The equation that broke the Internet

The Casio fx-9860 GII SD returns the correct answer to the equation as given. The next row provides 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.

Proof incorrect solution

We substitute x for 2(1+2) from the given equation:

If

6÷x = 9

then

6÷9 = x

there is a Common Denominator of 3 so we reduce the fraction...

2÷3 = x

Now we compare the value given in the equation to the value we find for x.

2÷3 != 2(1+2)

Therefore:

6÷2(1+2) != 9

Without fail, 2(1+2) will always equal (2*1 + 2*2), this is the Distributive Law.

Proof of correct solution

Encyclopedia Britannica on parenthetical expressions:

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)=?

6÷(2*1 + 2*2)=?

6÷(2 + 4)=?

6÷(6)=?

6÷6=1

The fatal error with solvers who emphasize the left to right rule.

6÷2(3)

6÷2*(3)

Processing rules

In the majority of incorrect proofs the respondent will cite PEMDAS or BODMAS, we all agree that parenthesis or brackets are processed first. 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.

PEMDAS

P parenthesis

E exponentiation

M multiplication

D division

A addition

S subtraction

–Wolfram PEMDAS

Parenthetical expressions

Wolfram Mathworld on 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.

–Wolfram Parenthesis

Distributive law

Encyclopedia Britannica on parenthetical expressions:

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

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

Explicit vs Implicit (implied) multiplication

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

Explicit operations have explicit operators (* ÷ + - ) which are delimiters that divide equations into separate terms. Implied multiplication is notation that informs us that the value of a variable, bracketed function or exponent are connected and not separate terms. Implied multiplication is everywhere, 1x = x and x/1=x. Any number times 1 is that number and any number divided by 1 is that number.

Implied multiplication is also implicit multiplication.

im·plic·it

adjective

1. implied though not plainly expressed.

2. essentially or very closely connected with; always to be found in.

im·plied

adjective

1. suggested but not directly expressed; implicit.

ex·plic·it

adjective

1. stated clearly and in detail, leaving no room for confusion or doubt.

–Oxford

6÷2(1+2) has two EXPLICIT operators, division and addition. However, IMPLIED multiplication tells us to IMPLICITLY multiply (2 * (2+1)), normally in the P step of PEMDAS. Then in the division step, we must apply division to the (entire) implicitly connected term. We cannot apply division to the coefficient (2) and then not apply division to the factor (2+1).

Incorrect: (because division is not EXPLICITLY applied to (1+2) )

6

--------- * (1+2)

2

Correct: (when EXPLICIT division is applied, it is applied to the entire term.)

6         (1+2)

--------- ÷ ---------

2           1

Correct: (For multiplication, inverting the denominator will maintain its value.)

6           1

--------- * ---------

2         (1+2)

As we can see, we cannot add an EXPLICIT operator to 6÷2(1+2) without maintaining the IMPLICIT relationship that was stated with IMPLIED multiplication. 6÷2(1+2) == 6÷(2*(1+2)).

The equation is not ambiguous

6÷2(1+2) is not equal to 6÷2*(1+2)

Standing alone, 6 will always equal 6 and 2(1+2) will also always equal 6, so dividing these two terms will always equal 1.

We would not write 2(a+b) as (2(a+b)), nor can we re-write the given equation as 6÷2*(a+b). We can however, rewrite the full equation to include an implicit multiplier if we take care to maintain the grouping with additional bracketing 6÷(2*(a+b)). Granted, the obelus (÷) is an archaic symbol for division, computers use the forward slash for division 6/(2*(a+b)).

• 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 without additional bracketing, in other words, 6÷2(1+2) is not equal to 6÷2*(1+2).

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(1+2)=?

3÷(1+2)=?

3÷(3)=?

3÷3=1

When we reduce by the common denominator, this does not complete the division operation, division must be applied to both the factor and the coefficient. Division is sustained.

• In the notation above, a multiplier of one is always implied for any number. For example, x == 1x.

Correct solution written as a fraction

In the denominator, both the coefficient and the factor must be divided in the following manner.

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.

Entering the equation into a programming language, or low quality calculator, requires the explicit multiplication symbol and outer parenthesis to maintain the grouping.

6÷2(1+2) == 6÷(2*(1+2))

Other calculators, including Google and Wolfram will simply strip the parenthesis and solve a different equation, 6÷2*3.

This is because within the programming, the open parenthesis (bracketing) triggers a different function within the programming, an open parenthesis tells the compiler to find the innermost parenthesized term and work outwards. Thus to get the correct answer from inferior calculators the input must be formalized with correct bracketing. I.e. 6÷(2*(1+2))

Spreadsheets

Entering the equation as =6/2(1+2) into a cell in a LibreOffice Calc spreadsheet will result in Err:509 (Missing operator), Google Sheets also returns an error (Formula parse error). This forces the user to format the equation using explicit multiplication. To avoid the left to right problem, the compiler must be instructed in advance to prepare for the function with an extra pair of parenthesis =6/(2*(1+2)).