Open main menu

Difference between revisions of "The equation that broke the Internet"

 
(64 intermediate revisions by the same user not shown)
Line 12: Line 12:
 
}}
 
}}
  
'''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 -- {{highlight|there is only one number that 6 can be divided by to result in 9, and that number is 2/3.}}
+
'''6÷2(1+2)=?''' is the algebra question that many Internet 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 -- {{highlight|there is only one number that 6 can be divided by to result in 9, and that number is 2/3.}}
  
 
==Proof incorrect solution ==
 
==Proof incorrect solution ==
We substitute x for 2(1+2) from the given equation:
+
We substitute x for 2(1+2) from the given equation and backtest the equation with the result of 9:
  
 
If  
 
If  
Line 25: Line 25:
 
:6÷9 = x
 
:6÷9 = x
  
there is a Common Denominator of 3 so we reduce the fraction...
+
there is a common divisor of 3 so we reduce the fraction...
  
 
:2÷3 = x  
 
:2÷3 = x  
  
Now we compare the value given in the equation to the value we find for x.
+
Now we compare the value given in the equation to the value we found (above) for x.
  
 
:2÷3 '''!=''' 2(1+2)
 
:2÷3 '''!=''' 2(1+2)
Line 37: Line 37:
 
:6÷2(1+2) '''!=''' 9
 
:6÷2(1+2) '''!=''' 9
  
<p style="text-align:center;"> Without fail, 2(1+2) will always equal 2(2*1 + 2*2), this is the Distributive Law.</p>
+
<p style="text-align:center;"> {{highlight|Without fail, 2(1+2) will always equal (2*1 + 2*2), this is the '''Distributive Law.'''}}</p>
  
== The Fundamentals of Algebra (1983) ==
+
== Proof of correct solution ==
 +
''Encyclopedia Britannica'':
 +
<blockquote>
 +
::'''''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.''
 +
::&ndash;[https://www.britannica.com/science/distributive-law Encyclopedia Britticana]
 +
</blockquote>
  
<blockquote>
+
Therefore, in 6÷2(1+2)=?, we start with the <span style="color:green;"> Distributive Law</span> in the <span style="color:green;">Parenthesis</span> step of the Order of Operations.
{{Quote|text=
 
''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 <u>must be multiplied by the coefficient.</u>''' This is done in the following manner.''
 
::''5(a + b) = 5a + 5b'' <br/>
 
::''3a(b - c) = 3ab - 3ac''
 
  
&ndash;[https://books.google.com/books?id=BabtEFxgZ2AC "Technical Shop Mathematics / Edition 2"], by John G. Anderson, ISBN-13:9780831111458, Industrial Press, Inc., 02/28/1983, Page:138
+
::'''6÷2(1+2)=?'''
}}
+
::'''6÷<span style="color:green;">(2*1 + 2*2)</span>=?'''
</blockquote>
+
::'''6÷(2 + 4)=?'''
 +
::'''6÷(6)=?'''
 +
::'''6÷6=1'''
  
 
==Processing rules==  
 
==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.
+
In the majority of incorrect proofs the respondent will cite PEMDAS or BODMAS, we all agree that parentheses or brackets are processed first.  However, some respondents incorrectly resolve the [P]arentheses or [B]rackets of PEMDAS - BODMAS, for some reason, these solvers believe that they can resolve within the parentheses, but neglect to multiply the adjacent and dependent coefficient.
 +
 
 +
=== PEMDAS ===
 +
<blockquote>
 +
 
 +
::P parenthesis
 +
::E exponentiation
 +
::M multiplication
 +
::D division
 +
::A addition
 +
::S subtraction
  
::'''Note''' Even more puzzling is that incorrect solvers will cite PEMDAS and then suspend it, doing the addition within the parenthesis before multiplication. Rather than correctly distributing the multiplication across the parenthetical term. I.e., 2(1+2) == (2 * 1 + 2 * 2).  (See the Distributive Law below.)
+
::&ndash;[https://mathworld.wolfram.com/PEMDAS.html Wolfram PEMDAS]
 +
</blockquote>
  
 +
=== Parentheses ===
 
''Wolfram Mathworld'' on parenthetical expressions:
 
''Wolfram Mathworld'' on parenthetical expressions:
 
<blockquote>
 
<blockquote>
Line 61: Line 76:
 
::[...]
 
::[...]
 
::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.''
 
