Thats my post on it as promised
Yes, 0.9 recurring is equal to 1.
1/3 = 0.333...
x3
1 = 0.99999...
That is perfectly fine.
Another way to look at it is that 0.9 recurring is 9/10 + 9/100 + 9/1000 + ... and then use the formula for the sum of a geometric series. It's a really nice bit of maths, so in case you don't know it I'll quickly go through it:
q: what is 1/5 + 1/25 + 1/125 + ... + 1/(5^10) ?
a: well, we know it is a number (i.e. this sum is finite as there are only finitely many terms), so call it S (for "sum").
Then S/5 = 1/25 + ... + 1/(5^10) + 1/(5^11), so
S-S/5 = 1/5 - 1/(5^11), and hence S = (5/4)*(1/5 - 1/(5^11), which after simplifying becomes (1-1/(5^10))/4.
There is a formula for this, but the instructive thing is to remember how to derive it, as above. It's not hard at all.
The infinite case is a bit trickier. How do we know that 1/5 + 1/25 + 1/125 + ... is actually a number, i.e. that it isn't infinite? This might seem unimportant, but consider instead the following situation:
S=1+1/2+1/3+1/4+1/5+... (this is called the harmonic series, and is very interesting). It turns out that this tends to infinity, so it is a logical fallacy to say "call this number S" and then use it as if it were a real number. You can get into all kinds of problems (and people did) by making this mistake.
It turns out that geometric series (i.e. those in which each term is a fixed multiple of the term before) always converge which the ratio is <1. The way to see this is to look at the first N terms, and use what we did above. So 1/5 + ... + 1/(5^N) = (1-1/(5^N))/4 as above, and this always less than 1/4, so 1/5 + 1/25 + ... doesn't tend to infinity. Exactly the same thing works for a+ar+ar^2+ar^3+... always converges when -1<r<1 (in the negative case you have to check that the sum doesn't tend to minus infinity either).
Now we know that, we can do the same trick as above:
S=9/10+9/100+... so S/10=9/100+9/1000+... and hence
(1-1/10)S=9/10, i.e. S=1.
I remember quite well being confused by this when my teacher mentioned it first. Here's a question to ask your self: what is 1-0.9999... ? It is clearly not negative. And it's less than 1/10, as 0.9999... is greater than 0.9. And it's less that 1/100, as 0.9999... is greater than 0.99 = 1-1/100. In fact, for any positive integer n you care to name, it is less than 1/(10^n), as 0.9999... is greater than 0.(n 9s). What non-negative number is less than 1/(10^n) for all n? Well, it can only be 0.
The maths of the real numbers (i.e. anything with a decimal expansion, so that includes integers, rational numbers (p/q), solutions of equations (root 2, sqrt(1+sqrt(2)), and even numbers that aren't roots of (polynomial) equations (pi, e, uncountably many others)) is very interesting. It turns out that this concrete construction, via infinite decimals, is not the most useful one. It makes it hard to prove things. There are three other characterisations of the real numbers, which are all equivalent.
(a)any increasing sequence which is bounded above (i.e. doesn't tend to infinity) tends to a limit. (monotone sequences axiom (monotone means strictly increasing or strictly decreasing))
(b)any non-empty set of real number which is bounded above has a least upper bound. (least upper bound axiom)
(c)an infinite number of points in a interval of finite length must have a subsequence which tends to a limit. (Bolzano-Weierstrass axiom) (strictly, the interval must contain it's end-points)
So to prove things in the real numbers, you choose on of the above axioms and use that and all the facts you know about the rationals (you're working in the smallest field containing the rationals such that your axiom is true). It is something you have to learn in the first year of a maths degree to prove that each axiom is equivalent. That would take too long for me to explain now, but it is worth seeing why these don't work in the rationals.
The thing that's hard to grasp the first time you see this is that when we say "tends to a limit" we mean "there is a point in the field we are considering which this sequences tends to".
(a)take succeedingly better approximations for pi. so 0,1,2,3,3.1,3.14,3.141,etc... This is increasing, bounded above (by 4, say), but if it did tend to a limit then that limit would have to be pi, and pi isn't a rational number.
(b)just take the set of all points I outlined above.
(c)again, the set above works, as it is contained in [0,4].
Thats my post on it as promised
Fantastic, well did you not say then that perhaps physicists look at the real world and know nothing about numbers?
It's human nature, unfortunately because some things in maths look deceptively simple people think they can apply their generalizations they've used before in their limited context.
