If you don’t like the notation (merely because you don’t understand it, I must say), you may consider its equivalent which is (\frac{1}{10} \times \frac{1}{10} \times \frac{1}{10} \times \cdots). That’s a “valid” numerical representation, isn’t it?
I’ve already stated it but you conveniently ignored it opting instead to focus on what you don’t understand.
It seems like you’re one of those rare people who can understand Ecmandu
There is no last (1) in (0.\dot01). The product (\frac{1}{10} \times \frac{1}{10} \times \frac{1}{10} \times \cdots) is infinite. The end is merely in the symbol.
My post about 1/10 ever being expressed implied your argument of 1/101/101/10… etc…
I thought I could have used short hand, but it doesn’t bother me that we’re trying to be precise here, so, I apologize for the shorthand.
So here’s the deal: “carrying” is when you carry numbers from right to left. This is first grade math.
If I have a number like 99+2… the 2 plus the last 9 is 11. The one gets deposited in the solution column and the 10 gets carried over to the left in the tens column as a 1. 1+9 equals 10, which you put in front of the one that you already have the the solution column, to equal 101.
Sure. But how is that relevant? That was my question.
Why do you mention carrying?
It’s obvious that you’re talking about the standard algorithm of adding numbers together. The question is: why?
Are you trying to use the standard algorithm to add (0.999\dotso) and (0.000\dotso1) together? If so, why?
Do you think that you can prove that there is no difference between (0.999\dotso) and (1) by proving that the standard algorithm of adding numbers together can’t be used to add (0.999\dotso) and (0.000\dotso1) together?
Actually I’ve known all this math since over half my life ago, and I have been regularly keeping up with advanced level maths to this day for fun - I actually really enjoy it.
The problem here though is if others aren’t close enough to the same level, they aren’t going to recognise this and will just assume it’s just another bogus claim from some randomer on the internet because they don’t know any better.
I’m dealing with a self-professed non-mathematician here and it shows. All he has left anymore is “you just don’t understand”, and when I say/prove that I do, he can just respond with incredulity and keep repeating the accusation. That’s all this thread is now - non-mathematicians claiming long-time competent mathematicians don’t understand, whilst repeating the same schoolboy mistakes over and over. Other competent mathematicians here recognise these mistakes too, and keep advising the non-mathematicians that they’re wrong and I’m right. But our old-friend “the backfire effect” is just making them double down harder because admitting you’re wrong is hard for people since it requires emotional maturity and intellectual honesty to overcome the cognitive dissonance. That’s why there’s no longer any point in me trying to help them anymore - they don’t want it.
I only recently learned all the MathJax code to express the math nicely on this forum though - I already thanked Magnus for his only contribution so far of bumping the post from a couple of years ago that mentioned it and linked to its documentation. I also unequivocally accept the utility of exploring what happens when you take something that doesn’t work mathematically, and treat it as something that does work - as I’ve mentioned several times by now. All I’m doing is explaining exactly why it doesn’t work mathematically - I’m not even against doing it anyway because that’s where new math can be born, but there’s still a couple of people left who find even that too much to accept.
See, I’m humble enough to give credit where credit is due, and I’m simply honest about people being wrong when they are.
There’s plenty of topics on this forum that I leave alone because I don’t know enough about them - you won’t find me reading up last minute on those topics and claiming expertise there, and you’ll only find me claiming expertise when I actually have it, such as here. There’s far better mathematicians out there than me, but the dissenters here most definitely aren’t one of them.
Yes, that’s where we agree. Where we disagree is whether or not infinities come in different sizes. You think that they don’t (that’s where you agree with SIlhouette.)
I’m not asking you, because you’re not showing rationality in dealing with me.
It’s okay for you to tell us all about yourself and how you’re not a mathematician, but when I inform someone who isn’t you that I am - that’s a little too much for you…
And I’m only just bringing up myself now because I’ve finished trying to help you with the topic and I’m explaining something to someone else. But any excuse to take a swipe at me since you failed whilst we were on topic, right? So low.
FYI, this “standard algorithm” of addition that you’re trying to bring into doubt is known as “addition”. You’ll even blame the literal math itself as lacking because it can’t find this non-existent gap that your non-mathematician intuitions can’t let go of. If you want to detract from one of the fundamental bases of mathematics, I suggest doing what even the best mathematicians that have ever lived still haven’t done to this day and “improve” addition itself just to bring your abstract and metaphorical notions (for which you say you don’t have a rigorous proof because you don’t think it’s necessary) into literal existence.
