I went for 0.9... but it's both.

We've had this discussion on the maths board, we never settled it because we got bored, but I'll offer you a proof that 0.9999... = 1.


First we show that for any 2 real numbers a,b where a<b there must exist a number c expressable as a non recuring decimal such that a<c<b.



since a<b we can conclude that b-a > 0.

and therefore there must exist 1/(b-a) > 0

Now define a number N = 2^p * 5^q
where p and q are positive whole numbers and N > 1/(b-a)


N > 1/(b-a)
=> 1/N < b-a

N = 2^p * 5^q
=> 1/N * m is expressable as a non recurring decimal for any whole number m.

now set m as the greatest whole number such that m/N <= a

by definition of m
(m+1)/N > a

as m/N <= a
b - (m/N) >= b-a
b - ((m+1)/N) >= (b-a) - 1/N > 0 (as 1/N < b-a)


b - ((m+1)/N) > 0
=> b > ((m+1)/N

and hence a < ((m+1)/N < b

and so there exists a number ((m+1)/N expressable as a non recuring decimal such that a<((m+1)/N<b.


So if 0.9... < 1 there must exist a non recurring decimal c such that

0.9.... < c < 1

clearly this is not the case so 0.9... = 1.