
#1
Feb2209, 03:01 PM

P: 10

why do irrational numbers exist? I am well familiar with the proof that irrational numbers exist, but why do they?




#2
Feb2209, 03:13 PM

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Are you using the word "why" in some unusual way? If so, you really should have said that up front.... 



#3
Feb2209, 03:38 PM

P: 122

Well, your question does seem odd, but my guess is that you want to ask a philosophical question.
Let me ask you a question. Do rational numbers exist? how do you know this? 



#4
Feb2209, 04:42 PM

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Irrational Numbers
For me, the intuitive answer is "because there aren't nearly enough rationals to 'fill in' all the gaps".




#5
Feb2209, 04:48 PM

P: 122

Well, I don't understand why people think rational numbers exist and some numbers don't. It's just easier to think that all numbers are mathematical constructs and real numbers are simply, yes, way to fill in gaps between rational numbers.
I wonder what the op thinks of complex numbers. 



#6
Feb2209, 05:06 PM

P: 288

Oh jeez, I'll try to tread softly in this thread.
It's actually a very interesting question the OP is getting at, and one I've often thought about myself. How much information do we need to have about a number before we can consider the number to be welldefined? Are all numbers which provably exist welldefined under our definitions of welldefinedness of a number? Is there a definition of the welldefinedness of numbers? One can also talk about whether numbers are computable or not. It's interesting that the real numbers are most incomputable... what does this mean? What can even be meant by incomputable number? I think it's an interesting discussion. To the OP: do you think that sqrt(2) exists, and in what sense do you think this? I mean, we both know that there is a proof that it is not rational. It's a relatively tame irrational number. Why do you feel the way you do? I could enjoy this conversation. 



#7
Feb2209, 05:34 PM

P: 10

no, i don't think sqrt(2) exists. This is my reason: sqrt(2) is just a symbol for it's decimal representation which is 1.414213562..., and the decimal places continue on infinitely.
So, if we will never reach the last digit in the decimal places for sqrt(2), how can we multiply it by itself. It's the same logic that goes with the fact that we can't multiply any number by infinity. For example, 0 x infinity is an ideterminate form, because although we know logically that you will get 0 if you keep multiplying 0s, we will never finish multiplying 0 infinitely many times so we say that it is undefined. In other words, sqrt(2) by definition is a number that you multiply by itself in order to get 2. However, we will never be able to get that number so it should be undefined for the same reason that infinity is undefined. 



#8
Feb2209, 05:50 PM

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What about this: I define "foo" as an ordered pair (a, b) where (a, b) = (c, d) iff (a  c)(b  d) = 0 and the operations plus and times are defined by (a, b) + (c, d) = (a + c, b + d) and (a, b) * (c, d) = (ac + 2bd, bc + ad). Do "foo"s exist? How about "bar"s, where "bar" is an ordered pair (a, b) where (a, b) = (c, d) iff ad = bc and the operations plus and times are defined by (a, b) + (c, d) = (ad + bc, bd) and (a, b) * (c, d) = (ac, bd)? Do "bar"s exist?\ Maybe "baz", where a "baz" is (a) where (a) = (b) iff a  b = 0 and the operations plus and times are defined by (a) + (b) = (a + b) and (a) * (b) = (ab). Do "baz"s exist? 



#9
Feb2209, 06:16 PM

P: 122

what about sqrt(2) km? Does that exist? 



#10
Feb2209, 06:47 PM

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It might be good to point out that while asking if a number is "welldefined" it makes no sense to focus entirely on the decimal representation of the number, as timjones007 does in #7. The decimal representation of a number is just that a representation and has little to do with the properties of the number itself.




