Is 0.999... exactly equal to 1?

Is 0.999... exactly equal to 1? This is in a fact a central problem in math(despite what you assume as insignificant) connected to existence of infinitesimals and foundations of modern math.

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 , real analysis, any field which denies infinitesimals(practically all modern math)
Some selected Wikipedia quotes:
<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?
some food for for thought:
0.999... isn't an integer(like 1) and its floor function(rounding down) leads to floor(0.999...)=0 with any amount of 9's

Now for the absurd premise:
2:: 1-0.999...=0
If two numbers are exactly equal their difference is zero x=y x-y=x-x=0
2.1::
1-0.999...=0
0.999... is a series 0.9+0.09+0.009+... with ratio 1/10(each next meber is 1/10 of the previous one).
1- series=0  the difference between the series and 1 must be zero
1-(0.9+0.09+0.009+...)=0  now we can remove brackets:
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::  10/10-9/10-9/100-9/1000-...=0

6:: which equivalent to division by 10 at each step of the process
so 10/10-9/10= 1/10 which is the same as (10/10)/10
and 1/10-9/100=1/100 which is the same as (1/10)/10 etc(since  x- (x*(9/10))=x*(1/10)=x/10 (taking 9/10 of a number is equal to division by 10 and every next member is 9/10 of previous result)
giving (((10/10)/10)/10)/10/...=0
now the first term 10/10 is 1 and it can be isolated
7:: 1/10/10/10/10/10...=0 the divisors can be combined due(x/a/b/c=x/(a*b*c) giving:
8:: 1/(10*10*10...)=0
Which can be canceled from the left due x/y=z -> x=z*y

9:: (1/(10*10*10...))*(10*10*10...)=0*(10*10*10...)
the left side has exact same sequences, which cancel to give (1/x)*x=1

10:: 1=0*(10*10*10...)
The right side is obviously equal to zero, but lets prove it

11:: 1=0*10*10*10*...
the first multiplication gives 0*10=0,which multiplied by the second gives also zero,which means

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)
 every number is zero. since if a=b x*a=x*b with 1*x=0*x which is x=0
Since all numbers(x can stand for any variable) are proven to be zero,
the is no point in arithmetic, measurements and science itself(which is math based)...thats what we get for believing 1=0.999...

But wait isn't there proofs for 1=0.999...? They are all defective:
Common "proofs" of 1=0.999...
1:: 1/3=0.333... than 3/3=0.333...*3 then 1=0.999...
the proof assumes that 0.333... is exactly equal to 1/3 but its not
1/3 division never completes, always leaving a remainder at the end so 0.333... approaches 1/3 with each "3" but never reaches the value of 1/3
Same thing proof based on 1/9=0.111..., they all rely on circular logic that assumes 1/frac=0.mmm... and then proves (with another fraction) 1/frac=0.mmm...

2:: digit manipulation proof
x=0.999...
10x=9.999...
10x-x=9.999...-0.999...
9x=9
x=1
The proof relies on the treating different lengths of "..." as the same
here is example #1
x=0.9
10x=9
10x-x=9-0.9
9x=8.1
x=0.9
example #2
x=0.99
10x=9.9
10x-x=9.9-0.99
9x=8.91
x=0.99
example #3
x=0.999
10x=9.99
10x-x=9.99-0.999
9x=8.991
x=0.999
AS you can see there is always a difference, which never disappears
 the trickery is using 9.(999... with N-1 places) and (0.999... with N places) as equivalent. The difference is very subtle, but in math it matters.


3:: proof using infinite series convergence formula
http://en.wikipedia.org/wiki/Geometric_progression#Infinite_geometric_series
http://en.wikibooks.org/wiki/0.999.../Proof_by_the_geometric_series_formula
the formula use is sum=a/(1-r) where a is first member and r is the ratio
giving  (9/10)/1-(1/10)=1

The proof uses a formula distinct from formula for finite series, that ignores infinitely small numbers. When this formula was "invented" infinitesimals were discarded from the finite series formula, because they believed that limit of series(an upper bound) is equal to its sum.
Here is the proper formula:
http://en.wikipedia.org/wiki/Geometric_progression#Derivation
Here is the "infinite formula"
http://en.wikipedia.org/wiki/Geometric_progression#Infinite_geometric_series
Notice the latter discards r^n as "zero", while in fact its an infinitesimal

