BODMAS as taught in Europe is the same but just clearer. Brackets first, Orders, Division, Multiplication, Addition and Subtraction. All in that order.

So

Brackets (2+1), and inside there’s just Addition, so now (3)

Division 6/2 =3

multiplication 3*3

= 9

If you were to apply PEMDAS in the same way, you’d end up with Multiplication before Division, and result of 1, but that’s not how PEMDAS works, as I understand it (not being American). Thus the confusion. BODMAS is just clearer.

The above arguments on maths texts are fine, but don’t correlate to basic BODMAS teachings in primary schools. Ask any kid who knows BODMAS and they’ll give you 9. They are unlikely to know a fractional calculus approach. Many computers and calculators follow the same convention, hence 9. Here’s one based on teaching BODMAS to kids, result… 9 http://www.mathsisfun.com/algebra/operations-order-calculator.html?i=6%2F2(2%2B1)

And there’s the word, convention. 9 or 1, the answers are both right or wrong depending on convention.

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]]>(.01307385*Power(.01499508,2)/((.01307385 – .00653693) * Power((.01499508 – .00653693),2)) – .22)

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]]>Kind Regards,

mm

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]]>According to Measure and Integral: An Introduction to Real Analysis

By Richard Wheeden, Richard L.

μ/2(k+1) =

μ

——-

2(k+1)

Let μ=6, k=2. What is the result?

No math text I have ever seen writes: a/bc = (a/b)c. “bc” is a composite quantity. So directly entering such a thing into a computer (today), may result in an incorrect solution, thus using parentheses might be required when using calculator.

What is a composite quantity?

Bonnycastle’s introduction to algebra

By John Bonnycastle, James Ryan, John Francis Jenkins

…3abc is a composite quantity, the factors of which are 3, a, b and c.

You can’t divide by the (3) in the above quantity, and then multiply by abc, because you changed the quantity.

1/2y = (1/2)y is NOT a “standard” by any math text that I have seen.

An Introduction to Real Analysis: The Commonwealth and International Library

By Derek G. Bal

ε/2K =

ε

—

2K

Let’s suppose ε=6 and K=2+1 here. What is the result ?

Let’s not forget

Introduction to Real Analysis

By Bartle and Sherbert (3rd Edition)

where 1/n(n+1) =

1

——–

n(n+1)

Let’s apply this to, for example, :

6/n(n+1) and let n=2. What is the result ?

What notation do they use to denote a fractional coefficient?

F(x) := (2/3)x^(3/2)

Mathematical Constants? Same concept:

π/2n means π/(2n). We don’t need the ( ) to clutter things up. It comes with experience of doing mathematics at this level.

And many more examples from the books:

g’ (1/2nπ) 3, then 0 < n² / n! < n/(n – 2)(n – 1) < 1/(n – 3)

Notice how they don't pull some BS move where:

n/(n – 2)(n – 1) = n / (n-2) x (n – 1) = n(n – 1) / (n – 2)

I think everyone gets the point now. What book says 1/n(n+1) = 1 / n x (n+1) = (n+1)/n?

None. That's how many.

If no books use this notation, would you consider it STANDARD ?

What books uses 1/2y = (1/2)y instead of 1 / (2y)?

None.

A Radical Approach to Real Analysis By David M. Bressoud:

Does he write 1/2(200/199) when they mean (1/2)(200/199)?

NOPE. He writes it as:

(1/2)(200/199)

What about (1/2)x ? Just that. Not 1/2x.

Do you believe authors of calculus books or keyboard mathematicians? That is for you to decide.

Bonnycastle's introduction to algebra

By John Bonnycastle, James Ryan, John Francis Jenkins

Division:

RULE: Set the dividend OVER the divisor in the MANNER OF A FRACTION, and reduce it to its simplest form, by cancelling letters and figures that are common to each term.

Examples (from the book)

6ab ÷ 2a =

6ab

—– = 3b

2a

and wikipedia that says 1/2y = 1 / (2y) is exception to the "standard", yet I can't find a single text that uses 1/2y = (1/2)y.

So according to the authors of published university calculus books, if you want 9 to be your answer, then the correct notation is:

(6/2)(2+1) = 9

Otherwise, we can deduce from 1/n(n+1), that 6/n(n+1) = 1 ; if n=2.

Dave Pinal showed all the steps to "pe(md)(as)", However in step 2, he "added a multiplication sign for 'further clarifcation'."

This of course changes the interpretation, as seen above.

1 / 2y = 1 / (2y)

but

1 / 2 * y = y / 2

I can also show the steps.

First. rewrite as a fraction like every math books shows you:

If 6/2(a+b) =

6

—–

2(a+b)

then 6/2(2+1) =

6

——–

2(2+1)

Step 1: Parentheses

6

—-

2(3)

Step 2: Division OR Multiplication. I need to add here, that there is NO NECESSITY to go LEFT TO RIGHT. This is by the axioms of the number system. You collapse equations according to precedence.

I will demonstrate:

Since mult. and div. are equal precedence, choose either one first. Say, division.

6 / 2 = 3/1, giving:

3

—- then division again, (since you removed the multiplicand by division) resulting in 1.

3

Or, multiplication first: 2(3) = 6, giving:

6

—- then divide, resulting in 1.

