Wikipedia:Reference desk/Archives/Mathematics/2007 December 3

= December 3 =

Does .999... hold a distinct location on the number line?
Does anyone know if .999... has its own distinct location on the number line, seperate from 1. If it does not I believe that would mean that either no repeating decimal has a distinct location or that there are odd gaps in the continuum. If yes, I believe that would mean that .999... is exactly equal to 1 in all mathematical calculations, but has an inherently different value, which I believe is the case. Can anyone confirm or clear this up for me? —Preceding unsigned comment added by Southcrossland (talk • contribs) 15:22, 3 December 2007 (UTC)


 * Not sure if this helps, but we have an article on .999... - this "denotes a real number equal to 1". I would conclude that there would be no reason to make a distinction between them on a number line, but deep math isn't my strong suit. Friday (talk) 15:36, 3 December 2007 (UTC)


 * No, 0.999... and 1 occupy the same location on the number line (unless you move the goalposts and work in a non-standard number system that allows non-zero infinitesimals). I don't follow how you reach your conclusion that "either no repeating decimal has a distinct location or that there are odd gaps in the continuum". Gandalf61 15:54, 3 December 2007 (UTC)

If multiple decimal expressions can hold the same value, than the nature of our number system as consisting of digits of set incremental value seem destroyed, yet mathematically I understand how this occurs.
 * But that ("digits of set incremental value") is not the nature of our number system, rather just a convenient way we use to represent them - and yes, it does have some features which may seem counterintuitive. -- Meni Rosenfeld (talk) 17:37, 3 December 2007 (UTC)


 * 4, 2+2, 2×2, and 0!+1!+2! all occupy the same spot in the sequence of natural numbers, because, although they are different representations of a number, they represent the same number. Likewise, 1/3 and 0.333... occupy the same spot on the real number line, because they are different representations of the same number. The intuition that different infinite sequences of digits, when interpreted as representations of a real number, necessarily represent different numbers, is wrong, and unjustified. On the contary, if 0.999... and 1 did not occupy the same spot, that would mean that there are truly odd gaps in the continuum, since there is a real number between every two distinct numbers, but there obviously can't be a number between these two. (There is just no space between them, since 1−0.999... = 0.000...; also, since 2×0.999... = 1.999... = 0.999... + 1, we see that an attempt to construct a number "halfway" between 1 and 0.999... is 1/2(1+0.999...) = 0.999... again.)  --Lambiam 19:03, 3 December 2007 (UTC)

Yes, I understand different forms of representation. However, I was refering specifically to decimal expressions. Nobody has trouble with accepting formula based equalities or fractional representations. But within a specific system where each digit is meant to have a particular value and be of a particular distance from the previous and subsequent digit, it is hard to reconcile different decimal expressions being equal. It seems to undermind the soundness of our system that any number with repeating 0's or 9's share a single value with another expression, while all other numbers do not. Although I do suspect that ultimately this is logically reconcilable. —Preceding unsigned comment added by 192.154.91.225 (talk) 19:53, 3 December 2007 (UTC)


 * Basically, the 0.999... issue is, like many things in mathematics, unintuitive at first glance, but works, and breaks a lot of other things if you try to "fix" it. For example, if you try to make them different numbers somehow, then you'll find you'll break subtraction, or division, or the ability to tell when one number is bigger than another. If you really want, consider it an equivalence relation, where you simply identify the "symbol" 0.999... with the "symbol" 1, and treat them as equal from there (in the same way, fractions form equivalence relations, such that 1/2 = 2/4 = 3/6 = ..., another way that you have different symbolic representations of the same value, even in a single representational system). Confusing Manifestation (Say hi!) 21:54, 3 December 2007 (UTC)
 * I think the 'unintuitive' character of this result has a lot to do with how real numbers and decimal expansions are taught at school, i.e. (in the UK) more or less not at all. As a result, people end up absorbing the idea that a real number is a decimal expansion, rather than is represented by one. As a result, they quite rightly have great difficulty grasping the simple idea that two different decimal expansions might happen to represent the same real number, while having no problem at all with such claims as "3/5=6/10", or even "3/3=1" which is not far from 0.999...=1. Unfortunately, I'm not yet mad enough to prescribe a proper course in real analysis at high school, so I'm not sure how the current situation can be improved. Algebraist 23:02, 3 December 2007 (UTC)

Math Help
I'm having trouble solving the following problems in my math homework:


 * Problem 1. FACTOR: (x+3)^2-9. I got to ((x+3)+3)((x+3)-3), and I know that (x+6)(x) doesn't work, so I was wondering if I was getting anywhere on this. I know the answer is x^2, but (x)(x) doesn't give me the original problem.