 Here is the idea
Infinite formula: a/(1-r)
Finite formula: a*((1-r^(n+1)))/1-r which lead to a- a*(r^(n+1))/1-r
because  a*r^(n+1) where n increases to infinity becomes infinitely small, it is discarded in the infinite formula giving rise to this error.
The problem here is depicted in the second box in
http://en.wikipedia.org/wiki/Geometric_progression#Infinite_geometric_series
Where r^(n+1) is assumed to be equal to zero(spoiler:its not, but it gets infinitely close)

4::proof using anything that relies on the real number system or
standard analysis( all those higher math proofs) are a form of circular logic.
Thats because real number system and standard analysis rely on the  archimedean property that there are no infinitesimals(the number we get from 1-0.999... which is infinitely small  and cannot be expressed in decimal since 0.000...1 isn't a valid real number) which was accepted as true during the definition of real number system and formation of standard analysis.
The modern definition of a limit also explicitly denies infinitesimals.
All high math has developed in two directions.
1.from ancient infinitesimal calculus to non-standard analysis(which allows 1!=0.999... btw) http://www.encyclopediaofmath.org/index.php/Infinitesimal_calculus
http://en.wikipedia.org/wiki/Infinitesimal
http://en.wikipedia.org/wiki/Infinitesimal_calculus
http://en.wikipedia.org/wiki/Non-standard_calculus
http://en.wikipedia.org/wiki/Non-standard_analysis

2. modern math which is almost exclusively standard analysis which explicitly denies that infinitesimals exist and replaces infinitesimals with a new definition of imprecise limit.=( http://en.wikipedia.org/wiki/(%CE%B5,_%CE%B4)-definition_of_limit  )
http://en.wikipedia.org/wiki/Archimedean_property
http://en.wikipedia.org/wiki/Real_analysis
http://en.wikipedia.org/wiki/Real_number
http://en.wikipedia.org/wiki/Construction_of_the_real_numbers

"Arguments to authority"
Your teacher,professor or fellow math nerd will argue that they :"know more math" than you, but all they know is standard analysis which forbids infinitesimals.
Tell them about the ("non-standard")branch of mathematics they have neglected for so long, Infinitesimal calculus and its modern offshoots
http://en.wikipedia.org/wiki/Infinitesimal
http://en.wikipedia.org/wiki/Infinitesimal_calculus
http://en.wikipedia.org/wiki/Non-standard_calculus
http://en.wikipedia.org/wiki/Non-standard_analysis

"Argument against precision"
Some say "but in the real world 0.999... is practically 1".
Not so in math. If its not absolutely equal its different. If they insist
try to reason about floor function(rounding down to the nearest integer) with the 0.999... which should yield 0 as it begins with 0.9 rather than 1.0.

"Argument against existence of infinite decimals"
Some would argue since 0.999... isn't possible to1 write down exactly it does not exist and 1=0.999... is not a valid equality.
The 0.999... is shorthand for infinite sequence 0.9+0.09+0.009... which does exist. Infinite series are impossible to "write down" since they contain infinite members. The "..." can be abstracted as N 9's after the decimal point, with arbitrarily large N without using infinity explicitly.

"Argument of representation"
0.999... is defined to be a representation of 1(since modern math rejects infinitesimals). This can be countered by the following observation.
1 is an integer, 0.999... is a real number(and not an integer, only a real number).  If two representations are the same single number x,then x is both not integer and integer which is absurd. Some will say 2/2 which is a rational number isn't an integer but is equal to 1 due n/n=1:
they forget 1 is also rational because integers are a subset of rationals and 1  is a real number because integers are subset of reals. The reverse isn't true however:
all reals aren't rationals, only some of them. and all rationals aren't integers, only some of them.
0.999... however isn't an integer(its value extends beyond the decimal point) or a rational(it can't represented as a fraction x/y where x and y are integers), its exclusively a real number.
Some would say but 1.000...(a real number) also extends beyond the decimal point, and its equal to 1. The zeroes in 1.000... are redundant 1.0=1.00=1.000... since they don't represent any quantity:every number can be written with extra zeroes after the decimal point without changing its value. 7.500000=7.5
Zeroes after last non-zero number aren't even useful. 1.000... isn't the same class as 0.999... because while 0.9!=1(every nine counts) 1.0=1(number of zeroes is irrelevant). value of 1.000... doesn't extend beyond the decimal point(zeroes can't change it, unlike 9's which add accuracy to 0.999...).