$$ |a-b| = |a|-|b| $$
I think I might missing something with absolute values. Can I split a binomial into two separate absolute values like above?
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1you can do it always when $a\ge b>0$ – haqnatural Aug 28 '15 at 20:37
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1so if both a and b are positive and a is greater than or equal to 0 – lindsaymacvean Aug 28 '15 at 20:44
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1Or if $a\leqslant b\leqslant0$. – Did Aug 28 '15 at 20:49
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1Thankyou guys, I assume there is no reason why the first constraint cannot be 0 such as $a \geqslant b \geqslant 0 $ – lindsaymacvean Aug 28 '15 at 20:59
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No, you cannot. For example, Let $a=1$ and $b=-2$. We have $|a-b| = |1-(-2)| = |3|=3$. On the other hand, we have $|a|-|b| = |1|-|-2| = |1|-|2|=1-2=-1$. Clearly, $3 \neq -1$.
It is, however, true that $||a|-|b|| \le |a-b|$. In fact, we can extend this:
$$||a|-|b|| \le |a-b| \le |a|+|b|$$
Fly by Night
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1if you include Battani comment above then this would be correct. – lindsaymacvean Aug 28 '15 at 20:46
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1The inequalities are valid for all $a$ and $b$. Your original post did not include the conditions $a \ge b > 0$, and I can only answer the question you ask. – Fly by Night Aug 28 '15 at 20:49
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1No it does not contain those conditions in my question, nor have I subsequently added them. I believe the constraints help answer my question as it seems there are some cases where this equality works as described by Battani and now Did. This is what was confusing me originally. Your answer is correct in the general sense so I think you should add in the special constraints before I can mark as correct. – lindsaymacvean Aug 28 '15 at 20:55
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1Ah I just wrote this out on a white board with a table of combinations {1,0,3,5} and understand what you mean now. What I was asking for is actually already in the second part of your answer. Thank you. – lindsaymacvean Aug 28 '15 at 21:32
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No. Breaking $|a-b|$ up like that is no good. Try $a = 1$ and $b = 2$. Your equation is true if $a \geq b > 0$.
Mike Pierce
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