Why is this "dis-proof" of mathematics wrong?

Question

This was presented to me as a joke, but I can't find any flaw in the logic. I know it's there, but I just can't place it.
Assume 1=1. Then, by the definition of division, 1=3/3. So 1=3x(1/3). So 1=3x0.333...(continuous). So 1=0.999...(contiuous). So 1 does not =1. You can then repeat ad infinitum and, by infinitesimal steps, eventually get 1=any number. Since the mathematics is based on the identity principle and the uniqueness of numbers, all mathematics is disproved.

by "3x1/3", I mean three times one-third. Pardon my notation.

...and then I misquote myself, I need more sleep.

Namenot, but 1 and .9999... are not equal. They are asymptopically close, but still distinct.

For "infinitesimal steps" part. I beleive that means that you could set .999...=1x.999... and go thru the same process to get .999... squared, which is slightly further from 1.

And infitesimal steps adding up to real differences is the basis of the number line. There may be an infinite number of fractions between 1 and 0, but eventually those negligible fractions add up to 1.

Answer 1

"So 1 does not =1"

False. 1 DOES equal .9999...., therefore saying that because 1=.9999... 1≠1 is invalid.

"can then repeat ad infinitum and, by infinitesimal steps, eventually get 1=any number"

Methinks that thou dost not understand the meaning of infinitesimal. An infinitesimal is a number so small that ANY number of infinitesimals put together will not equal 1. Neither will any number of them put together equal 1/2, 1/4, 1/1,000,000,000,000,000,000,000,000, or any other real number. So you can't take two different real numbers and connect them by infinitesimal steps. And in any case, which infinitesimal steps are you talking about? ∀n, changing the nth digit of .9999... will result in a number that is less than 1 by exactly 1/10^n - not an infinitesimal. No, you can't change the ∞th decimal place, because there is no ∞th decimal place - there are only an infinite number of finite decimal places (this is basically the same as saying that every natural number is finite, even though there are an infinite number of them). The only step that is smaller than any positive real number is .9999.... to 1, and it is precisely because that step is smaller than any positive real number that we know it must not be a step at all, because if it were a step, we would be able to take half a step, and in this case we can't.

Edit: "For "infinitesimal steps" part. I beleive that means that you could set .999...=1x.999... and go thru the same process to get .999... squared, which is slightly further from 1."

Here's an instructive exercise: try actually squaring .9999... - the whole infinite decimal expansion, using formal methods. You can use a variant of the normal grade-school algorithm that works from left to right - you will have the problem that each digit is provisional, and thus might be affected by borrows and carries (which is why the grade-school algorithm works from right to left, so you don't have to erase provisional results), but the method will work on infinite decimals just as well as finite ones. Since you want to finish in a finite period of time, once you start seeing a pattern in the decimals, use induction to show that EVERY digit after the nth will have a certain value (9), and that no carries will affect any digits prior to the nth. When you actually complete the multiplication (which is fairly complex), you find that .9999...²=.9999..., and thus .9999...² is no further away from 1 than .9999... itself was (which we would expect, since .9999... IS 1).

As a simpler example to help you get started (I don't intend to walk you through the multiplication itself, as that would take several pages), consider the addition of .9999... to itself. We start by adding the first two digits to get 8 with a carry of 1, giving us 1.8. We then add the next two digits, giving us 8 carry 1 again - the carry changes the 8 to a 9, and we have 1.98. Similarly with the third digits, we have 1.998. We must ask, are any of these digits non-provisional? Suppose after the nth step, we have 1, n-1 nines, and a trailing 8 in the final position. Then after the n+1th step, we have 1, n nines, and a trailing 8 in the final position. In particular, the n+1th step did not change any of the digits strictly prior to the nth (after the decimal point). However, because this is the same situation as we had before, except with an inceremented n, the n+2th step will not change any of the digits prior to the n+1th, and therefore certainly won't change any in the nth. Thus we have that the n+1th step doesn't have any affect on digits prior to the nth, and that if the n+1th step doesn't have any effect then neither will the n+2th. Then by induction, for all N ≥ n+1, the Nth step won't change any decimals prior to the nth. Therefore, ∀n, all steps after the nth are irrelevant to the value of any decimal prior to the nth. In particular, all steps after the nth will not change the n-1th place. Ergo, the value of the n-1th digit is the value of the n-1th digit after n steps of the algorithm. However, the n-1th digit after the point is always 9 after n steps of the algorithm. It follows that every digit after the point is 9 and the value of .9999... + .9999... is 1.99999... . Note however that this is exactly what we would get if we added 1 to .9999.... This is expected, because 1=.99999....