 * Problem 2. FACTOR: x^12-1. I got (x+1)(x-1)(x+1)(x-1)(x-1)(x^2+x+1) and I was wondering whether this made any sense.


 * Problem 3. FACTOR: (x+3)^3+27. I got ((x+3)+3)((x+3)^2+(3x+9)+9) and I was also wondering if this made any sense.

I appreciate the help and any assistance you can give me. A touch in the right direction if these answers are wrong would be helpful. I'm not asking for answers, but for help.

S♦s♦e♦b♦a♦l♦l♦o♦s ( Talk to Me  ) 21:10, 3 December 2007 (UTC)
 * On problem 1. remember that $$(x+3)^2$$ does not equal $$x^2+3^2$$. You can't distribute the exponent over the addition (only over multiplication). Your factoring is right, it's your simplification of the original problem to check your work where you went wrong.
 * For problems like number 2, I usually do the even factors first, so you get $$(x^6+1)(x^6-1)$$. Continue from there. A quick way to check for gross errors in factorings like this... count up the highest powers of x in the factors and see if you get the right total power of x to match the original problem. If you multiply your work back together you'll find you've only got $$x^7$$.
 * For problem 3, you're close but not quite. Remember that $$a^3+b^3=(a+b)(a^2-ab+b^2)$$. Do you see your mistake? Also, I'm guessing that you're supposed to simplify. It doesn't make sense to multiply out 3(x+3) but leave the rest unsimplified. Donald Hosek 21:19, 3 December 2007 (UTC)
 * I have corrected a few misprints in Donald's reply. -- Meni Rosenfeld (talk) 21:39, 3 December 2007 (UTC)
 * OK, but I'm still confused. Is the answer for problem one x(x+6)after $$(x^2+6x+9)-9$$ is simplified?
 * Is the second one then $$(x+1)(x^2-x+1)(x+1)(x^2-x+1)(x+1)(x^2-x+1)(x-1)(x^2+x+1) $$;or can I just leave it as $$(x^6+1)(x^3+1)(x^3-1)$$?
 * Is the third one then $$(x+6)(x^2+3x+9)$$ It doesn't seem to work out when I do $$(x+6)(x^2-3x+9)$$, though.
 * Thanks again for your support! S♦s♦e♦b♦a♦l♦l♦o♦s  ( Talk to Me  ) 21:48, 3 December 2007 (UTC)
 * 1 and 3 are correct. In 2, $$(x^6+1)(x^3+1)(x^3-1)$$ is correct but you should probably factor it further; you have a mistake in your factorization of $$x^6+1$$. -- Meni Rosenfeld (talk) 21:55, 3 December 2007 (UTC)
 * Wait, which one for 3?
 * I don't see the mistake in 2. S♦s♦e♦b♦a♦l♦l♦o♦s  ( Talk to Me  ) 21:56, 3 December 2007 (UTC)
 * Oh, didn't notice you gave two expressions for 3. The first one is correct, but you don't need me for that - you can just expand everything and see that it is the same.
 * As for 2, what did you get is the factorization of $$x^6+1$$? Expand it. Does it end up as it should? -- Meni Rosenfeld (talk) 22:01, 3 December 2007 (UTC)
 * I see: is it $$(x^2+1)(x^4-x^2+1)$$? S♦s♦e♦b♦a♦l♦l♦o♦s  ( Talk to Me  ) 22:11, 3 December 2007 (UTC)
 * Yes it is. $$x^4-x^2+1$$ can be further factored into $$(x^2-\sqrt{3}x+1)(x^2+\sqrt{3}x+1)$$, but since this involves nonintegers you may not have to do it. -- Meni Rosenfeld (talk) 22:47, 3 December 2007 (UTC)
 * Corrected your expression. Of course, we can factor further over the complex numbers, but that can't be what's expected (if only because it's too easy!). Algebraist 22:51, 3 December 2007 (UTC)
 * Of course these are pretty standard factorizations for Algebra I/Algebra II/College Algebra (more the latter two than the first, but given the time in the school year, I'm guessing that this is either a strong High School Algebra I course or possibly high school Algebra II). In that context, we're factoring over $$\mathbb Z[x]$$ so all those odd cases are just confusion for the original poster, as entertaining as they might be for us. Donald Hosek 23:44, 3 December 2007 (UTC)
 * But doing 2 properly is so beautiful! I suppose the OP will have to wait till university like everyone else... Algebraist 00:43, 4 December 2007 (UTC)

