The goal I'm kicking in the pants today is from List 40: 100 Mathematical Theories To Learn, and comes to you thanks to a set of rather...um...heated discussions I have with friends about the fact that 0.(9), which denotes 0.9 repeated (0.9 with an infinite number of 9's following...) is in fact equal/equivalent to 1.
Typically, when I assert the fact that, yes, 0.999999... = 1, I am met with various cries of "Get the heck out of here" or "No...No it isn't...", mixed (of course) with varying levels of obscenity. However occasionally I am faced with a different type of opposition - instead of decrying the fact itself, they respond along the lines of "Oh yes, that - of course it is, and here is the proof as to why..."
What they then proceed to thrust into my face is, in fact, not a proof of this fact at all. Well, rather, some of them do offer me a proof, but the mathematics behind it is flawed. So, if you don't believe that 0.(9) = 1, or if you think that using the "1/3 = 0.(3)" method is the best way to quash resistance to this fact, then read on, and be informed!
0.(9) is a number that is expressible as what is called a Geometric Series - this basically means that we can take 0.9999... and split it up into a series of numbers where successive terms have a common ratio between them. In the case of 0.(9) the common ratio is a power of 1/10 - example:
9*(1/10) = 9/10
(9/10)*(1/10) = 9/100
(9/100)*(1/10) = 9/1000
0.9999... = 9/10 + 9/100 + 9/1000 + 9/10000 + ...
Now, even though this sum is infinite (that is, if you tried to add it together, you'd be going forever because there are always new terms to add) there is a mathematical nicety in the fact that if the terms of the sum are approaching zero (which ours are) then the sum is finite (I will offer a proof of this later.)
The sum of such a series is given by:
S = (a)/(1 - r)
Where S is the sum of the series
Where a is the initial term of the series
Where r is the common ratio
So we have:
a = 9/10 = 0.9
r = 1/10 = 0.1
Which all lead us to...
0.9999... = 0.9 + 0.09 + 0.009 + 0.0009
= a/(1 - r)
= 0.9/(1 - 0.1)
= 0.9/0.9
0.9999... = 1
So, there you have it - a pretty simple proof to what is, to most people, quite an astonishing fact. What follows now is just the proof of convergence I promised earlier (that is, the proof that the sum of a series is, in fact, equal to a/(1 - r))
So, if you're not interested in reading that (and you've made it this far...) then I say to you the following: congratulations for making it through the post; thank you for sharing my first achievement with me; and later days Internet-Crowd!!
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PROOF OF CONVERGENCE
The sum of a geometric series can be written as the sum of a geometric progression of the form:
SUM[k=0, inf] ar^k = a + ar + ar^2 + ar^3 + ar^4 + ...
Which we will write, then as
lim[n->inf] (SUM[k=0,n] ar^k)
If we look, first at:
SUM[k=0,n] ar^k = a + ar + ar^2 + .. + ar^n
If we multiply both sides by (1 - r) we obtain:
(1 - r)(SUM[k=0,n] ar^k) = (1 - r)(a + ar + ar^2 + ... + ar^n)
(1 - r)(SUM[k=0,n] ar^k) = a - ar + ar - ar^2 + ar^2 + ... - ar^n + ar^n - ar^(n+1)
(1 - r)(SUM[k=0,n] ar^k) = a - ar^(n+1)
Dividing both sides by (1 - r) now:
SUM[k = 0, n] ar^k = (a - ar^(n + 1))/(1 - r)
Substituting that back into our original equation we get:
lim[n->inf] (SUM[k=0,n] ar^k) = lim[n->inf] (a - ar^(n + 1))/(1 - r)
We can separate the fraction to become:
lim[n->inf] (a/(1 - r) - ar^(n + 1)/(1 - r))
We can then use the fact that lim[n->inf] a - b = lim[n->inf] a - lim[n->inf] b to obtain:
lim[n->inf] (a/(1-r) - lim[n->inf] ar^(n + 1)/(1 - r)
Now, we are restricting the problem domain here to values of 'r' that are strictly less than 1 (if you remember, in our example, r = 0.1, which is less than 1)
When r < 1, and as n approaches infinity, r approaches zero, thus
lim[n->inf] ar^(n + 1)/(1 - r) = a*0/(1 - r) = 0
Which gives us:
lim[n->inf] a/(1 - r) - 0 = lim[n->inf] a/(1 - r)
And because there are no terms of n in this, it is easily calculated to be:
a/(1 - r)
So, we have just proven that:
a + ar + ar^2 + ar^3 + ar^4 + ... = a / (1 - r)
in case you didn't believe me blindly, in which case - good for you! And congratulations on making it to the true end of my very first accomplished post!
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