::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.''
::&ndash;[http://mathworld.wolfram.com/Parenthesis.html Parenthesis]
+
::&ndash;[http://mathworld.wolfram.com/Parenthesis.html Wolfram Parenthesis]
 
</blockquote>
 
</blockquote>
  
''Encyclopedia Britticana'' on parenthetical expressions:  
+
===Distributive law===
 +
''Encyclopedia Britannica'' on parenthetical expressions:  
 
<blockquote>
 
<blockquote>
 
::'''''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.''
 
::'''''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.''
 
::&ndash;[https://www.britannica.com/science/distributive-law Encyclopedia Britticana]
 
::&ndash;[https://www.britannica.com/science/distributive-law Encyclopedia Britticana]
 
</blockquote>
 
</blockquote>
 
+
<!--
 
Therefore, in 6÷2(1+2)=?, we start with the Distributive Law in the <span style="color:green;">Parenthesis</span> step of the Order of Operations.
 
Therefore, in 6÷2(1+2)=?, we start with the Distributive Law in the <span style="color:green;">Parenthesis</span> step of the Order of Operations.
  
 
::'''6÷2(1+2)=?'''
 
::'''6÷2(1+2)=?'''
::'''<div style="color:green;">(2*1 + 2*2)=?</div>'''
+
::'''<span style="color:green;">(2*1 + 2*2)</span>=?'''
 
::'''6÷(2 + 4)=?'''
 
::'''6÷(2 + 4)=?'''
 
::'''6÷(6)=?'''
 
::'''6÷(6)=?'''
Line 87: Line 103:
  
 
::<s>'''<div style="color:red;">3*(3)</div>'''</s>
 
::<s>'''<div style="color:red;">3*(3)</div>'''</s>
 +
-->
 +
=== Parenthetical expressions ===
  
== Implicit vs implied multiplication ==
+
<blockquote>
 +
{{Quote|text=
 +
''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 <u>must be multiplied by the coefficient.</u>''' This is done in the following manner.''
 +
::''5(a + b) = 5a + 5b'' <br/>
 +
::''3a(b - c) = 3ab - 3ac''
  
Implicit 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. Apparently, this rule has been forgotten with the advent of modern calculators.
+
&ndash;[https://books.google.com/books?id=BabtEFxgZ2AC "Technical Shop Mathematics / Edition 2"], by John G. Anderson, ISBN-13:9780831111458, Industrial Press, Inc., 02/28/1983, Page:138
 +
}}
 +
</blockquote>
  
6÷2(1+2) has one implicit operator and therefore has two terms.
+
== Explicit vs Implicit (implied) multiplication ==
 +
<div style="text-align:center">
 +
{{Quote|text='''''Implied multiplication''' has a higher priority than explicit multiplication to allow users to enter expressions, in the same manner as they would be written.''<br/>
  
<div style="width:100px; float:left;">
+
&ndash;[https://epsstore.ti.com/OA_HTML/csksxvm.jsp?nSetId=103110 Implied Multiplication Versus Explicit Multiplication on TI Graphing Calculators]
::6         
+
}}
:---------
 
:2(1+2)     
 
 
</div>
 
</div>
<div style="clear:both;"></div>
 
If we introduce a second implicit operator 6÷2*(1+2) the equation now has three terms. Due to the Order of Operations, adding an implicit operator changes the value of the equation.
 
  
<div style="width:200px; float:left; color:red;">
+
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.''
::6           
+
 
:---------  * (1+2)
+
Implied multiplication is also implicit multiplication.
::2       
+
 
 +
{{Quote|text=
 +
'''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.
 +
 
 +
&ndash;Oxford
 +
}}
 +
 
 +
6÷2(1+2) has two EXPLICIT operators, division and addition. However, IMPLIED multiplication tells us to IMPLICITLY multiply, [2(2+1)], in the P step of PEMDAS.
 +
In the division step, we must apply division to the (entire) implicitly connected parenthetical expression.  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 the entire parenthetical expression'')
 +
<div style="color:red;">
 +
<div style="width:60px; text-align:center; float:left;">
 +
6           
 +
----   
 +
2       
 
</div>
 
</div>
<div style="clear:both;"></div>
+
<div style="width:60px; text-align:center; float:left;">
Where a calculator or programming language requires implicit operators, we maintain the value of the equation by adding an additional grouping pair of parenthesis  6÷(2*(1+2)) for inline notation.
+
<br/>
<div style="width:100px; float:left;">
+
<nowiki>*</nowiki>
::6         
+
<br/>
:---------
 
:(2*(1+2))     
 