Ecmandu, I am saying (1 = 0.\dot9).
You were literally just explaining that the difference between the values never arrives because the infinite zeroes never ever get to that mythical 1 “at the end of infinity” (a contradiction in terms)…
In the same way that this difference turns out to not actually exist, (1 + “0.\dot01”) will just be (1) too.
Has nobody actually bothered to check what all the professional mathematicians have concluded on this subject?
I’ll assume not, because if you do a quick search you’ll find they all side with me that (1 = 0.\dot9). It’s almost as if I wasn’t kidding when I said I knew what I was talking about, and I haven’t just been posturing all along like the dissenters here.
But I guess all the amateurs here can reject that because they know better, huh? This is just an appeal to authority after all, so it can be ignored if it doesn’t match your misunderstandings right?
So you agree that the length of the lines is the same even though there are fewer 90-inch segments than 1-inch segments. But that would imply that if you counted the segments in each, you would finish counting the 90-inch segments before the 1-inch segments (that’s what “fewer” means). That in turn implies that at least the number of 90-inch segments is finite.
That’s sorta kinda how we derive the definition. We work with line segments and say: “Hey, why don’t we refer to the distance between one end and the other as the line’s ‘length’”. If you want to expand the definition to mean something different with infinite lines, you’re on your own.
Yeah, 'cause heaven forbid we confuse conceptual matters with empirical ones. You’d get science!!!
Does so!
I’m sure it was convenient for you.
Really???
Exactly! You’re arguing my point.
You’re right about that, but variables (unknowns) can be treated like units, and visa-versa. 3X means 3 times the quantity X, but X here can be thought of as a unit, as in 3 times some object we call X (where this object is a set of X things). It works the other way around too: units can be treated like variables: 3cm = 3 x (100mm). You’re right that we know how much a centimeter is and how much a millimeter is, but if I tell you 1cm = n gwackometers, it suddenly becomes unknown (you don’t know how much a gwackometer is). So 1cm = an unknown number of gwackometers. The fact that you can divide a centimeter into any number of arbitrarily long segments means that it is an unknown in the sense that we don’t know how many of these arbitrary segment there are.
The central insight in algebra is that it doesn’t matter what the variables (or units) stand for–they might as well be unknowns–its the rules for manipulating them that matter. And that was my point about plugging (\infty) into algebraic equations–it doesn’t matter what it stands for, it’s just a symbol–manipulate to your hearts content. You don’t end up prooving anything.
Whatever it is you think this “similar” length is with respect to rays, you are once again on your own (and strictly speaking, we’re talking about terms like “shorter” and “longer”).
“Infinite quantity” isn’t really a quantity. That’s just a phrase we use to talk about infinity. If you want to say you know what (\infty) stands for in the equation (namely infinity), then you’re simply doing something you can’t do. It doesn’t stop it from being a symbol which you can plug into the equation and manipulate according to the rules of algebra. I can do the same with the symbol for male:
(\frac{{6}\times{}}{3}) = ({2}\times{})
Does this prove anything about males? Do you think it proves males are quantities?
Then you’re saying (\infty) is both the unit and the quantity at the same time. I take back what I said. I don’t agree that you can do this. You can still manipulate the symbols according to the rule of algebra, but you’re doing nothing but shuffling symbols around. You’re not actually deriving any insight into the nature of infinity.
The only time I’ve seen (\infty) used in an equation is:
(\frac{1}{0}) = (\infty)
…and (\infty) in this case explicitly means “undefined”–as in “you can’t do that!” You’re grade school teacher taught you not that dividing by zero gives you a really, really, really big number, but that you can’t divide by zero.
Ok, Magnus, you’re insisting that we treat (\infty) as an infinite quantity. Fine. I actually agree, (\infty) does mean “an infinite number of things”. But then I deny your right to use it in math. You’re trying to sneak it into math by, simultaneously, treat it as a unit–which, as I said, you can do but not while also treating it as a quantity. What number does 3cm equal? That is, how do you get rid of the units so that you’re left with only a number? 3cm = X. What is X? ← That’s what you’re trying to do with (\infty). You’re trying to say 3(\infty) = X, and since (\infty) is an infinitely large number, you can plug that number into (\infty), multiply it by 3, and solve for X.
What about the formula E = mc^2? We know what c is (speed of light), but what is m? It’s a symbol for mass, but do you know what the mass is?
}}{3}) = ({2}\times{