Induction on .999...

Date: 10/19/2000 at 08:19:40
From: Blake Thomson
Subject: Further questions on .999... = 1 

I have been having an argument with my physics teacher over the fact 
that point nine recurring (.999... or PNR) equals 1. I showed him the 
proof on your site and he pointed out a fact that I hadn't noticed.

How can you multiply PNR by ten if you can't get to the beginning? For 
example, to multiply 215 * 3, you would start off on the right, 
multiplying 3 by 5, then 1, then finally 2. Now how can you multiply 
PNR by ten if the farthest right value cannot be reached?

Thank you for any clarification you can give.

Blake

Date: 10/19/2000 at 13:18:25
From: Doctor TWE
Subject: Re: Further questions on .999... = 1 

Hi - thanks for writing to Dr. Math.

Check out our FAQ on the subject at:

   Why does 0.9999... = 1?
   http://mathforum.org/dr.math/faq/faq.0.9999.html   

This, and the links therein, explain several different ways of proving 
this interesting fact.

I wanted to comment on your (or your teacher's) idea that "you can't 
get to the beginning." We multiply values from right to left only as a 
matter of convenience, so we don't have to "backtrack" each time we 
carry. It is equally valid to multiply from left to right. This is 
often done when we want to get a quick estimate for the answer, then 
get more accuracy later. The LEFTMOST digits contribute the most to 
the answer, so to get "in the ballpark," we can multiply them first. 
I'll demonstrate with a concrete example, then a general argument.

Let's use your example of 215 * 3. I can multiply them as follows; 
first, I'll multiply the 3 by the 2 (actually, by 200) and put the 
answer in the hundreds place, like this:

       215
     *   3
       ---
       6

Then I'll multiply the 3 by 1 and put the answer in the tens place:

       215
     *   3
       ---
       6
        3

Finally, I'll multiply the 3 by 5 and put the answer in the units 
place:

       215
     *   3
       ---
       6
        3
        15

When I add these up, I get the correct answer:

       215
     *   3
       ---
       6
        3
        15
       ---
       645

Notice that along the way, I got pretty good estimates for my final 
answer. After the first step my total was 600 (not a bad estimate, 
off by less than 10%), after the second step I had 630, and after the 
third step, I got 645. Going right to left, my totals after each step 
would be: 15 (not a very good estimate of the final answer), 45 (still 
not very good) and finally 645.

In the general case, consider multiplying a 3-digit value ABC by some 
value X, where A, B and C are digits. What we really have, then, is

     (100*A + 10*B + 1*C) * X

Conventionally, we solve this starting with the units digit as:

     X*C*1 + X*B*10 + X*A*100

But using the commutative property of addition, this is equal to:

     X*A*100 + X*B*10 + X*C*1

showing that the order doesn't matter. If I wanted to, I could start 
in the middle, for example:

     X*B*10 + X*A*100 + X*C*1

I'd just have to be careful not to miss any digits or do any ones 
twice.

Without being able to start with the most significant digit, we could 
never find a value like 2*pi, because pi (like all irrationals) is an 
infinite non-repeating decimal. When we say 2*pi is approximately 
6.2831853, we can do so because we started multiplying at the end and 
not the beginning.

One final note: How can we be sure that the multiplications by 10 
continue as we expect (i.e. they continue shifting the digits 1 place 
to the left) and that the subtractions of successive digits produce 
zeroes infinitely? These steps can be proven using a technique called 
mathematical induction. That's too complex to explain here, but if you 
search our Ask Dr. Math archives for the word "induction" (type it 
without the quotes), you'll find many questions and answers about it. 
Our searcher is at:

   http://mathforum.org/mathgrepform.html   

I hope this helps. If you have any more questions, write back.

- Doctor TWE, The Math Forum
  http://mathforum.org/dr.math/   

Date: 10/20/2000 at 08:51:29
From: Blake Thomson
Subject: Induction concerning point nine recurring

I recently sent you a letter regarding the proof that .999... = 1 in 
the FAQ. Specifically, I asked how was possible to multiply PNR by ten 
when you could not get to the right-hand side of it. This was 
explained quite neatly.

