I have been reading books in the algebra, and I noticed that some books use terms "usual multiplication" and "usual addition". Do they carry different meaning that multiplication and addition? If "usual" means something, is there such things as "usual division and subtraction"?
[Math] difference between the “usual multiplication” and multiplication
abstract-algebraarithmetic
Related Solutions
It is false that your PEDMSA rule (without the left-to-right rule) suffices to correctly evaluate any arithmetic expression.
$(6/3)/2 \color{red}\ne 6/(3/2)$.
$2^{(1^2)} \color{red}\ne (2^1)^{^2}$.
This shows that you still need the left-to-right rule when processing each arithmetic operation, and need the inside-out rule for exponentiation. It is true that with these rules in place, PEDMSA works. But the justification given by other existing answers is incomplete. To handle arbitrarily long arithmetic expressions, you will need induction. Also take note that exponentiation in some programming languages like Python have right-to-left associativity, so 2**1**2 is interpreted as 2**(1**2); this is obviously because there is no superscript in source code.
Here is the proof that PEDMSA+LTR+IO (namely PEDMSA with the left-to-right and inside-out rules) works for all arithmetic expressions involving $+,-,*,/$ and superscript-exponentiation and brackets, but no unary negation.
Take any expression $E$. By induction we can assume that PEDMSA+LTR+IO works for expressions that have fewer operations than $E$. Here parentheses are counted as operations. We shall analyze what happens when we apply PEDMSA+LTR+IO to $E$. We have the following cases:
$E$ has a parenthesis: The inner expression has fewer operations than $E$, and is hence evaluated correctly. After that, the resulting outer expression now has fewer operations than $E$, and hence we get the correct answer.
$E$ has no parentheses but has exponentiation: $E$ is reduced in the same way as conventionally (inside-out for chained exponentiations), so we get the correct answer.
$E$ has division and no higher-precedence operation: Let $p$ be the position of the first $/$ in $E$. By considering the $*$ operations immediately to the left of $p$, if any, it is clear that $E$ is of the form $\cdots [x * \cdots] y / z \,\cdots$, where there is no $*$ or $/$ immediately before $x$, and there is only a chain of $*$ between $x$ and $y$, and here the square-brackets in $[x * \cdots]$ denote that that subexpression is optional (it will not be there if whatever operation immediately to the left of $p$ is not $*$). Then the value of $E$ is $\cdots \Big( \big( [x * \cdots] y \big) / z \Big) \,\cdots = \cdots \Big( [x * \cdots] \big( y / z \big) \Big) \,\cdots$, and so our reduction results in an equal expression with fewer operations, and hence we get the correct answer.
$E$ has multiplication and no higher-precedence operation: $E$ is reduced in the same way as conventionally (leftmost $*$ first), so we get the correct answer.
$E$ has subtraction and no higher-precedence operation: $E$ is of the form $[x + \cdots] y - z \,\cdots$, where there is only a chain of $+$ between $x$ and $y$, or possibly $[x + \cdots]$ is empty (if the first operation is $-$). The value of $E$ is $\Big( \big( [x + \cdots] y \big) - z \Big) \,\cdots = \Big( [x + \cdots] \big( y - z \big) \Big) \,\cdots$, and so our reduction yields an equal expression with fewer operations, and hence we get the correct answer.
$E$ has only addition: Our reduction yields the correct answer because any reduction does.
I shall leave it as an exercise to extend this theorem and proof to cater for unary negation. One must think very carefully about how to amend the evaluation rules to correctly handle expressions like:
$-3/-1*--4--2*---5$
If you do not consider this valid, then you must decide and precisely specify what expressions are valid.
You may notice that subtraction can be seen as a special case of addition, i.e. $$a-b=a+(-b)$$ Thus by commutative law, $$a-b=a+(-b)=(-b)+a=-b+a$$ It is certainly not $b-a$.
Similarly, division is as a special case of multiplication, i.e. $$\frac{a}{b}=a\cdot(\frac{1}{b})$$ Where $\frac{1}{b}$ means the inverse element of $b$ with respect to multiplication.
Therefore, $$\frac{a}{b}\cdot{c}=a\cdot\frac{1}{b}\cdot{c}$$ However, $$\frac{a}{b\cdot{c}}=a\cdot\frac{1}{b\cdot{c}}=a\cdot\frac{1}{b}\cdot\frac{1}{c}$$ And they are certainly not equal.
Edited: Since you are motivated to pursue advanced math, I have a little idea to share.
