Thread: 62(1+2) = ?
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Old January 15th, 2013, 11:30 AM   #309 (permalink)
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Hi all. I researched this topic well for about 2 weeks now and have come to the following conclusions. I will summarize what I have said in other forums with respect to the notations, then I will address other points.

if you want to say 0.5x, then you HAVE to write (1/2)x with parentheses or, x "all over 2" with a horiztonal fraction bar, or write x/2. I have never seen (1/2)x before I researched this equation, but since searching online, I HAVE seen fractional coefficients written this way, only because computers are limited to the horizontal typing space.
x/2 = (1/2)x = 0.5x
1/2n = 1/(2n) This sort of notation is used especially with pi, ln, or e. We have never had to say 1/(2pi). It was simply 1/2pi, or 1/2e^2.
I have always used ab/cd to mean (ab)/(cd) and I topped almost all of my calculus classes since high school through university.(moot point, I know)
Just to re-iterate, to use 6/2 as a fraction, parentheses are REQUIRED. Every book will tell you this.

Now consider the Identity Law:
a = 1a = 1(a)
We know there is ALWAYS an 'invisible' 1 as a ceofficient of a variable if no other number is there. Therefore:
a/a = 1, and if a is also 1a, then a/1a = 1. Blindly using 'pemdas', some folks would do this:
a/1a = a/1*a = a*a = a^2. I hope this drives home the silliness of this calculation.

Now, on to my second point:
consider: factoring, simplifying equations, and the distributive property.
Lets start with the number 6.
6 = (4+2). There is a common factor here: 2. So let's factor it out of both terms.
(4+2) = 2(2+1). The outside 2 remains a part of of the 2 inner terms at all times. It cannot be used in an operation by itself without the rest of (4+2). The reverse of factoring is distribution, so, 2(2+1) = 6. This has to be true always. The argument I have seen to this is that (6/2) can be distributed. This is true ONLY is 6/2 is in parentheses, otherwise, the 6 and 2 are separated by a division slash, and the 2 is a factor of 2+1.

So, let's prove the initial equation:
6/6 = 1
6/(4+2) = 1
6/2(2+1) = 1

the same can be done for other factors:
6/6 = 1
6/(3+3) = 1
6/3(1+1) = 1
Distribution is actually a part of "Simplifying Equations" and is not bound to the order of operations as "multiplication", since it is in fact "removing parentheses by distributing". This can be googled and several references found.
Simplifying 2(2+1) + 3(2+1) = 5(2+1). We "combined like terms" here, by adding, and did not perform the "parntheses" part of order of operations, nor did we multiply, which is also higher priority than adding, because we only simplified.

Lastly, I hear the argument that "This is strictly numbers and you don't use algebra rules since there are no variables". That is the most asinine arguement I have heard yet. All axioms, laws, and properties use variables, meaning that they hold true for "any number", hence the proofs with variables.
I welcome thoughts on this, in an intellectually formed response. I am tired of the 'flaming' that goes on by imbciles on some other forums with rebuttals like "it is 9. go back to grade 3 you moron", or "google says it is 9", when google changes the equation to (6/2)*(2+1), and wolfram contradicts itself with 2n/2n = 1, and 6/2n = 3/n, but then says 6/2(2+1) is 9. wolframs "terms" state that any answer should be verified with common sense and accuracy should also be verified.
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