However, the nice person who helped me out said it could be proven 
that one can multiply PNR by 10 by using a process of induction. I 
tried to search the archives, but the results I found there were not 
very satisfactory because it seemed to me that each induction question 
was actually problem-specific.

To cut short, I would like to know the induction proof for multiplying 
PNR by ten.

Cheers for any help you can give!

Date: 10/23/2000 at 16:02:47
From: Doctor TWE
Subject: Re: Induction concerning point nine recurring

Hi again Blake! Thanks for writing back!

In general, we use proof by induction whenever we want a proof that 
involves an infinite sequence or series, in this case .999...

Inductive proofs require two steps:

Step 1 (the basis step): Prove it for some starting value, like n = 1.

Step 2 (the inductive step): Prove that if it's true for n = k, then 
it is true for n = k+1.

For the inductive step, we assume that it is true for n = k, and we 
usually use this assumption in the proof itself.

A word of caution: because of their self-referential nature, inductive 
proofs are usually hard to follow; it's easy to get lost in the 
details, losing track of what n, k, and k+1 are supposed to be.

In our case, we want to show that when multiplying the nth digit by 
10, we get a 9 in the (n+1)st digit, no carry to the (n+2)nd digit, 
and a 0 in the nth digit (counting from the right.) We want no carry 
to the (n+2)nd digit so that we don't have to adjust the previous 
(greater place value) digits multiplied because of a "retroactive 
carry." We want a 0 in the nth digit so that it will not produce 
further carries when multiplying the next digit (smaller place value) 
by 10. Mathematically:

     10 * (9*10^n) = 9 * 10^(n+1)

We want to show that:

   a) the nth digit is 0
   b) the (n+1)st digit is 9
   c) the (n+2)nd digit is 0

To minimize the confusion, I'm going to use n = -1 as my basis step, 
and show that if it is true for n = k, then it is true for n = k-1 as 
my inductive step. This is the "negative" of the standard, but it will 
eliminate the need for using -k's and -(k+1)'s, etc. in my equations.

Step 1: Show that it is the case for n = -1

     10 * [9*10^(-1)]  =  9 * 10^(-1+1)
         10 * (9*0.1)  =  9*10^0
             10 * 0.9  =  9*1
                    9  =  9

   a) the -1st digit (the tenths place) is 0: True
   b) the 0th digit (the units place) is 9: True
   c) the 1st digit (the tens place) is 0: True

So we've proved the basis. Now for the inductive step.

Step 2: Prove that if it's true for n = k, then it's true for n = k-1.

Let's let d = the initial (k+1)st digit. This digit is 0 from 
conclusion (a) of the previous step. Then:

     10 * (9*10^k) + d  =  9 * 10^(k+1)
     9 * (10*10^k) + 0  =  9 * 10^(k+1)
          9 * 10^(k+1)  =  9 * 10^(k+1)

a) the kth digit is 0: True - 9*10^(k+1) has a 0 in the kth place.
b) the (k+1)st digit is 9: True - 9*10^(k+1) has a 9 in the (k+1)st 
place
c) the (k+2)nd digit is 0: True - 9*10^(k+1) has a 9 in the (k+2)nd 
place

So, if 10 times the kth digit is 9*10^(k+1), then 10 times the (k-1)st 
digit is 9*10^k, and thus 10 times the (k-2)nd digit is 9*10^(k-1), 
and thus...

Therefore 10 * 0.999... = 9.999...

Q.E.D.

Depending on the skepticism of the intended audience, you may have to 
use a similar inductive proof for the step:

       9.999...
     - 0.999...
       --------
       9.000...

For most audiences, the following will do:

     Let x = 0.999... 
     then 9.999... = 9 + x

     so 9.999... - 0.999...  =  (9 + x) - x  = 9

However, some audiences (like your teacher, perhaps) will not be 
satisfied that algebraic operations work on "infinite strings" (i.e. 
they won't be convinced that x - x = 0 when x is an infinite repeating 
decimal.)

I hope this helps! If you have any more questions, write back again!

- Doctor TWE, The Math Forum
  http://mathforum.org/dr.math/