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:6÷2(1+2) '''!=''' 9
<p style="text-align:center;"> Without fail, 2(1+2) will always equal (2*1 + 2*2), this is the '''Distributive Law.'''</p> == Proof of correct solution ==''Encyclopedia Britannica'' on parenthetical expressions: <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.''::–[https://www.britannica.com/science/distributive-law Encyclopedia Britticana]</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. ::'''6÷2(1+2)=?'''::'''6÷<span style="color:green;">(2*1 + 2*2)</span>=?'''::'''6÷(2 + 4)=?'''::'''6÷(6)=?'''::'''6÷6=1''' So the fatal error with solvers who emphasize the left to right rule, is in incorrectly handling the Parentheses step, disregarding the Distributive Law. ::<s>'''<div style="color:red;">6÷2(3)</div>'''</s> Some are even compelled to add an explicit multiplier, not given in the equation. ::<s>'''<div style="color:red;">6÷2*(3)</div>'''</s> ::<s>'''<div style="color:red;">3*(3)</div>'''</s>
==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.
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::'''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.)
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=== PEMDAS ===<blockquote> ::'''Note''' Even more puzzling is that incorrect solvers will cite PEMDAS and then suspend it, doing the addition within the P parenthesis before ::E exponentiation::M multiplication::D division::A addition::S subtraction ::–[https://mathworld. Rather than correctly distributing the multiplication across the parenthetical termwolfram. Icom/PEMDAS.e., 2(1+2) == (2 * 1 + 2 * 2). (See the Distributive Law below.)html Wolfram PEMDAS]</blockquote>
=== Parenthetical expressions ===
''Wolfram Mathworld'' on parenthetical expressions:
<blockquote>
::[...]
::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.''
::–[http://mathworld.wolfram.com/Parenthesis.html Wolfram Parenthesis]
</blockquote>
===Distributive law===''Encyclopedia BritticanaBritannica'' on parenthetical expressions:
<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.''
::–[https://www.britannica.com/science/distributive-law Encyclopedia Britticana]
</blockquote>
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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.
::<s>'''<div style="color:red;">3*(3)</div>'''</s>
-->
=== The Fundamentals of Algebra (1983) ===
<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''
–[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>
== Implicit vs implied multiplication ==
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.
6÷2(1+2) as written has one implicit operator outside of the parenthesis and therefore has two terms.
<div style="width:100px; float:left;">
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 expression full equation 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)).
* 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).