As an aside, I very rarely try to formally add or multiply infinite decimal sequences in their entirely, for the simple reason that it IS so complex, and in most cases finite prefixes out to, say, 20 digits are more than sufficient for practical work (and for theoretical work, it's easier to avoid decimals altogether and use fractions). But it can be done, and it always gives you A valid decimal representation of the correct result (although for terminating decimals, which have two valid representations, which one of those particular representations you end up with may depend on your order of operations. This need not be alarming if you recognize that they are in fact the same number, any more than ending up with 3/6 instead of 1/2 should be for someone working in fractions).

"And infitesimal steps adding up to real differences is the basis of the number line."

No, I'm afraid that's complete bullshit. The basis of the real number line is that every cauchy sequence of rational numbers has a limit in R, or equivalently, that every nonempty set of reals bounded above has a least upper bound in R. Note that the latter proves the nonexistence of infinitesimals - suppose that there was at least 1 positive infinitesimal. Then the set of positve infinitesimals is nonempty. It is clearly bounded above (by, say, 1), and therefore has a least upper bound. Call this least upper bound x. Either x is an infinitesimal or it isn't. If it is, then so is 2x, which (since x is positive) is strictly greater than x, and so x is not an upper bound for infinitesimals. If it isn't, then neither is x/2, which is strictly smaller than x, and so x clearly isn't a _least_ upper bound. Either way, we have a contradiction, and therefore our initial assumption, that positive infinitesimals exist, must be false. Q.E.D.

"There may be an infinite number of fractions between 1 and 0, but eventually those negligible fractions add up to 1."

Huh? 1/2+1/3+1/4 adds up to 13/12, which is more than 1, and only uses a finite number of fractions. I don't see... AH... youre thinking of integration, not the real number line. At one point, calculus (not the real numbers per se) made heavy use of infinitesimals, however this was considered nonrigorous, because infinitesimals would alternately be considered as numbers (when taking derivatives and integrals), and as nonnumbers (after the derivative was taken, any term that continued to contain an infinitesimal was simply forgotten). Eventually, a rigorous foundation for calculus was developed in terms of limits, which made the use of infinitesimals completely unnecessary, thus leading to their being unceremoniously dropped from the number system. (certain people have tried to ressurect them in the form of non-standard analysis, using the hyperreal numbers, but that has nothing to do with the real numbers). Presently, the foundation of both calculus and the reals is based on limits, not infinitesimals. And in any case, at no point was it thought that you could add any finite number of infinitesimals and get a real number (an infinite number of infinitesimals, yes, that was how integrals were once formulated - but never any finite number). And insofar as .9999... is interpreted to be the decimal expansion of a real number, .9999... =1

Answer 2

The "flaw" in this reasoning is the ol' nemisis of the terms "infinite and infinity."

1/3 = 0.3333333........to infinity

0.3333....+ 0.3333.... + 0.3333.... = 0.9999....
Therefore, 1/3+1/3+1/3=0.9999.....
But it doesn't - it clearly equals 1

Unless 0.9999......to infinity = 1

Infinite numbers do not "prove" that 0.999.....
equals 1. First, because the term "infinite" is a man-made and defined term - infinite anything does not occur in reality. Secondly, when we try and define infinite in real terms, it just doesn't compute. For example, what do we get if I add 1 (or any number, for that matter) to your infinite number? A new and larger infinite number? Does that mean the original number isn't infinite any more? How about if we square the infinite number? Wow, that product must be a super infinite number, eh?

It is a common misconception that if we divide a number by zero, the answer is an infinite number. By that reasoning, the "proof" would reveal that a zero number of infinites would be infinity - and that is rediculous. A zero number of ANYTHING is still zero. If zero divided by 0 is 1 (soulds logical - zero goes into itself one time?) But, again, we hit a snag - the "proof" reveals that zero times zero equals 1 - again, this just ain't so.

In conclusion - don't divide by zero, don't use the term "infinite" to "disproof" mathematics and don't pee off the front of a fast moving train.

On the other hand, if you give me a penny for my thoughts, and I give you my two cents worth - where the hell does the other penny go?

Answer 3

1 does equal 0.999... continuous. Different expressions of the same number.

Somewhere I have a book with a better puzzle. It takes you through some equations, then seems to prove that 1=2 ! The trick is that at one point, the denominator of a fraction was x-1, when x itself was also 1... meaning it had divided by zero, which is against the rules because you get nonsense answers.

Answer 4

1 and .999... *ARE* equal! Numbers with terminating decimal expansions can also be written in forms with infinitely many 9's. So,
.25=.2499999....
also. The numbers are the same; there are simply two ways of representing them.

Answer 5

Just use algebra

You have x and x/3. Multiply x/3 by 3 and get 3x/3 or x.

It equals 1, end of story.

Answer 6

1=3x(1/3) would be 3/1 * 1/3 that is 3/3

Answer 7

wellll, 1/3 is really equal to .33 in your case, its just the closest we can get. 1/3rd is just, 1/3rd.

Answer 8

1 does equal .99999..... This doesn't disprove anything, it only proves that identity.