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Infinity

If we look at the integers 1 and 2 we can see that there is obviously an end to the decimal points as we have to reach 2. Even if we skip multiple millions of decimal places, we still need to reach it. Now if we start counting the decimal points, we can't actually ever reach 2. The decimal points can range from 0.1 to 0.0000000000000001 (etc) and essentially go one forever, making the numbers infinite. So how come we can reach 2 then? I'm sorry if this sounds like a stupid question, it's just been bugging me for a while :) Thank you

Note by Bianca Ayling
3 years, 8 months ago

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the intended height that it can never reach... Rabia Basri · 3 years, 8 months ago

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@Rabia Basri Thank you, this is what I believe. See we can go on about counting the numbers all day, but what about them just being there? In between 0 and 1 there is 0.000001, 0.000002 etc... the digits past the decimal point are infinite, or so they tell us, but then they wouldn't reach the next number? 0, 0.000000000000000000000001 ...... 0.9999, 1 The zeroes could go on forever! But if they did, we could not get to 1. I've pretty much just repeated myself there... But yes, I understand that when we simply count 1, 2 , 3 etc... we are counting 'real' numbers, and that makes sense to me, but getting to the numbers, from the previous, doesn't. Wow... I'm bad at wording my thoughts.... Bianca Ayling · 3 years, 8 months ago

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Actually the set of real numbers is "well ordered" but not "countable".You can never travel along the real number line because you never know what comes next.There are infinite real numbers between any two real numbers."Well ordered" refers to the fact that if two real numbers are given;say a & b you can predict that a<b/a>b/a=b.But we cannot count real numbers. In fact isn't this where calculus comes into play?The concept of limiting values for a function actually refers to catching up values that a function f(x) tends to reach as x tends to reach a particular value.If you notice , set of natural number itself is a function,f(x)=x!L.O.L. In fact there are sets which are neither countable,nor well ordered,nor finite.Like set of complex numbers. Natural numbers is on the other hand countable.Though you must not forget that it is an infinite set. In a nutshell,the question of reaching 2 by travelling along number line from 1 is,I am afraid, irrelevant.Just forget the fact that real numbers means travelling along real number line. Just imagine it to be a set. A big B'day balloon with too many real numbers in. Too many to count ;) Biswaroop Roy · 3 years, 8 months ago

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@Biswaroop Roy The set of complex numbers is well-ordered. Take a well-ordering of the reals \(P\); we can obtain a well-ordering of the nonnegative reals \(P'\) by removing all the negative numbers (might not be correct!), and then a well-ordering \(P''\) of the reals in \(\left[ 0, \frac{\pi}{2} \right)\) by taking the arctan of every number, and then a well-ordering \(Q\) of the reals in \([0, 2\pi)\) by scaling all numbers by \(4\). We can then define a well-ordering of the complex numbers in their polar coordinate form \((r, \theta), r \in [0, \infty), \theta \in [0, 2\pi)\) except when \(r=0\) where \(\theta = 0\), such that \((r_x, \theta_x) < (r_y, \theta_y)\) if and only if \(P'(r_x) < P'(r_y)\) or \(r_x = r_y\) and \(Q(\theta_x) < Q(\theta_y)\). Ivan Koswara · 3 years, 8 months ago

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@Biswaroop Roy Thank you so much! Bianca Ayling · 3 years, 8 months ago

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I don't quite get what you mean. Basically you're saying that there are an infinite number of numbers between 1 and 2? Or something else... Ivan Koswara · 3 years, 8 months ago

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@Ivan Koswara Sorry, it's not really clear... If there are infinite numbers between 1 and 2, how come we can reach 2? Bianca Ayling · 3 years, 8 months ago

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@Bianca Ayling Why can't we do so? It depends on how you count. If we count on the natural numbers, it's obvious that \(2\) follows \(1\). If we count on the real numbers, this is what I believe your question is:

If we enumerate the real numbers not less than \(1\) as \(e_1, e_2, \ldots\) such that \(e_i < e_j\) whenever \(i < j\) and \(e_1 = 1\), we cannot have \(e_n = 2\) for any \(n\) since there are infinitely many real numbers between \(2\) (namely \(1.9, 1.99, 1.999, 1.9999, \ldots\)) and so \(n\) is infinite; is it true?

If that's your question, then it is true. The problem is when we attempt to enumerate the real numbers in that way, so the concept of "counting the real numbers" is meaningless. The concept of "counting" only exists in sets with the well-ordered property; that is, for any subset of a set with well-ordered property, there exists a "smallest" element. In the case of natural numbers, it has the well-ordered property, so it makes sense to count \(1, 2, 3, \ldots\). But in real numbers, or even in the integers, there is no well-ordered property with the usual \(<\) relation, so it doesn't make sense to "count the real numbers" or "count the integers".

Just for fun, we can redefine the \(<\) relation so that for the integers, \(0 < 1 < -1 < 2 < -2 < \ldots\), and then it has the well-ordered property, and it makes sense to count the integers as \(0, 1, -1, 2, -2, \ldots\), although it's probably of little value. We can even make the rational numbers to have the well-ordered property. However, we cannot make the real numbers to have the well-ordered property; this touches upon another concept of infinities, that the set of natural numbers has the same number of elements as the set of rational numbers but less than the set of real numbers.

