Steps one goes through to arrive at an answer, whatever "answer" may mean in the context of a problem. I sincerely hope this is not where the conversation is going.What is a process equation?
Steps one goes through to arrive at an answer, whatever "answer" may mean in the context of a problem. I sincerely hope this is not where the conversation is going.What is a process equation?
Yes, I was completely wrong. I apologize. But what I was getting to is what you are saying, that there is some number in between, unlike 0.23 and 0.22999...
Yes, but I already claimed why the proofs are not at least complete. So there is an error. Let me use an example:
this is a classical proof. But I would argue on the second line. The point is can you prove that the second line is correct??Code:x = 0.(9) 10x = 9.(9) 10x - x = 9.(9) - 0.(9) 9x = 9 x = 1
But I would argue on the second line.?Code:10x = 9.(9)
Argue what? Anything but that isn't multiplication by 10. Does it matter that there are infinite places in x? No. As I said, any number has an infinite expansion, but you still move the . to the right.
sum(i = 0 ... inf){9 * 10^-i) = 9 / (1 - .1) = 10. See here. I quote Euler, therefore I win.
The multiplication exists only as a form of summing a certain times the same number. I believe that is its definition. How do you sum decimals? You start from the last digit and go on. You say there is no last digit, it expands forever. I am stuck.
Or are you telling me that the last digit is 9, therefor the sum has to have a last digit of 8, which is not true since you say it is 9?
You don't have to start from the last digit; you can start from the first corresponding digit pair too, with a little backtracking. Either way, this does not really matter, in my opinion: the fact that we can err... float the decimal point to the right when multiplying by 10 in base 10 should hold even if you cannot fathom how to actually add the numbers on paper using the given representation.Originally Posted by C_ntua
Look up a C++ Reference and learn How To Ask Questions The Smart WayOriginally Posted by Bjarne Stroustrup (2000-10-14)
Repeated summing is far from the definition of multiplication. How do you do 1/2 * 3/4 with repeated addition? Summing is a special case, not the rule.
If there was a last digit, it wouldn't necessarily be a 9, but there isn't a last digit.
Applying the method of decimal addition I remember from grade school:
The 1.111... is where I "carried."Code:.999... + .999... 0.888... 1.111...
True...
You don't necessarily have to start at the last digit, if you're willing to do more work, you can start at the first one (which, for simplicity of concept, we can do with numbers that expand in non-zeros)...
1.888888... + 2.333333... =
sum the 1 and 2, we get 3.
Sum the 8 and 3, we get 11, keep the second 1, add the last 1 to the last sum (to be 4).
Sum the 8 and 3, we get 11, keep the second 1, add the last 1 to the last sum (this time to be 2).
Sum the 8 and 3, we get 11, keep the second 1, add the last 1 to the last sum (again to be 2).
Sum the 8 and 3, we get 11, keep the second 1, add the last 1 to the last sum (again to be 2).
Sum the 8 and 3, we get 11, keep the second 1, add the last 1 to the last sum (again to be 2).
And it then keeps going on. We get
4.222222....
There's no problem with this, it's all good. One might say "but it goes on forever, so you can't know what comes out to be", but this isn't true. As soon as things stop changing (that is, we enter the calculation loop of the "..."), we're done, because we at that point know everything we need to know. To prove it, ask me what the value at **any** given index/digit is and I can give you an answer. Try to use numbers that expand in non-zeros with the "normal" numbers that do expand in zeros. It always works without issue.
One can't use the argument of "you get stuck" or "you can't know", because we don't ever get stuck, and we do know everything about these numbers that there is to know. They can be used in **any** mathematical operation flawlessly without a problem or "error" arising.
Lets say index number 5. With your logicTo prove it, ask me what the value at **any** given index/digit is and I can give you an answer.
Sum the 8 and 3, we get 11, keep the second 1, add the last 1 to the last sum.
So you have: 4.22221
So how exactly do you get that any digit will be 2? It rather seems that there will always be a digit that will be 1....
1/2 * 3/4 = 1*3 / 2*4 = 3 / (4+4)Repeated summing is far from the definition of multiplication. How do you do 1/2 * 3/4 with repeated addition? Summing is a special case, not the rule.
Wrong, you add one additional 1. If you exclude that 1 you will get 8 as the last digit. In any case, the carrier will NOT apply to the sum that "created" the carrier, so with this logic you always have an 8.Code:.999... + .999... 0.888... 1.111...
