Order AND Parentheses Free Notation?

In mathematics and logic expressions often get written with the operation in infix notation... we write a+b, a^b, a v b. As logicians and mathematicians know this means that we either have to establish an order of operations when we have more than operation involved, such as Exponents, Multiplication, Division, Addition, Subtraction for arithmetic, or use parentheses, or have ambiguous expressions. Since at least the time of Lukasiewicz, mathematical and logical scholars have known that we can stop excusing our dear Aunt Sally. In Lukasiewicz notation instead of writing (5+2)*4, we can simply write *+5 2 4. In reverse Lukasiewicz notation we can write 5 2+4*. I highly recommend that you read the link there, and translate some common logical or mathematical expressions into reverse Lukasiewicz notation. Since these notational systems establish that we can write mathematical and logical expressions in prefix notation (Lukasiewicz notation), infix notation, and postfix notation (reverse Lukasiewicz notation), can we write mathematical and logical expressions in a way such that the operations involved can come *anywhere* in the expression *without* having a fixed order as declined languages do?

For variables or constants, suppose we write a unique subscript below each of them. For an n-ary operation suppose we write n sequenced subscripts which correspond to the subscripts in the order of variables or constants it operates on. E. G. if we write x_a y_b+_ab, then we indicate that the operation "+" operates on x and y. We can then immediately write x_a+_aby_b as well as +_abx_ay_b. So, suppose we have a function F such that a b cF=b c*a+. Then we could write 4_c8_dF_cba6_a7_e+_de2_b. So, do you believe that this gives us an order free notation for the operations in a logical or mathematical expression? Please answer this question and write a response down on paper before reading my further comment.

Lately I've read Susan Bassein's An Infinite Series Approach to Calculus . She presents an interesting proof, in my opinion, that the positive square root of 2 cannot get expressed exactly in terms of rational numbers. At least once I could understand her language.

Here, we'll define a 2nd power root of n as any number x such that x²=n. The positive square root, denoted sqr(n) hereafter, consequently comes as a positive number which satisfies this equation. Due to the way exponentiation works there only exists one positive number here. With that definition in mind, we'll proceed to show that if b, and c are integers, then sqr(2) ≠ b/c.

First since sqr(2) ≠ b/c, we have that 2 &ne (b/c)². So, 2c² &ne b². We can also derive sqr(2) ≠ b/c from 2c² &ne b², so if we show 2c² &ne b², then the conjecture follows. We'll now develop some seemingly trivial lemmas.

Lemma 1: the product of two odd numbers d, e is another odd number.
Since d=2y+1, and e=2z+1, we have de=(2y+1)(2z+1)=4yz+2y+2z+1=2(yz+y+z)+1. Since yz+y+z equals another integer, 2(yz+y+z)+1 is odd.

Lemma 2: the product of two even numbers, f, g, is another even number.
Since f=2w, g=2v, fg=(2w)(2v)=2(2wv). Since 2wv equals another integer, fg is even.

If one does something even more seemingly "trivial" here, something else can become apparent.

Lemma 3: the square of an even number h is also even.
Since h=2z, h²=(2z)²=2²z²

Now suppose that b in 2c² &ne b² is odd. So, b² is odd also. Since 2c² is even this implies this inequality as correct.

Suppose that b is even, and c is odd. Then, the inequality holds since 2c² is even also and can only get divided by 2 once, and b²=(2n)²=2²n which implies it can get divided by 2 twice.

Now suppose that b in 2c² &ne b² is even and c is even also. So, both b² and c² are even also. All even numbers can get written as a product 2^xo, where o denotes an odd number. Consequently, b and c can get divided by 2 x times, or we could say that b and c have x even factors. So, now treating y as any integer, y²=[(2^x)o]²=(2^x)²*o². So, we have 2^2x*o²=2^x*2^x*o². Thus, b and c have x+x=2x even factors. Since this holds, 2c² has an odd number of even factors. So, the inequality holds.

Therefore, for all integers b, c, sqr(2) ≠ b/c.

As far as I can tell, there's no hidden reductio ad absurbum lurking here, but it might come as trickier to spot than I believe.