 
</div>
 
</div>
<div style="clear:both;"></div>
 
  
==The equation is not ambiguous ==
+
<div style="width:60px; text-align:center; float:left;">
===6÷2(1+2) is not equal to 6÷2*(1+2)===
+
<br/>
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.
+
(2+1)
 +
<br/>
 +
</div>
  
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 expression to include an implicit multiplier if we take care to maintain the grouping with additional bracketing for use with calculators and programming languages 6÷(2*(a+b)). Granted, the [https://www.wyzant.com/resources/blogs/14831/the_obelus_for_division obelus (÷)] is an archaic symbol for division, computers use the forward slash for division 6/(2*(a+b)).
+
<div style="width:60px; text-align:center; float:left;">
 +
<br/>
 +
=
 +
<br/>
 +
</div>
  
* 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).
+
<div style="width:60px; text-align:center; float:left;">
 +
<br/>
 +
3 * (1+2)        
 +
<br/>   
 +
</div>
 +
</div>
 +
</div>
 +
<div style="clear:both;"></div>
  
{{Quote|text='''''Implied multiplication''' has a higher priority than explicit multiplication to allow users to enter expressions, in the same manner as they would be written.''<br/>
+
'''Correct:''' (''when EXPLICIT division is applied, it is applied to the entire term.'')
 +
<div style="color:green;">
 +
<div style="width:60px; text-align:center; float:left;">
 +
6         
 +
---- 
 +
1     
 +
</div>
 +
<div style="width:60px; text-align:center; float:left;">
 +
<br/>
 +
÷
 +
<br/>
 +
</div>
  
&ndash;[https://epsstore.ti.com/OA_HTML/csksxvm.jsp?nSetId=103110 Implied Multiplication Versus Explicit Multiplication on TI Graphing Calculators]
+
<div style="width:60px; text-align:center; float:left;">
  }}
+
2(1+2)       
 +
----  
 +
1   
 +
</div>
  
==Correctly solving left to right==
+
<div style="width:60px; text-align:center; float:left;">
 +
<br/>
 +
=
 +
<br/>
 +
</div>
  
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
+
<div style="width:60px; text-align:center; float:left;">
 
+
2 * (1+2)        
::''6÷2(1+2)=?''
+
----  
::''3÷1(1+2)=?''  
+
6 * 1    
::''3÷(1+2)=?''
+
</div>
::''3÷(3)=?''
+
</div>
::''3÷3=1''
+
</div>
 +
<div style="clear:both;"></div>
  
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.
+
'''Traditional solution:'''
* In the notation above, a multiplier of one is always implied for any number. For example, x == 1x.
 
  
==Solution written as a fraction ==
+
<div style="color:green;">
In the denominator, both the coefficient and the factor must be divided in the following manner.
+
<div style="width:60px; text-align:center; float:left;">
 
 
<div style="width:100px; text-align:center; float:left;">
 
 
6           
 
6           
 
----   
 
----   
 
2(1+2)       
 
2(1+2)       
 
</div>
 
</div>
<div style="width:100px; text-align:center; float:left;">
+
<div style="width:60px; text-align:center; float:left;">
 
<br/>
 
<br/>
 
=
 
=
Line 156: Line 237:
 
</div>
 
</div>
  
<div style="width:100px; text-align:center; float:left;">
+
<div style="width:60px; text-align:center; float:left;">
 
3           
 
3           
 
----   
 
----   
(1+2)       
+
1*(1+2)       
 
</div>
 
</div>
  
<div style="width:100px; text-align:center; float:left;">
+
<div style="width:60px; text-align:center; float:left;">
 
<br/>
 
<br/>
 
=
 
=
Line 168: Line 249:
 
</div>
 
</div>
  
<div style="width:100px; text-align:center; float:left;">
+
<div style="width:60px; text-align:center; float:left;">
 
3           
 
3           
 
----   
 
----   
Line 174: Line 255:
 
</div>
 
</div>
  
<div style="width:100px; text-align:center; float:left;">
+
<div style="width:60px; text-align:center; float:left;">
 