And what is exactly the proof that we can do so for a decimal number with infinite digits? When you normally float you end up with one less digit, or the digit 0. Which is not the case here. So this is not exactly consistence, I don't find it really obvious in the sense that it doesn't need further proof or analysis.float the decimal point to the right when multiplying by 10 in base 10 should hold even if you cannot fathom how to actually add the numbers on paper using the given representation.
I would actually say that all the above rather disprove the point rather than prove it.
Furthermore, if I told you "you have two decimal numbers, one has an integral part of 1 and the other of zero, which one is the bigger?" usually you would answer "the one with integral part 1". And this would be true for all decimal numbers with finite digits. Or actually all decimal number with the exception of trailing 9s. So where I am getting at is that the the infinite 9s doesn't follow anyway all the rules. Why do we assume that it just follows the rule of "multiplying by 10 you just float the digits"?
It's called a "rhetorical question."
It sums to 1.999... because there is no last digit. If there was a last, it would end in 8, but there isn't, so it doesn't.
EDIT: I didn't answer specifically. I know there the 9s continue in 1.999..., because the carry continues as long as the string of 9s in .999... The string of 9s is infinite, so the carry is also infinite. The idea of there being an 8 at the end is incorrect because of the axiom of real numbers that all digits in a decimal expansion must have a definite position. We can say the digit a_x = 9 for all a in 0.a_0a_1a_2a_n..., but, we have no way to specify the index of the 8 "at the end of infinity".
Last edited by User Name:; 03-13-2011 at 05:54 PM.
This is not a repeated sum. As proof of the idea that multiplication is repeated sums, I would accept that you added up 1/2 a total of 3/4 times. I'm genuinely interested in you demonstrating this idea as you stated it -- that multiplication is repeated addition, for reals other than integers.1/2 * 3/4 = 1*3 / 2*4 = 3 / (4+4)
But let's do it your way anyway!
If you want to multiply a number like 0.9... You could imagine multiplying each term in a equivalent series:
9/10 * 10 + 9/100 * 10 + 9/1000 * 10 and continuing. I don't know where you get the idea that there would be a last 8, because all you do when you multiply a number by its base is shift places. It's plenty easy to think of doing this for every 9 in 0.9...
Just a quick thought without much thinking behind it:
For a real number m in base-10 with n digits in decimal part and o digits in integer part: m = sum(i = o .. -n)(a_i * 10^i). Isn't 10m = sum(i = o .. -n)(a_i * 10^{i + 1}) and isn't this true even if n = inf?
There are inconsistencies in the Real Numbers system that have been studied by others. I have not looked yours up in detail. I'm merely picking your quote for truth.
The Real numbers are dubbed by some as self-inconsistent. I'm unsure as to the real meaning of that wording. I suspect it tries to refer to the fact that these inconsistencies are what keeps the number system from falling apart when dealing with infinity. Basically the number system falls apart when dealing with these quantities. However, the rules and axioms that have been established, while inconsistent, keep the number system consistent with the classical approach to calculus that we know as Limits.
Meanwhile, non classical approaches exist for calculus, and for the study of Real numbers, that support the notion of a precise definition for an infinitesimal that is non zero. And this changes everything. What some here seem to keep implying is that this does not happen in R, when it does! While hyperreals do extend the number space, other methods employed by non-standard analysis don't. That is the case of IST. It's not an extension of R. It's an extension to ZFC, the set of axioms governing pretty much all of modern mathematical thinking.
So what can be said at best is that 0.999... equals 1 in classical calculus employing Limits. But definitely not in R where other approaches demonstrate them to be different. This is not a limitation of the number system. This is merely a limitation of the methods employed and their mathematical foundations. And it's also one of the foundations of my claim that 0.999... equals 1 can't be proved.
In fact the importance of an infinitesimal has lead to a sort of friction between renowned investigators. So I find it particularly curious that by the time we reach page 14 of this debate, we are still hearing some students of maths in here fully convinced they can provide proofs on subjects that are at a very minimum minimum arguable and far from being universally accepted. Or that may be subject to change in the future.
I can only conclude because they are stuck with classical calculus as their one and only tool to handle real numbers. They see the inconsistency, but they trust their axioms. That's the type of indoctrination that happens at our schools (I apologize if I sound rude or pretentious. That's not what I mean).
In fact, on many places, exposing these inconsistencies or daring to reject established axioms (regardless of still studying them, employing them and showing good knowledge of them) is grounds for a negative note. I wouldn't be surprised if some teachers would flunk you for even daring mentioning non-standard analysis in their classroom.
Originally Posted by brewbuck:
Reimplementing a large system in another language to get a 25% performance boost is nonsense. It would be cheaper to just get a computer which is 25% faster.