From Newsgroup: comp.lang.c++

This updated file should solve most doubts I encountered. Hope, useful to you (of course, not in official exam if that is your 'real') https://sourceforge.net/projects/cscall/files/MisFiles/RealNumber-en.txt/download
+-------------+
| Real Number | ('computational' may be added to modify terms used in this file +-------------+ if needed)
n-ary Fixed-Point Number::= Number represented by a string of digits, the
string may contain a minus sign or a point:
<fixed_point_number>::= [-] <dstr1> [ . <dstr2> ]
<dstr1>::= 0 | <nzd> { 0, <nzd> }
<dstr2>::= { 0, <nzd> } <nzd>
<nzd> ::= (1, 2, 3, 4, 5, 6, 7, 8, 9) // 'digit' varys depending on n-ary
Two n-ary fixed-point number (same n-ary) x,y are equal iff their
<fixed_point_number> representation are identical.
Real Nunmber(ℝ)::= {x| x is finitely represented by n-ary <fixed_point_number>
and those that cannot be finitely represented }
Note: Numbers that is not finitely representable cannot all be explicitly
defined, this is the property of real number based on discrete symbols
(like quantum?). E.g.
A= lim(n->∞) 1-3/10^n = 0.999...
B= lim(n->∞) 1-2/2^n = 0.999...
C= lim(n->∞) 1-1/n = 0.999...
...
IOW, by repeatedly multiplying 0.999... with 10, you can only see 9,
the structure of the rear end of 0.999... is not seen.
Since <fixed_point_number> is very definitely real and infinity is
involved, theories that composed of finite words cannot be too
exclusive about such a ℝ. 'Completeness' is impossible.
Note: This definition implies that repeating decimals are irrational number.
Let's list a common magic proof in the way as a brief explanation:
(1) x= 0.999...
(2) 10x= 9+x // 10x= 9.999...
(3) 9x=9
(4) x=1
Ans: There is no axiom or theorem to prove (1) => (2).
Note: To determine whether a repeating decimal x is rational or not, we can
repeatedly subtract the repeating pattern p(i) from x.
If x-p(1)-p(2)-...=0 can be verified in finite steps, then x is
rational. Otherwise, x is irrational, because, if x is rational, the
last remaining piece r(i)= x-p(1)-p(2)-... must exactly be the
repeating pattern p(i). But, by definition of 'repeating', r(i) cannot
be pattern p(i). Therefore, repeating decimal is irrational.
Real number is just this simple. The limit theory is a methodology for finding derivative, nothing to do with what the real number is (otherwise, a definition like the above must be defined in advance to avoid circular-reasoning). +-------+
| Limit |
+-------+
Limit::= lim(x->a) f(x)=L
http://www.math.ntu.edu.tw/~mathcal/download/precal/PPT/Chapter%2002_04.pdf
http://www.math.ncu.edu.tw/~yu/ecocal98/boards/lec6_ec_98.pdf
https://en.wikipedia.org/wiki/Limit_(mathematics)
https://en.wikipedia.org/wiki/Limit_of_a_function
L is defined as the limit (a number) while x approaches a (f(a) may not
be defined, although, while f is continuous, L=f(a)). L is a defined value,
not "if something infinitely close ... then equal" (no such logic).
Ex1: A= lim(n->∞) 1-1/n= lim(n->0⁺) 1-n= lim 0.999...=1
B= lim(n->∞) 1+1/n= lim(n->0⁺) 1+n= lim 1.000..?=1
Ex2: A=lim(x->ℵ₀) f(x), B=lim(x->ℵ₁) f(x) // ℵ₀,ℵ₁ being proper or not is
// another issue here. But problematic
// for "finally equal" interpretation.
Limit defines A=B, does not mean the contents of the limit are equal. If the
"x approaches..., then equal" notion is adopted, lots of logical issues arise.
Note: The equation of limit may be questionable
lim(x->c) (f(x)*g(x))= (lim(x->c) f(x))*(lim(x->c) g(x)):
Let A=lim(n->∞) (1-1/n)= 1
A*A*..*A= ... = lim(n->∞) (1-1/n)^n // 1=1/e ? +--------------------------------------+
| Restoring Interpretation of Calculus | +--------------------------------------+ http://www.math.ntu.edu.tw/~mathcal/download/precal/PPT/Chapter%2002_08.pdf https://en.wikipedia.org/wiki/Derivative
Assume calculus is basically the area problem of a function: Let F compute the the area of f. From the meaning of area, we can have:
(F(x+h)-F(x)) ≒ (f(x+h)+f(x))*(h/2) // h is a sufficiently small (test)offset
<=> (F(x+h)-F(x))/h ≒ (f(x+h)+f(x))/2 // the limit(h->0) of rhs is f(x)
Expected property of F: (1)Error |lhs-rhs| strictly decreases with the tiny
(test) offset h (2)When h=0, lhs=rhs.
Because the h in the lhs cannot be 0, the basic problem of calculus is
finding such a F (or f) that satisfies the expected porperty above...Thus,
D(f(x))= lim(h->0) (F(x+h)-F(x))/h = f(x)
Note: Hope that this interpretation can avoid the interpretation of infinity
/infinitesimal, and provide more correct foundation for some theories
, e.g. Zeno paradoxes, repeating decimal,...,and more (exponiential,
Cantor set, infinite series...). ------------------------------------------------------------------------------- --- Synchronet 3.20a-Linux NewsLink 1.114