#11
Feb2209, 07:33 PM

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#12
Feb2209, 10:25 PM

P: 288

"no, i don't think sqrt(2) exists. This is my reason: sqrt(2) is just a symbol for it's decimal representation which is 1.414213562..., and the decimal places continue on infinitely. "
So you take the definition of a number as its decimal representation? This would take a little elaboration to take into account the (very valid) objection raised by CRGreathouse. For instance, you could say that a number is well defined if its decimal representation repeats with a string of digits of finite length L for all places N at least N_0 to the right of the decimal. This covers repeating decimals (1/9 = 0.1111... letting N_0 = 1 and the string of digits being "1", and 1/10 = 0.1000... is covered letting N_0 = 2 and the string of digits being "0", etc.) Obviously, the choice of 10 as the base doesn't make any difference... you could allow this to vary as well. But the real problem with that is that you're taking the properties of rational numbers and saying that's what makes a number welldefined. Does that make sense? I mean, if we are trying to show that irrational numbers are not well defined, it's a little selfserving to equate a property of rational numbers with welldefinedness. Savvy? "So, if we will never reach the last digit in the decimal places for sqrt(2), how can we multiply it by itself. " Well, the problem with this is that, as CRG said, 1/10 = 0.100... and this technically also goes on forever... perhaps a better way of saying what you're thinking is that you have a finite set of rules with which you can always generate the next digit in the decimal (or some other sensible) representation. For instance, 1/10 is welldefined because I can say "tenths' place 1, all other places 0" and you can use the two rules to write out the number to any desired number of digits. Does this sound alright, tim? The only snag with that, of course, is that sqrt(2) is also well defined by this definition of welldefinedness. Consider this: sqrt(2) can be found as follows: sqrt(n):: x := 0 p := n // could be made more efficient, but who cares? for p = n to p_min begin while x <= n begin x = x + 10^p end x = x  10^p end Let's see this operating on n = 2. x = 0. x = 100, p = 2. x = 0, p = 2. x = 10, p = 1. x = 0, p = 1. x = 1, p = 0. x = 2, p = 0. x = 1, p = 0. x = 1.1, p = 1 x = 1.2, p = 1 x = 1.3, p = 1 x = 1.4, p = 1 x = 1.5, p = 1 x = 1.4, p = 1 x = 1.41, p = 2 etc. As you can see, this will always allow you to find the nth decimal digit in a finite number of steps... so you would need a stricter definition than the one I provided to exclude sqrt(2). 



#13
Feb2209, 10:27 PM

P: 288

What does everybody else think about what it takes to define a number? Do numbers have to have a value? If so, and you know a number exists for which you cannot possibly find its value... does this mean anything?




#14
Feb2209, 11:56 PM

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(The properties don't have to be complete  though the definitions for common number systems like the integers or the reals are complete in the appropriate sense) *: Or type or class or language or whatever foundational gadget you want to use. Once you actually have an actual, concrete list of properties to work with, you can usually answer simple questions relatively easily. e.g. it's fairly straightforward to show that * in the rational numbers, 2 doesn't have a square root. (what would its factorization be?) * in the real numbers, 2 does have a square root. (construct it as the least upper bound) * for fields the question is undecidable  some fields do and some fields don't have a square root of 2. (as shown by the previous two examples) 



#15
Feb2309, 04:46 AM

P: 41

You are right! sqrt(2) does not exist. And in fact also 1 does not exist. 1 is the multiplication of 13/7 and 7/13 now 13/7 and 7/13 are just symbols for their decimal representations which are 13/7 = 1,85714285...... 7/13 = 0,53846153...... and the decimal places continue on infinitely. So, if we will never reach the last digit in the decimal places for these two numbers, how can we multiply them together??? In other words, 1 is a number that you get multiplying 13/7 by 7/13. However, we will never be able to get that number so it should be undefined for the same reason that infinity is undefined. 



#16
Feb2309, 06:05 AM

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This is an interesting question and I think it is one that has been discussed since ancient times.According to David Wells (the penguin dictionary of curious and interesting numbers) pi is the only irrational and transcedental number that occurs naturally.People here have been using root 2 as an example and I have been trying to think of an example where this number can be given a unit.Suppose we were told that a square had an area of root 2 metres squared.Does this mean anything when such a square cannot be consructed or have I picked on a dopey example?




#17
Feb2309, 07:15 AM

P: 288

Dadface:
That could be a dopey example. What about the distance between opposite corners of a square of area 1? Hurkyl: I see what you're saying, but I think the problem we're all having is in communicating. I agree that you're absolutely right about numbers... a very clear and thoughtful exposition. However, I think that the OP means to talk about the value of numbers, not their properties... to know what the number is, not whether it is there or not. I mean, 2 *is* an integer, but how big is 2? We can get to 2 using a finite number of logical steps. Is sqrt(2) a real number? The OP didn't think so, but perhaps after my last post he will agree that sqrt(2) must exist as well... since we can get as close as we like to it on a whim. But in what sense do the numbers which we cannot find values for have these values  even if we know the number must exist? I apologize that the discussion is a little vague. I'd love to give you an example of such a number, but obviously I can't... I don't know, maybe the reason this topic isn't more mainstream is that it's a rabbit hole, makes no sense, and has no good answer. 


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