I assume you have intuitively understood the arithmetic about nature numbers. It is based on this simple idea: the one-to-one correspondence between (abstract) nature numbers and (concrete) things. Then you easy find that:
$(+, 1)$ The commutative law of addition $$a+b=b+a$$
$(+, 2)$ The associative law of addition $$a+(b+c)=(a+b)+c$$
$(\times, 1)$ The commutative law of multiplication $$a\cdot{b}=b\cdot{a}$$
$(\times, 2)$ The associative law of multiplication $$a\cdot(b\cdot{c})=(a\cdot{b})\cdot{c}$$
$(\times, 3)$ The identity element of multiplication, i.e. $1$ $$a \cdot 1=1 \cdot a=a$$
$(+, \times)$ the distribution law of multiplication to addition $$a \cdot (b+c)=a\cdot {b}+a \cdot {c}$$
A possible intuitive approach has been shown by Paul Sinclair. And you also have noticed that addition and multiplication can be defined as recursive operation. That is, $$a+b:=a+\underbrace{1+1+1+...+1}_{b \text(terms)}$$ And $$a\cdot{b}:=\underbrace{a\cdot{a}...\cdot{a}}_{b \text(terms)}$$
You also have a nature idea to introduce the inverse operation, i.e. substitution and division. But you may find that $3-5$ is illegal in this stage (it has no definition), and neither is $\frac{3}{5}$.
But that is not hard for you, just noticed that you can define $\frac{3}{5}$ as a ratio: There certainly can be a thing $k$ such that $k\cdot{5}=3$, then we can denote $k$ by $\frac{3}{5}$. Here come the rational numbers! And you may just find
$(\times, 4)$ The reverse element of multiplication $$\text{For every number $a$ there is a number $b$ such that $a\cdot{b}=b\cdot{a}$}$$
And we can easily find that by definition, $b=\frac{1}{a}$. (the "number" here means nature number and positive rational number.)
However you are not satisfied. Sometimes you need to find a way to denote "nothing", so you need
$(+, 3)$ The identity element of addition, i.e. $0$ $$\text{For all $a$, $a+0=0+a=a$}$$
As far as I'm concerned, historically, in a relative long time, $0$ or similar notations were initially but for this kind of convenience.
A similar idea for convenience is negative number, it was originally used to express debt. With introducing negative numbers, we can derive:
$(+, 4)$ The reverse element of addition $$\text{For every number $a$ there is a number $b$ such that $a+b=b+a=0$}$$
And we can also easily find that by definition, $b=-a$.
Indeed, It took people a lot of time to accept the philosophical aspect of "nothingness", or to understand how debt times debt will be income (this was not philosophical confusion initially, but a result of wrong metaphor; however, it finally became one, but that is another story.)
But for mathematics, the most horrible thing is division by 0, it is certainly not legal. There is no help for it, so we have to ban it. Indeed, this is the only difference between addition and multiplication in abstract sense. If you consider about a more general case such as real numbers, the original, recursive idea will failed, or, at least, no more intuitive. Thus you have to define addition and multiplication in a more abstract way, that is, introducing the axioms of addition and multiplication, i.e. $(+,1)-(+,4),(\times,1)-{(\times,4)},(+,\times)$ I've mentioned before. If you compare $(+,1)-(+,4)$ with $(\times,1)-(\times,4)$, you will find they are nearly the same.
A set equipped with two operation satisfying those axioms will be the field which trb456 mentioned. (To say it clearly, the result of those operations must always within that set, and the identity element of addition can't be the same with that of multiplication.) We need this term for other reason: because there are a lot of other structures satisfying this definition. And modern algebra, though I am totally not familiar with it yet, seems like a study of mathematical structures.
This story is obviously not the real history of numbers and their arithmetic, and I just want to introduce a half-intuitive approach to understand those things. I omitted a lot of details, or maybe I just got things wrong. But I hope you enjoy it.
Best Answer
"Usual multiplication" in this sense refers to the elementary definition of multiplication which is when a number $a$ is added to itself multiple times (we'll call the frequency $b$), which is thought as repeated addition: $a \times b = \underbrace{a + \cdots + a}_{\text{$b$ times}}$. However, just "multiplication" in this sense refers to the combining of matrices, vectors or any other sort of quantities under certain rules (probably used most in abstract algebra) to get a certain product. Basically, "usual" refers to elementary. I'm sure "usual subtraction" and "usual division" do exist like "usual addition" and "usual multiplication" because they all refer to the elementary use of these operations. Do note that in abstract algebra, we only use the addition and multiplication operations because subtraction and division can be rewritten in addition and multiplication respectively.
So, the bottom line is..."usual multiplication" is elementary multiplication, which is NOT the same thing as mutliplication in abstract algebra. Same goes for addition.