I have followed a lot of the comments regarding the subject expression and the controversy created. Many agree that the expression is ambiguous but some settle for either PEMDAS and the order of operations from left to right while others use juxtaposition to resolve the group first. A lot of the confusion results from the use of the obelus (:-) for division with some noting that it is a syntax error and one should use either the slash (/) or more preferably the vinculum (---) instead. However, I have seen a different way of solving the expression that uses the uncontroversial rule for how to solve the division of fractions. i.e, multiply the first fraction by the reciprocal of the second fraction. To wit: the expression 6:- 2(1+2) uses the obelus to indicate division. But all can agree that any value is equal to itself if divided by 1, so if we write the expression as 6 over 1 :- 2(1+2) over 1 then it becomes 6 over 1 x 1 over 2(1+2)[the reciprocal]. From here it does not matter which operation is chosen first, i.e to either resolve the denominator first: 6 over 6 or to divide the group factor (in this case 2) into the numerator it becomes 3 over 3, and both methods result in 1 as the answer. Any comment on this?

I have followed a lot of the comments regarding the subject expression and the controversy created. Many agree that the expression is ambiguous but some settle for either PEMDAS and the order of operations from left to right while others use juxtaposition to resolve the group first. A lot of the confusion results from the use of the obelus (:-) for division with some noting that it is a syntax error and one should use either the slash (/) or more preferably the vinculum (---) instead. However, I have seen a different way of solving the expression that uses the uncontroversial rule for how to solve the division of fractions. i.e, multiply the first fraction by the reciprocal of the second fraction. To wit: the expression 6:- 2(1+2) uses the obelus to indicate division. But all can agree that any value is equal to itself if divided by 1, so if we write the expression as 6 over 1 :- 2(1+2) over 1 then it becomes 6 over 1 x 1 over 2(1+2)[the reciprocal]. From here it does not matter which operation is chosen first, i.e to either resolve the denominator first: 6 over 6 or to divide the group factor (in this case 2) into the numerator it becomes 3 over 3, and both methods result in 1 as the answer. Any comment on this?