And finally, that's all what I believe to be true; I haven't proven rigorously a few of the implications I use. I appreciate comments, of course. Ivan Koswara · 3 years, 8 months ago

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@Ivan Koswara Turns out I'm wrong! By the well-ordering theorem, every set can be well-ordered, including the real numbers. So we can order the real numbers in some way so that it's well-ordered. So I use a different argument:

By counting a set, we implicitly create a bijection between the natural numbers (or the first few natural numbers) with another set. For example, if we count the primes \(2,3,5,7,11,\ldots\), we create the bijection \(1 \to 2, 2 \to 3, 3 \to 5, 4 \to 7, 5 \to 11, \ldots\). Same with the integers I put above: \(1 \to 0, 2 \to 1, 3 \to -1, 4 \to 2, 5 \to -2, \ldots\). But the real numbers have more elements than the integers, so we cannot "count" the real numbers. (This is why the set of natural numbers is called to have countably infinite number of elements but the set of real numbers is called to have uncountably infinite number of elements; the natural numbers can be counted but the real numbers can't.)

And if we restrict ourselves to the set of rational numbers (which is countable!), we can create a well-ordering of the rational numbers. It's quite tough; this is one ordering of the nonnegative rational numbers that I know, although there are many others: \(0, \frac{1}{1}, \frac{2}{1}, \frac{1}{2}, \frac{3}{1}, \frac{1}{3}, \frac{4}{1}, \frac{3}{2}, \frac{2}{3}, \frac{1}{4}, \ldots\), putting \(\frac{a}{b} < \frac{c}{d}\) (in their simplest forms) when \(a+b < c+d\) or \(a+b=c+d\) and \(b < d\). (To get the rational numbers, simply insert the negation of a fraction right after each fraction.) This also means \(2\) is right after \(1\), but \(3\) is two numbers after \(2\), and \(4\) is two numbers after \(3\), and \(5\) is four numbers after \(4\)... But that's an ordering.

So how we count a set depends heavily on how we order the elements of the set. In some cases we cannot even count (real numbers); in other cases the ordering might not be obvious (rational numbers). Ivan Koswara · 3 years, 8 months ago

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@Ivan Koswara Thank you for answering my question. Sorry about not being clear before. This is a really good explanation, thank you so much! Bianca Ayling · 3 years, 8 months ago

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IS IT TRUE THAT INFINITY= -1/12?? IF YES, HOW? Shreya R · 3 years, 5 months ago

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As we all know the numbers (1 2 3 4 5 ...) are called as Natural Numbers.We have learned from our History that these number i.e. numbers 1-9 were first discovered numbers in our number system.As the 0 was discovered by Aryabhatt,the counting starts going on and on and we get to know that there is no limit of formation of numbers and thereby we named it as infinity(the one which is not finite). Before the discovery of 0 , peoples does'nt has confusion what you are having because there was nothing to be put between two numbers.As because 0 & decimal was'nt invented. So basically, we are following the first discovered pattern of number system i.e. we count as (1 2 3 4 5...)... There is no logic why we are not counting number as 1,1.00001 ....and so on.. We are just following the pattern which we have from our ancient history.... Keshav Naarayana · 3 years, 8 months ago

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This isn't really a formal answer, just my thoughts: It's true that there are infinite numbers between any two real numbers (such as 1 and 2). But in my point of view, the set of real numbers is just a way of arranging numbers in a way that we can divide them or multiply them in any way we want. This is, any "real" ratio is acceptable. So, imagine an object which is absolutely malleable. You can bend it in any way you want, at any point of the object. With the real set of numbers, you can (again) divide or multiply in any way you want, without being necessarily "aware of" every single element, but knowing they are potentially infinite. I'm speaking in these terms (comparing reals to an object, talking about potential infinity) because your question actually sort of "goes out" of the pure mathematics. You talk about counting, about reaching, about skipping... anyway, all of those actions are defined in a certain period of time. And math (I dare say) is sort of "timeless". Lets suppose, for the sake of argument, there actually WAS a way of counting every single real number from 1 to 2. That would imply two possible scenarios: That time was infinit, or that time were infinetly divisable. If that were the case, then you could actually "translate" that into a mathematical function. In the first case, the function could be, for example, \( f(t) = \frac{2t+1}{t+1}, t\in N \). In the second case, it could simply be \( f(t) = t, t\in R \). Notice that in the first case t is a positive integer, while in the second case t is real. All I meant to say is that your question directly implies a notion of "physical reallity", which is, in the end, the place where numbers were inspired from to exist. Because mathematics by themselves don't need "counting". We just need to be able to divide a number as much as we want, because (in my point of view) numbers only make sense when they're compared to each other. As 1 to 2 is the same as 0.00005 to 0.0001, if you think of them as ratios. But number 1 in itself, could be anything (in a more metaphysical way). I know that I wasn't clear, but I hope you at least kind of understood what I tried to say. Sasha Brenner Socas · 3 years, 8 months ago

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