1,815
edits
Changes
:2÷3 = x
Now we compare the value given in the equation to the value we find found (above) for x.
:2÷3 '''!=''' 2(1+2)
:6÷2(1+2) '''!=''' 9
<p style="text-align:center;"> {{highlight|Without fail, 2(1+2) will always equal (2*1 + 2*2), this is the '''Distributive Law.'''}}</p>
== Proof of correct solution ==
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.
'''Incorrect:''' (''because division is not EXPLICITLY applied to (1+2) the entire parenthetical expression'')
<div style="color:red;">
<div style="width:60px; text-align:center; float:left;">
::''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 coefficientof the parenthetical expression.
==The equation is not ambiguous ==
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 grouping parenthetical expression with additional bracketing 6÷(2*(a+b)).
=== Obelus ===
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 value of the groupingparenthetical expression.
'''6÷2(1+2) == 6÷(2*(1+2))'''