Are All Infinitely Long Repeating Numbers Even?
Date: 06/06/2000 at 17:53:50 From: Richard Huggins Subject: All infinitely long numbers of repeating form are even? Take any infinitely long repeating series: x = 1234123412341234... This one repeats after a section of length 4. Multiply the number by 10^(section length), in this case 10^4. Thus we have: 10000x = 12341234123412341234... and x = 1234123412341234...... Subtract the two: 9999x = 1234000000000000000000000000... Clearly the right-hand side is even, because it ends in zero, i.e. "9999x" is even. But since 9999 is odd, therefore x must be even. Thus all infinitely long numbers of a repeating form can be shown to be even. Does this make sense? Is it correct? How could a series consisting entirely of odd numbers (ex. 1357135713571357....) be even? Thanks in advance for any response. Richard Huggins
Date: 06/07/2000 at 18:03:38 From: Doctor Rick Subject: Re: All infinitely long numbers of repeating form are even? Hi, Richard. Interesting question! If you do the subtraction with any finite number of terms in the sum, you will see that the difference does not end in zero; it does not have to be even. 1243124312430000 - 124312431243 ---------------- 1242999999998757 Even if you have a huge number of digits, you can't ignore those last digits just because they are "infinitely small" compared to the number. However small they are, they still determine whether the number is even or odd. What makes your argument interesting is its resemblance to the argument that 0.999... (with the digit 9 repeated infinitely) = 1. See our Dr. Math FAQ on " 0.9999... = 1" at: http://mathforum.org/dr.math/faq/faq.0.9999.html The simple form of that argument looks like this: x = 0.9999... 10x = 9.9999... 10x - x = 9.9999... - 0.9999... 9x = 9 x = 1 Very much like yours; however, it's important to recognize what's going on behind the scenes here. Students are right to question this argument; we need to know more to be sure it's valid. The number 0.99999, to pick a finite number of 9s, is really a shorthand way of writing 9/10 + 9/100 + 9/1000 + 9/10000 + 9/100000 The number 0.9999... is a shorthand way of writing the limit of the sequence 0.9, 0.99, 0.999, 0.9999, ... as the number of 9s increases without end We can write this in more formal mathematical notation as n Limit [ Sum 9/10^i ] n->inf. i=1 For any finite number of decimal places (say, n = 5), we have: As n increases, the sum approaches a limit, which is 1. By this we mean that you can pick any number, however tiny, and I can find a value of n such that the difference between the sum (0.999... with n or more 9's) and 1 is less than the number you chose. That, in essence, is what we mean by a limit. The FAQ gives an equivalent proof to the simple form above, using the notation of limits and sums. It hinges on a theorem in the theory of limits, that you can interchange a limit and a sum *provided* that the limit exists. We can treat your argument similarly. Your "infinitely long repeating series" can be written formally as n Limit [ Sum M*10^(Ki) ] n->inf. i=1 where M is the repeating part (like 1234) and K is the number of digits in the repeating part. Your argument breaks down because the limit above does not exist. In other words, the series (sum) does not "converge." The nth term increases without limit: 1234, 12340000, 123400000000, ... If a series does not converge to a limit, you can't do operations on it as you can with numbers. Here's another way of saying it: you are subtracting two "infinite" numbers. Infinity is not a number, and you can't do ordinary arithmetic on it as if it were. See our Dr. Math FAQs on "Infinity and large numbers" and "Dividing by 0" for more discussion of this subject. But the better way of saying why your argument fails is that you are subtracting two series that do not converge, and this is invalid. I hope I've stimulated some thought about important mathematical concepts. Fallacious proofs are a great way to get us thinking, and this is a good one! - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
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