Digit Manipulation part 1

One popular proof for .999… = 1 than I enjoyed back when I was foolish enough to believe it true was through digit manipulation. Observe:

x = .999…

10x = 9.999… (both sides of the equation are multiplied by 10)

9x = 9 (“x” is subtracted from both sides)

x = 1 (both sides of the equation are divided by 9)

Ta da! What could be more foolproof than that? The starting premise is obviously correct, and the same thing is done to both sides of the equation until it spits out a conclusion that even a guy like me would have to admit is true. Right? Right???

Well, obviously not, otherwise I wouldn’t have written this blog. While it appears that the same x is being subtracted from both sides in the third line, it’s actually two different x’s. When you multiply a number by 10, you don’t simply add 9 to it (more on this in part 2)… you get that 9 before the decimal point by moving the decimal place over to the right one space. This is true of any real number… multiplying by 10 moves the decimal place over once. By doing this, there is one less digit to the right of that decimal point. That’s true of every finite number, and should also be considered true of infinite numbers. Therefore, when one subtracts x in the equation above, you’re subtracting an infinite series of 9’s from another infinite series of 9’s that is one digit shorter. If you want to see me do all of the actual math to show you that this checks out, view my YouTube video on the subject.

A final note: While I welcome criticism, please don’t tell me .999… multiplied by 10 still has the same number of 9’s after the decimal point unless you can demonstrate why this is so. I’ve already talked about how mathematicians accept different size infinities, and I explained in my first blog in this series why inductive reasoning suggests that if any real number multiplied by 10 has one less digit after the decimal point then it should still be true when dealing with infinite digits. If you’re going to argue with me, please present a point that I haven’t already refuted.

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About starcrashx

I love statistics. They drive my poker playing, my reasoning, and my research. As Penn Gillete said "Luck is probability taken personally". There's no such thing as luck... but I wish you positive chance.
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9 Responses to Digit Manipulation part 1

  1. Matthias says:

    for this you need to get in touch with limits… as a matter of fact, the decimal is moved one to left… but that really doesn’t matter since the mistake is 0 in the end… Take the series 10*\lim_{n\to \infty} \sum_{i=1}^n 9*0.1^i=\lim_{n\to \infty} \sum_{i=1}^n 9*0.1^{i-1}=\lim_{n\to \infty} \sum_{i=0}^{n-1} 9*0.1^i… it has shifted as you see…=\lim_{n\to \infty} \sum_{i=0}^{n} 9*0.1^i, that sounds hard to believe, but do you know that equality means for real numbers, that if you do subtract, the rational sequences, its limit is approaching 0. and this is rather easily seen.

    • starcrashx says:

      Again, I’ll cite this paper — https://www.filesanywhere.com/fs/v.aspx?v=8b6966895b6673aa6b6c — in which the author explains why Euler’s definition of limits is just rounded off. Like every mathematician I speak to about this concept, you have to understand that 1 divided by 10 to the infinite power is not zero, even if it’s as close to zero as you can get.

      • Jeremy says:

        1) All numbers have a fixed value
        2) .888…. is a number
        c) .888…. has a fixed value

        The value is already there in repeating decimals since they are numbers. They’re not functions, series, or sequences and therefore can’t tend towards anything.

  2. Jeremy says:

    Even if it has one less nine why should there be anything on the other side of it. All rational numbers are infinite in length or I’ve heard said. 1.2 is actually 1.20000… This is where all those zeroes come from when you multiply by high powers of 10. In .8888…. the eight repeat infinitely and when you multiply by high power of ten you can only get eights, There are no zeroes in there.

  3. Wani says:

    “A final note: While I welcome criticism, please don’t tell me .999… multiplied by 10 still has the same number of 9′s after the decimal point unless you can demonstrate why this is so.”

    Because adding a number to an infinite set does not change the amount of numbers in the set, and yes, I can prove it. I’ll use it via a proof from contradiction. Let’s assume that adding or removing a number from an infinite set does change the size of the set. We’ll look at the set of counting numbers (ie S = {1, 2, 3, …}). This has size N.

