Is 0.99999etc (so, with infinitely many nines after the decimal point) equal to 1?
This was the first question that was written about in this blog. The answer was and is: yes, 0.99999…=1.
You can look at this equality in a slightly different way and justify it by starting from 1 and find a decimal expansion of that number; I write `a' because some numbers have more than one such expansion. The one that you find is determined by the way you make it.
Take a number, x, in the interval (0,1). We make sequences of digits, a1, a2, a3, … and numbers, x1, x2, x3, … as follows.
There is i<10 such that i/10≤x<(i+1)/10. We write a1=i and x1=a1/10. The difference x-x1 is less than 1/10, so we can find j<10 with j/100≤x-x1<(j+1)/100. We write a2=j and x2=x1+a2/100. The difference x-x2 is less than 1/100, so we can find k<10 with k/1000≤x-x2<(k+1)/1000. We write a3=k en x3=x2+a3/1000.
We keep following this recipe: at stage n we have xn with 0≤x-xn<1/10n; so there is an an+1<10 such that an+1/10n+1 ≤ x-xn < (an+1+1)/10n+1. Using this we make xn+1=xn+an+1/10n+1.
According to the definition given in this post the sequence of xn's converges to x and we express this, rather suggestively, as x=0,a1a2a3…. Note well that for a mathematician this equality represents the whole story given above.
Sometimes you are lucky and you find only finitely many i; that happens when there is an n such that x=xn; in that case ai=0 for i>n. If you do not like that then you can adapt the recipe and make sure you will keep getting non-zero ai.
Given our number, x, in the interval (0,1) we make sequences of digits, b1, b2, b3, … and numbers, y1, y2, y3, … as follows.
There is i<10 such that i/10<x≤(i+1)/10. We write b1=i and y1=b1/10. We have 0<x-y1≤1/10, so there is j<10 with j/100<x-y1≤(j+1)/100. We write b2=j and y2=y1+b2/100. We have 0<x-y2≤1/100, so there is k<10 with k/1000<x-y2≤(k+1)/1000. We write b3=k and y3=y2+b3/1000.
We keep following this recipe: at stage n we have yn such that 0<x-yn≤1/10n; so there is a bn+1<10 such that bn+1/10n+1 < x-yn ≤ (bn+1+1)/10n+1. Using this we make n+1=yn+bn+1/10n+1.
The sequence of yn's converges to x too and we express this as above, rather suggestively, as x=0,b1b2b3…. There is a whole story behind this equality too.
The difference between the two recipes is that possibly xn=x for some n but that yn<x for all n. If xn=x never occurs then we find xn=yn and an=bn for all n and we find just one decimal expansion for x. However, if xn=x then the sequences go their separate ways: you can check that bn=an-1 in that case and also bi=9 for i>n. For example when x=1/2: then a1=5 and b1=4. So 1/2=0.5 and 1/2=0.4999….
The Dutch version of the question mentions the fractions 1/9, 2/9, … 8/9. For these both recipes yield the same expansion: i/9=0,iiiii…; the nice thing is that the second recipe also works for 1=9/9: you will find bn=9 for all n. The equality 1=0,999… is thus a direct result of the second recipe.
