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[6 / 0 / ?]

[1373732063] Is 0.999... exactly equal to 1?

No.56079 View ViewReplyOriginalReport
Reductio ad absurdum proof
1:: 1=0.999...
source: http://en.wikipedia.org/wiki/0.999...
academic consensus: nearly all academics agree
proofs: Many
field of math:real numbers
<The equality 0.999... = 1 has long been accepted by mathematicians and is part of general mathematical education.>
<Although the real numbers form an extremely useful number system, the decision to interpret the notation "0.999..." as naming a real number is ultimately a convention, and Timothy Gowers argues in Mathematics: A Very Short Introduction that the resulting identity 0.999... = 1 is a convention as well:

    However, it is by no means an arbitrary convention, because not adopting it forces one either to invent strange new objects or to abandon some of the familiar rules of arithmetic.[47]

One can define other number systems using different rules or new objects; in some such number systems, the above proofs would need to be reinterpreted and one might find that, in a given number system, 0.999... and 1 might not be identical. However, many number systems are extensions of —rather than independent alternatives to— the real number system, so 0.999... = 1 continues to hold. Even in such number systems, though, it is worthwhile to examine alternative number systems, not only for how 0.999... behaves (if, indeed, a number expressed as "0.999..." is both meaningful and unambiguous), but also for the behavior of related phenomena. If such phenomena differ from those in the real number system, then at least one of the assumptions built into the system must break down.
Infinitesimals>
<
    Q: Is it possible to create a new number system other than the reals in which 0.999... < 1, the difference being an infinitesimal amount?

    A: Yes, although such systems are neither as used nor as useful as the real numbers, lacking properties such as the ability to take limits (which defines the real numbers), to divide (which defines the rational numbers, and thus applies to real numbers), or to add and subtract (which defines the integers, and thus applies to real numbers).>

Intuitively we see that 1>0.9 and 1>0.99 and 1>0.999 and at any step 1>0.999... but what if we could prove it?
2:: 1-0.999...=0
If two numbers are exactly equal their difference is zero x=y x-y=x-x=0

3:: 1-0.9-0.09-0.009-...=0
The difference can be described as process of sequential subtraction from 1

4:: ((1-0.9)-0.09)-0.009-...=0
Each step can be reliably described as...

5:: (((1/10)-0.09=1/100)-0.009=1/1000)-...=0
equivalent to division by 10 at each step of the process

6:: (1-0.9=1/10-0.09=1/100-0.009=...)=0
the sequence is obviously based on division by 10 which is...
6.1::  1/1-9/10-9/100-9/1000-...=((((1/10)/10)/10)/10)/...=0
7:: 1/10/10/10/10...=0
a repeated division by tens which can be defined as...

8:: 1/(10*10*10...)=0
division by a infinite sequence representing all divisors combined x/a/b/c=x/(a*b*c)

9:: (1/(10*10*10...))*(10*10*10...)=0*(10*10*10...)
but we can cancel out the combined divisors from both sides! x/y=z x=z*y

10:: 1=0*(10*10*10...)
Canceling the left side is trivial (x/y)*y=x

11:: 1=0*10*10*10*...
the right side can be examined as a sequence with zero at start which leads to

12:: 1=(((0*10=0)*10=0)*10=0)*...
propagation of zeroes to the right and since 10*0=0 at any step, that means:

13:: 1=0 
reductio ad absurdum completed, but wait there is more

14:: x*1=(x*0=0) since x=y z*x=z*y
 every number is zero. Since all numbers are zero,the is no point in arithmetic, measurements and science itself...or there is a problem with 1=0.999...