The Simplifier

Example Three 3. Solve: {1 – [1 – 1 x (1 + 1) – 1 ] of 1} ÷ 1 + 1 This has nested brackets (i.e. bracket inside another bracket and we must start from the innermost bracket). Starting from innermost bracket: {1 – [1 – 1 x (1 + 1)   – 1] of 1} ÷ 1 + 1 = {1 – [1 – 1 x 1 – 1] of 1} ÷ 1 + 1 Apply the Precedence Rule also in the Innermost Bracket:  {1 – [1 – 1 x 1 – 1] of 1} ÷ 1 + 1 = {1 – [1 – 1 – 1] of 1} ÷ 1 + 1 Applying Precedent rule from left to right inside the bracket we have: {1 – [1 – 1 – 1] of 1} ÷ 1 + 1 = {1 – [0 – 1] of 1} ÷ 1 + 1 Furthermore we have: {1 – [0 – 1] of 1} ÷ 1 + 1 = {1 – [–1] of 1} ÷ 1 + 1 Next is to consider the of in the Innermost Bracket (if it exists): {1 – (–1) of 1 } ÷ 1 + 1 = {1 – (–1) } ÷ 1 + 1 Notice that we do not write two mathematical operators together without a bracket to separate them: e.g. instead of 1  –  –  1 we write 1 – ( - 1) . Next Step finalize the operations in bracket...