8÷2(1+3)

[Thomas Klemm]

I didn't learn algebra with PEMDAS. So I'm a bit confused.
Does this really mean addition before subtraction?
And multiplication before division?
How would you parse the following expressions?

\(1 - 2 + 3\)

\(
\begin{matrix}
a) & (1 - 2) + 3 & = & 2 \\
b) & 1 - (2 + 3) & = & -4 \\
\end{matrix}
\)

\(1 - 2 - 3\)

\(
\begin{matrix}
a) & (1 - 2) - 3 & = & -4 \\
b) & 1 - (2 - 3) & = & 2 \\
\end{matrix}
\)

\(1 \div 2 \times 3\)

\(
\begin{matrix}
a) & (1 \div 2) \times 3 & = & \frac{3}{2} \\
b) & 1 \div (2 \times 3) & = & \frac{1}{6} \\
\end{matrix}
\)

\(1 \div 2 \div 3\)

\(
\begin{matrix}
a) & (1 \div 2) \div 3 & = & \frac{1}{6} \\
b) & 1 \div (2 \div 3) & = & \frac{3}{2} \\
\end{matrix}
\)

If all your answers are \(a)\) then why on earth should implied multiplication make any difference?

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In my understanding the operations \(-\) and \(\div\) are left-associative.
The operations \(+\) and \(\times\) are associative.
And the power-operation is right-associative.

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Ok, I understand that maybe in the seventies it was hard to type something like the following in a paper:

\(
c_k = \frac{1}{2 \sqrt{\pi}} \cdots
\)

Thus instead of writing on a typewriter:

      1
c = ------ ...
 k      __
    2 -/pi


You'd rather write:

           __
c  = 1/2 -/pi ...
 k


But is that reason enough to come up with a special rule about implied multiplication?

---

The following expression is not ambiguous:

\(
\frac{1}{2\sqrt{3}}
\)

It only becomes a problem once you enter it into a calculator.

In Mathematica I'd enter it as:

1/(2Sqrt[3])

This leads to:

0.288675...

And yes, I'd be surprised if a calculator gives the same result if I enter: \(1\div2\sqrt{3}\).