<br/>
 
<br/>
 
=  1
 
=  1
 
<br/>
 
<br/>
 +
</div>
 
</div>
 
</div>
 
<div style="clear:both;"></div>
 
<div style="clear:both;"></div>
 +
 +
<!--
 +
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)).
 +
-->
 +
 +
==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 by expressing the common divisor first. However, division is sustained and must be applied to everything to the right of the explicit division operator.
 +
 +
::''6÷2(1+2)=?''
 +
::''3÷1(1+2)=?'' 
 +
::''3÷(1+2)=?''
 +
::''3÷(3)=?''
 +
::''3÷3=1''
 +
 +
When we reduce by the common divisor, this does not complete the division operation, it simply reduces the coefficient to one. Division must be applied to both the factor and the coefficient of the parenthetical expression.
 +
 +
==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) and maintain the same value. We can however, rewrite the full equation to include an explicit multiplier if we take care to maintain the parenthetical expression with additional bracketing 6÷(2*(a+b)).
 +
 +
=== Obelus ===
 +
Granted, the [https://web.archive.org/web/20210417145938/https://www.wyzant.com/resources/blogs/14831/the_obelus_for_division obelus (÷)] is an archaic symbol for division, it visually represents a fraction with one dot being the numerator and the other being the denominator. Computers use the forward slash for division because the standard keyboard does not have an obelus key. 
 +
 +
For those who claim that the obelus is outdated, why then is it present on virtually every modern calculator?
  
 
== The calculator problem ==
 
== The calculator problem ==
Line 185: Line 294:
 
The [https://www.desmos.com/scientific Desmos Scientific Calculator] handles our given equation correctly. The error checking appears to be built into the division key.
 
The [https://www.desmos.com/scientific 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.
+
Entering the equation into a programming language, or low quality calculator, requires the explicit multiplication symbol and outer parenthesis to maintain value of the parenthetical expression.
  
 
'''6÷2(1+2) == 6÷(2*(1+2))'''
 
'''6÷2(1+2) == 6÷(2*(1+2))'''
  
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 in programming languages, the compiler must be instructed in advance to prepare for the function with an extra pair of parenthesis 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)).
 +
 
 +
==See also==
 +
* [https://www.nytimes.com/2019/08/05/science/math-equation-pemdas-bodmas.html "That Vexing Math Equation? Here’s an Addition"]
 +
* [https://www.youtube.com/watch?v=hsZCtgFcL40 "How to Solve 8÷2(2+2) Using Implied Multiplication"]
 +
* [https://slate.com/technology/2013/03/facebook-math-problem-why-pemdas-doesnt-always-give-a-clear-answer.html "What Is the Answer to That Stupid Math Problem on Facebook?"]
 +
* [https://www.inc.com/dave-kerpen/this-basic-math-problem-is-breaking-internet.html This Basic Math Problem Is Breaking the Internet: How do you solve this simple arithmetic problem?]

Latest revision as of 21:14, 30 March 2022

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 algebra question that many Internet 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 and backtest the equation with the result of 9:

If

6÷x = 9

then

6÷9 = x

there is a common divisor of 3 so we reduce the fraction...

2÷3 = x

Now we compare the value given in the equation to the value we found (above) 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:

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*1 + 2*2)=?
6÷(2 + 4)=?
6÷(6)=?
6÷6=1

Processing rules

In the majority of incorrect proofs the respondent will cite PEMDAS or BODMAS, we all agree that parentheses or brackets are processed first. However, some respondents incorrectly resolve the [P]arentheses or [B]rackets of PEMDAS - BODMAS, for some reason, these solvers believe that they can resolve within the parentheses, but neglect to multiply the adjacent and dependent coefficient.

PEMDAS

P parenthesis
E exponentiation
M multiplication
D division
A addition
S subtraction
Wolfram PEMDAS

Parentheses

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

Parenthetical expressions

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)], in the P step of PEMDAS. In the division step, we must apply division to the (entire) implicitly connected parenthetical expression. 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 the entire parenthetical expression)

6


2


*


(2+1)


=


3 * (1+2)

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

6


1


÷

2(1+2)


1


=

2 * (1+2)


6 * 1

Traditional solution:

6


2(1+2)


=

3


1*(1+2)


=

3


3


= 1


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 by expressing the common divisor first. However, division is sustained and must be applied to everything to the right of the explicit division operator.

6÷2(1+2)=?
3÷1(1+2)=?
3÷(1+2)=?
3÷(3)=?
3÷3=1

When we reduce by the common divisor, this does not complete the division operation, it simply reduces the coefficient to one. Division must be applied to both the factor and the coefficient of the parenthetical expression.

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) and maintain the same value. We can however, rewrite the full equation to include an explicit multiplier if we take care to maintain the parenthetical expression with additional bracketing 6÷(2*(a+b)).

Obelus

Granted, the obelus (÷) is an archaic symbol for division, it visually represents a fraction with one dot being the numerator and the other being the denominator. Computers use the forward slash for division because the standard keyboard does not have an obelus key.

For those who claim that the obelus is outdated, why then is it present on virtually every modern calculator?

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 value of the parenthetical expression.

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)).

See also