What I'm asking is, when subtracting numbers is there any integers that make $a-(b-c) = (a-b)-c$?
Besides $0$ of course. Thanks! Sorry for the stupid question.
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Simplify either sides – lab bhattacharjee Mar 11 '14 at 06:04
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1You are asking when $a-b+c=a-b-c$. The answer will undoubtedly be clear to you. – André Nicolas Mar 11 '14 at 06:05
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3$c=0 {}{}{}{}{}{}{}{}$ – Euler....IS_ALIVE Mar 11 '14 at 06:10
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Hint: Expand everything out and try to cancel what you can. Work out what you have left and the variables that do not have an "equation" are free variables and can take on any value.
$$a-b+c=a-b-c\\ \implies2c=0\\ \implies c=0, \;a,b\in \mathbb Z$$
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