'+ 2 3' becomes 'add 2 and 3'
In any case, it's a silly objection.
+ 2 3
is naturally how you'd write addition if addition were a standalone Unix command!(You can't just remember what you learned in math class when writing a computer program, because now we have bitwise and, or, xor, left shift, right shift, equality, pointer dereferencing, assignment, logical and, or, modulo, pre-increment, post-increment, bitwise not, logical not, etc. So you have to learn it all again. And it's different for every language.)
Put everything up front, and suddenly you don't have to think about this anymore. (+ (expt 2 3) (/ 4 2)). The computer just does what you say and there are no rules to memorize.
Personally, I always use RPN calculators (dc is my favorite), because I don't trust the calculator to remember operator precedence. They usually do it right, but it's difficult to reason about. (The example would be "2 3 ^ 4 2 / + p" in dc.)
In Indian schools it used to be BODMAS, for Brackets Of Division Multiplication Addition Subtraction, for operator precedence, for which our teachers told us the mnemonic was badmaash, meaning a scoundrel or a villain, in Hindi. Guaranteed to make kids remember it. :)
Of, as in half or a quarter or ... of some other number.
I think FORTRAN copied it first. Which makes sense, because of the original scientific target market for FORTRAN, represented in its name derived from “formula translating system”. Translating “a sufficiently rich mathematical language”, as Backus put it, was an explicit goal as early as 1954.
C then (presumably) just followed the convention established by FORTRAN.
what is unnatural is the complicated case specific precedence rules mid notation requires. rpn still has precedence rules but because they are simple it appears that there are none and because they are universal there is no confusion as to where any new operators fit in.
I wonder if anybody has ever written a linguistic history of math notation and how it ended up the way it did.
(x-3)(x+2)-8 > 2x^2+3
x^2-x-8 > 2x^2+3
x^2+x+8 < -3
x^2+x+11 < 0
vs x 3 - x 2 + * 8 - x x * 2 * 3 + >
how would I simplify the expressions here? Maybe it's possible without transforming to infix but I can't do it.