av(d) was > d(av) and where they were equal. av(d) was never less than d(av).
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take a > 0 and b <= a and b >= 0 for an example.
a > 0 could be 25.
b <= a and b >= 0 could be 15.
numbers are a = 25, b = 15
av(d) = av(25-15) = av(10) = 10.
d(av) = av(25) - av(15) = 25-15 = 10.
they are equal as they should be since I picked the example from the catgory where av(d) = d(av).
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I picked at random a < 0 and b > 0
if we let a = -10 and b = 1, this should satisfy the requirements.
av(d) = av((-10)-1) = av(-11) = 11.
d(av) = av(-10) - av(1) = 10 - 1 = 9
av(d) > d(av) as it should be since I picked it from the category where av(d) > d(av).
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I picked from b = 0 for another spotcheck.
a should be able to be any value.
let b = 0 and let a = -33
av(d) = av(a-b) = av((-33)-0) = av(-33) = 33.
d(av) = av(a) - av(b) = av(-33) - av(0) = 33 - 0 = 33.
they are equal as they should be since they are in the category where av(d) = d(av).
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spot check with b < 0 and a > b.
pick b = -20 and a = -18.
av(d) = av(a-b) = av((-18)-(-20)) = av(-18+20) = av(2) = 2.
d(av) = av(18) - av(20) = 18 - 20 = -2.
av(d) > d(av) as it should be since that's the category it's in.
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last spot check with b < 0 and a <= b.
pick b = -20 again.
pick a = -22.
av(d) = av((-22)-(-20)) = av(-22+20) = av(-2) = 2.
d(av) = av(-22) - av(-20) = 22 - 20 = 2.
av(d) = d(av) as it should be since that's the category it's in.
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These samples check out.
The rest should check out as well.
You may spot check a few for yourself to see also.
Try to find a case where any of the statements is not true.
You shouldn't find any unless I screwed up.
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