6

Everyone running around these days, typing "LEFT to RIGHT" in giant letters is removing the underlying understanding of why left to right "works". I can demonstrate for any equation, that "Left to Right" is not required. That said, what is not allowed is to add parentheses around certain portions of an equation, which of course changes things drastically, which is what the associative property demonstrates.

Example:

8 ÷ 4 ÷ 2 = ?

The operands and their associated operators must remain together. Thus, I can do this operation in any order I want. I can divide by EITHER the 4 or the 2 first, then the other one:

8 ÷ 4 ÷ 2 = 8 ÷ 2 ÷ 4

That is because it is really: 8 (÷4) (÷2), which translates to:

8 * (1/4) * (1/2)

What you cannot do is: 8 ÷ 4 ÷ 2 = 8 ÷ (4 ÷ 2).

This is obviously an incorrect math statement since 8 ÷ (4 ÷ 2) = 8 ÷ 4 * 2 ; which is clearly not 8 ÷ 4 ÷ 2

Again, if you really understand the math basics, you'd know that "left to right" is only a tool for people who don't.

IMO, the major crux to the OP's argument is: "calculator's inline parsing convention".

We can't blindly enter things into calculators. Some knowledge of the input is required, as well as some knowledge of the computer's parsing convention. If we know that 1/2y is 1/(2y), but the computer will make it (1/2)y, then we add the necessary ( ). As the saying goes when entering info into a computer: Garbage in = Garbage out.

The OP showed the steps that the CALCULATOR uses to get 9, which is an explanation of why the calculator gets "9".

For hundreds of years, equations of the type 6/a(a+b) have been interpreted

6

——–

a(a+b)

and is still published in most (if not all) math, analysis, and algebra texts today.

Let a=2, and b=1.

Any Questions?

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]]>“any procedure or system of calculating it”

And any machine’s solution needs to be verified by a human for correctness. I am not talking about “how a machine would parse this equation”. I am talking about how humans have solved this for the last couple hundred years. Not sure how this will turn out on this forum, but let’s try:

Intro to Real Analysis by Bartle and Sherbert:

—————-QUOTE——-

X :=

1

— : n ∈ ℕ

2n

or more simply: X = 1/2n

———-End Quote———

You can verify that passage by googling the eBook.

The introduction to machines and programs have very much clouded the way things have been done historically. Wolfram|Alpha changed its code in 2013 and was solving it as above, before the change.

One of your arguments relies on : 2(2+1) = 2 x (2+1)

But doesn’t 2(2+1) also equal (2x(2+1)) ??

This is why the equation is ambiguous.

Since you also used machines as an argument, I strongly welcome your comment the following: Use wolfram|alpha and type in:

First example: cos2a

Second: cos2a/cos2a

Third: cos2a/b/cos2a/b

You can also find examples of algebraic division on a mathematician’s math page at Syracuse University:

http://cstl.syr.edu/fipse/algebra/unit2/parenth.htm

Eg: 6 / 2n = 3/n

Now let n=2+1 and solve both sides…

Remeber how we were taught to simplify algebraic division?

1 – Divide the coefficients of the like terms

2 – Subtraction the powers (exponents) of the like terms

Eg: 4a^2 / 2a = ??

Divide coefficients: 4 / 2 = 2

Subtract powers: 2-1 = 1

result: 2a

I stand by 1 as how I would solve it and how my colleagues would accept and understand it, however, after researching the topic, it is very clear how ambiguous it is :)

I look forward to your comments on these points.

King Regards,

your beloved mathman :)

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]]>Your explanation from the point view of 6/6 overlooked one math rule; Introduction of parenthesis or bracket.

While 6/(2(2+1)) = 1,

6/2(+1) = 9.

What helps our communications in math is the parenthesis or bracket.

6/6 = 6/(4+2) = 6/(2(2+1))! Here, as a math student one knows that the 2(2+1) is one as a denominator. However,

6/2(2+1) = 6/2x(2+1), which when either BODMAS or PEMDAS with LEFT TO RIGHT is applied you’ll get 9.

6/2x(3) = 3×3=9..

Now consider, 1/2n was shown in your comment.

This could be interpreted as (1/2) x n or 1 /(2n)

You’ll agree with me that both would give different answers. But in simple math operation, the expression if written the way it’s stated, i.e. 1/2n, any procedure or system of calculating it would take (1/2) x n. If the other is meant, it must surely be communicated with parenthesis or bracket, i.e. 1/(2n) any no mathematical method or procedure of calculation would mistake its answer.

Therefore beloved mathman, the expression 6/2(2+1) would simply be solved as 6/2*3=3*3=9.

Thank you.

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]]>.

6÷6=1..

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]]>LikeLike

]]>This equation is 6 over 2(2+1).

No one ever wrote a / bc when they meant ac / b

That is because they are different expressions.

Distributive property cannot change the final value,

and, when used, is 6 / (4+2).

One more thing, do the factoring for 6 + 3 = 9

Now factor 6/2 from 6 + 3 and we get:

(6/2)2 + (6/2)1 = 9

Notice how 6/2 MUST be in parentheses?

Now final step:

(6/2)(2+1)

The same simple idea applies to “Foiling” binomials:

x+1(x+2) = x + ((1)x + (1)2)

(x+1)(x+2) = (x+1)x + (x+1)2

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]]>6/2*(1 + 2) = 9

6/2(1+2)= 1

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