How do you simplify #n-{1-[n-(1-n)-1]}#?

2 Answers
Jun 7, 2018

#n-{1-[n-(1-n)-1]} = 3n-3#

Explanation:

Given:

#n-{1-[n-(1-n)-1]}#

Work from the inside, outwards, replacing each parenthesised expression by its simplest form and evaluating left unparenthesised compound expressions from left to right...

#n-{1-[n-(1-n)-1]} = n-{1-[n+(n-1)-1]}#

#color(white)(n-{1-[n-(1-n)-1]}) = n-{1-[n+n-1-1]}#

#color(white)(n-{1-[n-(1-n)-1]}) = n-{1-[2n-1-1]}#

#color(white)(n-{1-[n-(1-n)-1]}) = n-{1-[2n-2]}#

#color(white)(n-{1-[n-(1-n)-1]}) = n-{1+[2-2n]}#

#color(white)(n-{1-[n-(1-n)-1]}) = n-{1+2-2n}#

#color(white)(n-{1-[n-(1-n)-1]}) = n-{3-2n}#

#color(white)(n-{1-[n-(1-n)-1]}) = n+{2n-3}#

#color(white)(n-{1-[n-(1-n)-1]}) = n+2n-3#

#color(white)(n-{1-[n-(1-n)-1]}) = 3n-3#

Alternatively, if you wanted to find the result as quickly as possible, you could note that the whole given expression is a sort of sum, with the minus signs toggling the value of the individual terms between #+# and #-#. So you could look at all the terms involving #n#, noting whether they will end up as positive or negative, and add them together. Then do the same with the constant terms...

The first #n# is positive:

#color(red)(n)-{1-[n-(1-n)-1]}#

The second #n# is also positive, since it is affected by an even number (#2#) of minus signs:

#ncolor(red)(-){1color(red)(-)[color(red)(n)-(1-n)-1]}#

The third #n# is also positive, since it is affected by an even number (#4#) of minus signs:

#ncolor(red)(-){1color(red)(-)[ncolor(red)(-)(1color(red)(-n))-1]}#

So the final term in #n# is #3n#.

Similarly we can find that all the #1#'s are affected by an odd number of minus signs, resulting in a final constant term #-3#.

Jun 7, 2018

#3n-3#

Explanation:

Start with the innermost brackets and work outwards, one set of brackets at a time.
Remember that a negative in front of a bracket changes the sign when you multiply into the bracket.

#n-{1-[ncolor(blue)(-(1-n))-1]}" "larr# remove the round brackets

#=n-{1-[ncolor(blue)(-1+n)-1]}#

#=n-{1color(red)(-[n-1+n-1]}" "larr# remove square brackets

#=n-{1color(red)(-n+1-n+1]}#

#=ncolor(green)(-{1-n+1-n+1})" "larr# remove curly brackets

#=n-1+n-1+n-1" "larr# simplify

#=3n-3#