What is 6÷2(1+2)?

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The Casio fx-9860 GII SD returns the correct answer to the equation as given, and the answer that a less sophisticated calculator would return.

6÷2(1+2)=? is the pre-algebra question that a majority of respondents seem to get wrong. If you are in the camp that knows the correct answer is 1, read no further. If you believe the answer is 9, you should probably review this document -- there is only one number that 6 can be divided by to result in 9, and that number is 2/3.

Processing rules

In the majority of incorrect proofs, the respondent will cite PEMDAS or BODMAS, however these solvers incorrectly resolve the [P]arentheses or [B]rackets of PEMDAS - BODMAS, ignoring the Distributive Law. For some reason, they believe that they can solve within the parentheses, but ignore the adjacent and dependent coefficient.

Distributive law, in mathematics, the law relating the operations of multiplication and addition, stated symbolically, a(b + c) = ab + ac; that is, the monomial factor a is distributed, or separately applied, to each term of the binomial factor b + c, resulting in the product ab + ac.
Encyclopedia Britticana


Therefore, in 6÷2(1+2)=?, we start with the Distributive Law in the Parenthesis step of the Order of Operations.

6÷(2*1 + 2*2)=?
6÷(2 + 4)=?
6÷(6)=?
6÷6=1

Wolfram Mathworld also confirms that this is the way we handle parenthetical expressions:

1. Parentheses are used in mathematical expressions to denote modifications to normal order of operations (precedence rules)...
[...]
3. Parentheses are used to enclose the variables of a function in the form f(x), which means that values of the function f are dependent upon the values of x.
Parenthesis

So the fatal error with solvers who emphasize the left to right rule, is in incorrectly handling the Parentheses step, disregarding the Distributive Law.

6÷2(3)

Some are even compelled to add an explicit multiplier, not given in the equation.

6÷2*(3)

The Fundamentals of Algebra (1983)

Parenthetical Expression. The parenthesis was described in Chapter 1 as a grouping symbol. When an algebraic expression is enclosed by a parenthesis it is known as a parenthetical expression. When a parenthetical expression is immediately preceded by coefficient, the parenthetical expression is a factor and must be multiplied by the coefficient. This is done in the following manner.

5(a + b) = 5a + 5b
3a(b - c) = 3ab - 3ac

"Technical Shop Mathematics / Edition 2", by John G. Anderson, ISBN-13:9780831111458, Industrial Press, Inc., 02/28/1983, Page:138

The equation is not ambiguous

Standing alone, 6 will always equal 6 and 2(1+2) will also always equal six, so dividing these two terms will always equal 1. We would not write 2(a+b) as (2(a+b)), nor would we re-write the given equation as 6÷2*(a+b). Granted, the obelus (÷) is an archaic symbol for division, but this is not a postgraduate level equation, this is middle-school level math.

2(1+2) is understood to be a function, placing this function anywhere in a larger equation must always resolve to the same value. The equation, 6÷2(1+2)=? is illustrative to why we cannot substitute explicit multiplication for implied multiplication, in other words, 6÷2(1+2) is not equal to 6÷2*(1+2).

Implied multiplication has a higher priority than explicit multiplication to allow users to enter expressions, in the same manner as they would be written.
Implied Multiplication Versus Explicit Multiplication on TI Graphing Calculators

Correctly solving left to right

If we insist on a left to right solution, ignoring both PEMDAS and the Distributive Law, we can do this with the Least Common Denominator rule

6÷2(1+2)=?
3÷(1+2)=?
3÷(3)=?
3÷3=1

When we reduce by the common denominator, the division is sustained...

Example:

6


2(1+2)


=

3


(1+2)


=

3


3


= 1

The calculator problem

The Desmos Scientific Calculator handles our given equation correctly. The error checking appears to be built into the division key.


I realized that shorthand parenthetical notation could become a problem when I was taking a 'C' programming course in college. Unfortunately, we are still using code that was written for computers that had severe memory limitations.

Unlike humans, mathematics modules do not look ahead to identify a parenthetical expression embedded within an equation. In the 'C' language, to process our given equation, we need explicit operators adjacent to the first parenthesis; additionally, we must maintain the grouping of our function(s). Still to this day, many programming languages / spreadsheets cannot process a parenthetical expression without an explicit operator.

Entering the equation into a programming language, or low quality calculator require the explicit multiplication symbol and outer parenthesis.

6÷2(1+2) == 6÷(2*(1+2))

When our given equation is correctly conditioned to read, 6÷(2*(1+2)) the mathematics routine reserves additional memory to process the parenthetical expression(s) and then reassembles the equation according to the rules of the Order of Operations (PEMDAS). Conversely, in our given equation, 6÷2(1+2) the calculator processes 6÷2 and then discovers the parenthetical expression too late. Instead of branching to fix the mistake, or declaring an error condition, mathematics modules simply assume that the data-entry person is competent and adds explicit multiplication where implicit multiplication was given.

6÷2(1+2) != 6÷2*(1+2)


The calculator solution

Our given equation is two terms, 6 is always equal to 6 and 2(1+2) is also always equal to 6, dividing these two numbers must always equal 1. The equation is not ambiguous, it challenges the theory that multiplication and division are equal in terms of the Order of Precedence. First, let's review fundamental fact that x÷1 = x and that 1*x = x. When we enclose a term within parentheses, implicit multiplication assumed, if the multiplier is not given, the multiplier is the number 1 because 0 times any number is 0.

Example that ignores the Distributive Law:

6÷2(1+2)=?

becomes...

3÷1(1+2)=?


As we see above, the explicit division sustains, as does the implicit multiplication, we have simply reduced both sides of the equation by the common denominator of 2, leaving us with 3÷1(1+2)=?. When we try to divide three by one, the one is also a common denominator, finding a common denominator does not dismiss division, but if we dismiss the assumed one, we must also dismiss the implicit multiplication. Continuing from above, we are left with the following equation...

3÷(1+2)=?

3÷(3)=?

3÷3=1

Therefore, if the calculator module has already processed the division 6÷2 by the time it reaches the parenthesis, the coefficient and the accompanying implicit multiplication has been eliminated. The division must be sustained to apply to the remainder of the function, in left to right processing, it cannot be presumed that an implicit/implied operator always infers multiplication in computer mathematics modules. In this case, the processing must branch.