    Then, let’s add a 0 to the start, to give us S0 = {0, 1, 2, …}. Since we’ve just added a number to the start, this has size N + 1.
    Now, let’s add 1 to every number in set S0, which we’ll name S1 . This gives us S1 = {1, 2, 3, …}. This set is identical to set S. If we look up the nth value in S1, for any natural number n we choose, it’s exactly the same as the nth value in S. There is no difference whatsoever. Therefore, they must have the same size, so N = N + 1. Therefore, 0 = 1. This is a contradiction, thus, our assumption that adding or removing a number from an infinite set changes the size of the set must be wrong.

    This exact same logic applies to trying to say that, if x = 0.999…, 10x has less 9’s after the decimal place than x. If we say that, then we wind up with logical fallacies as above that state that 0=1.

  4. starcrashx says:

    It’s funny that you accuse me of arguing that N = N + 1. You’re the one that actually believes that adding a number to an infinite set can keep it the same size… that is, *you* are arguing that N = N + 1, not me. Are you even aware of that?

    The rule in mathematics is that if you multiply a number by 10, there is one less digit after the decimal point. The logical fallacy made when you assume that an infinite string is the exception is called “special pleading”.

    • Wani says:

      It’s not a fallacy if it is an actual exception, because that’s what you’re arguing. You can’t say “there’s one less 9 after the decimal point if you multiply 0.999…” because you’re saying that infinity – 1 = something not infinity. In which case, either you’re not talking about infinity, or you’re arguing that N = N + 1.

      Also, as another wee note, the definition of 0.999… with infinitely many 9s is that, for any natural number n, if you look up the nth decimal place, you find a 9. As you’ve stated, when you multiply by 10, you basically shift the decimal place one to the right, giving you 9.999… with infinitely many 9s after that. For a usual, non-recurring number, there’s a point, m where there’s only 0s after that and this m gets shifted back 1 decimal place. For a recurring decimal, this m does not exist, because for any decimal place n, there’s a 9 (Or a 7 if you’re dealing with 0.777… and so on). Therefore, there cannot be a point, m, where there are only 0’s after that, or else our recurring decimal isn’t recurring, it stops at some point. It follows that, for 10x (ie, 9.999…), it’s still true that for any natural number n, if you look up the nth decimal place, you find a 9. It seems counter intuitive, but this means there isn’t one less 9 after the decimal point, because for every natural number n, if you look up the nth decimal place, there’s a 9 in both x and 10x. If there’s one less 9 in 10x, there’s an nth decimal place where there’s a 9 in x but not 10x, and if that isn’t true, it means that both x and 10x aren’t infinitely recurring.

      That was probably far more rambly than I intended.

      • starcrashx says:

        It’s true that it’s not a fallacy if it’s an actual exception, but you have to explain why it is an exception if you wish to convince me to agree with you.

        Please don’t misrepresent me. I’m not saying that adding or subtracting a number from an infinite set will give you “something not infinity”… I’m arguing that it will give you an infinite set of a different size — which, if you believe that the infinite sets of real numbers and natural numbers are not equal, you agree is possible.

        Furthermore, I’m not arguing here or elsewhere that .999… has any digit that is not a 9. The “0” at the end is merely a placeholder to demonstrate that there is one less 9 after the decimal place.

        I understand where you’re coming from, but you’re still reifying infinite numbers. Obviously you’re never going to come to the same conclusion that I do if you keep treating them like physical realities. What I’m proposing is a form of model-dependent reality such as that put forth by Stephen Hawking, where we can use nonsense ideas like “invisible time” if it consistently works to explain ideas that can’t be understood otherwise. Infinity is a troublesome concept, which is why I came at it from patterns through inductive reasoning (taking numbers that we can handle and then expanding them infinitely).

      • Wani says:

        “Furthermore, I’m not arguing here or elsewhere that .999… has any digit that is not a 9. The “0″ at the end is merely a placeholder to demonstrate that there is one less 9 after the decimal place.” But if there’s a 0 there, then you’ve found an nth decimal place that doesn’t have a 9, which means we don’t have a recurring decimal. And what comes after the 0?

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