Karl's Calculus Tutor -- Box 1.5: Coached Proof

Coached Proof

Prove that all infinite decimals specify a Cauchy sequence.

This exercise will sharpen your notion of what a limit is. I will actually only coach you on an easier proof and then demonstrate the proof of the above assertion.

What you will prove is that the sequence that an infinite decimal represents has terms that get arbitrarily close to each other, one to the next. That is, if I tell you how close two neighboring terms ought to be, you can tell me how far into the sequence I have to go so that each of all the subsequent terms is at least that close to its successor.

First recall how we defined what is meant by an infinite decimal. It is that we take a series of fractions, the first digit over 10, the first two digits over 100, the first three digits over 1000, and so on.

So for example, we would have as 0.414213...:

    4   41   414   4142   41421   414213
     ,    ,     ,      ,       ,         ...
   10  100  1000  10000  100000  1000000

To prove the assertion, we must show how, given any delta, no matter how small, we can find a counting number, j, so that the jth term of the sequence will differ from the j+1st term by less than delta.

First, look at the only at numerators in the example given above. Let's use the symbols,

   N1, N2, N3, ...

and more generally, the jth numerator in the sequence, we symbolize as N_j.

If you take any numerator, multiply it by 10, then subtract it from the next numerator in the sequence, the difference is always less than how much? So write down an inequality that expresses that relationship in the general case. Hint: it will involve an expression in N_j and N_j+1. Label that, expression 1.

Now explain in a sentence or two why this relationship applies not only to the example given here, but to the numerator sequence given by any decimal.

Now what can you say about the denominators? Let's call the denominators,

   D1, D2, D3, ...

and more generally, the jth denominator in the sequence we symbolize as D_j.

Well, these follow a very definite sequence that is the same no matter what infinite decimal we are dealing with. Write down the equation that states the relationship between one denominator and the next. Write also an equation that gives, in general, the expression for the value of the jth denominator. Label both of those expressions 2 and 3.

Remeber that when you wrote expression 1, you multiplied one numerator by 10 and subtracted it from the next. Now consider the two fractions,

   Nj         Nj+1
        and       
   Dj         Dj+1

   Nj+1      Nj
         -     
   Dj+1      Dj

Write label that as expression 3. Now can you explain how you can choose a j big enough to make that difference as small (that is as close to zero) as you like? Good. You have proved that you can find two consecutive terms in the sequence that are arbitrarily close. Now, based upon the relationship you have discovered between neighboring terms, explain why all pairs of neighboring terms beyond that are even closer.

Are you satisfied with your proof? If you are unsure you can email it to me (being sure to mention "box 1.5" with any email) at hahn@netsrq.com and I will check it. If you have questions, I will try to answer them. But before you do, or if you didn't get it, perhaps you might try following the proof I am about to give, and then go back and try to do this exercise again.

I shall now use a similar method to prove the assertion that the sequence generated by an infinite decimal is always a Cauchy sequence. Recall that in a Cauchy sequence, if you tell me how close you want the terms to be together, then I can tell you how far you have to go into the sequence so that any pair of subsequent terms will be at least that close together -- not just pairs that are neighbors, but any pair of terms, both of which are at least as far into the sequence as I told you to look.

First, please note that the in example sequence given above for 0.414213..., each term is greater than the last. In math jargon, the sequence is "monotone increasing." You should be able to show, using arguments that you have already made in response to the coached proof, that the sequence generated by any infinite decimal is always monotone increasing. We shall use the "monotone increasing" property of the sequence at the end of this proof.

What I will do is, rather than proving something about the difference between the jth and j+1st term, I will prove the same thing about the difference between the jth and j+kth term. In other words, I will prove the same thing you just proved, but not just between neighbors. I will prove it for terms that are as far apart as you'd like (that is they are k terms apart, and you can choose as big a k as you'd like).

As you have already noted, the expression for the jth denominator is:

   Dj = 10j

From algebra we recall that you add exponents in order to multiply exponential terms. So to put the jth term and the j+kth term over a common denominator, I will have to multiply top and bottom of the jth term by:

   10k

You see, the denominator of the j+kth term is:

   10j+k

So, to put the jth term over a common denominator with the j+kth, we have:

   Nj × 10k      Nj × 10k
              =          
   10j × 10k       10j+k

Notice that when you do this, the jth numerator, N_j, multiplied by 10^k, is a series of j arbitrary digits followed by k zeros. N_j+k, on the other hand, is j digits that all match the first j digits of N_j, but are followed by k digits that may or may not be zeros. In the example we have been using, if we choose j=2 and k=3, we have:

   Nj = N2 = 41

   10k = 103 = 1000

   Nj × 10k = 41000

   Nj+k = N5 = 41421

Clearly when I take the difference between N_j multiplied by 10^k and subtract it from N_j+k, I'll get a number less than the 10^k. In the example we are using, we would have:

   41421 - 41000  =  421  <   103  = 1000

Now, putting it all together:

      Nj+k        Nj × 10k         10k        1
             -              <           =      
      10j+k        10j+k          10j+k       10j

Or in the specific case of 0.414213... with j=2 and k=3, we have:

      41421        41000         1000       1
              -             <           =     
     100000       100000        100000     100

But the general case is what proves the assertion. What it says is that no matter what k is -- that is no matter how far I skip ahead into the sequence from the jth term -- the difference between the j+kth term and the jth term is never any bigger than a term that is independent of k (that is it is independent of how far I skipped ahead). Why? Because the first j digits of the two numbers will always match. Not only that, I can make that term (that is the bound on the difference), which is

    1
      
   10j

as small as I like (that is, as close to zero as I like) by choosing j large enough. And because the original sequence is always increasing, this means that the difference between any two terms beyond the jth term can never be greater than the above value. Can you explain why? If so, you have completed the proof.

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