Checking nth power residuosity
Is there a way to check if some residue is really an nth power residue?

For example 4^x mod 31 produces [4,16,2,8,1]

To check for cubic residuocity you raise the residue to (31-1)/3 and mod it with 31 and see if this equals 1.

However in this case all of the values generated by 4 are already cubic residues.

So is there a way to check if the residues really are 3rd powers of 4 mod 31?

By that I mean could you check if they are 9th powers of something? or do some other test.

24.250.132.195 21:29, 3 December 2007 (UTC)ForgotMyLogin


 * Your question is not clear to me. Mod 31, the only 3rd power of 4 is $$4^3=2$$. -- Meni Rosenfeld (talk) 21:48, 3 December 2007 (UTC)


 * I used mod 31 because I thought this would simplify the problem. I was asking how I can check any residue generated by 4 to see if it is of the form 4^(3x) mod 31

24.250.132.195 21:57, 3 December 2007 (UTC)ForgotMyLogin
 * (ec)I don't understand your question very well. Yes, every power of 4 mod 31 is a cube mod 31, since 4 = 16^3. What do you mean by 'really are 3rd powers of 4'? Surely there is only one 3rd power of 4 mod 31, to wit 4^3 = 2? For the last part, I haven't thought about this stuff for ages, but I think the easiest way to more-or-less solve all such problems at once is to find a primitive root mod 31 (3 is one). Given this, the ninth powers mod 31 are exactly 3^9, 3^18, 3^27, 3^6, 3^15, 3^24, 3^3, 3^12, 3^21 and 3^0 (where the exponents are the multiples of 9 mod 30). Hopefully someone for whom number theory is a less distant memory will be able to give a better algorithm. Algebraist 21:59, 3 December 2007 (UTC)
 * (After reading clarification) Oh, all of {4,16,2,8,1} are of the form 4^(3x) mod 31. Algebraist 22:01, 3 December 2007 (UTC)
 * Yes that is true, but only 2 is 4^3x where the exponent is less then the order of 5. If you don't use a bound on the exponent then all the residues of 4 would be of the form 4^3x mod 31.  The larger exponents don't really count because when the exponent is modded with the order of 4 it becomes equivalent to 4^3x mod 31.  —Preceding unsigned comment added by 24.250.132.195 (talk) 22:08, 3 December 2007 (UTC)
 * Are you worried because (31-1)/3 = 10 is not divisible by 3? This just means that every third power is also a ninth power mod 31. Similarly, every number is a seventh power mod 31.  This sort of reasoning works mod p for any prime p.  Modulo prime powers p^k, as long as n is not divisible by p, a number is an n'th root mod p if and only if it is an n'th root mod p^k.  Modulo composite numbers, you just check mod each prime power.  When the power n and the modulus are not coprime, then things are a little more complicated for prime powers, but not too bad.  If you want to actually find the roots, there is even a relatively straightforward algorithm, but it does require the modulus to be factored. JackSchmidt 22:32, 3 December 2007 (UTC)
 * I am trying to find a congruence for the exponent; for example 4^x mod 31 = 2 where x would be 3z+0; 4^x mod 31 = 4 where x would be 3z+1. ect. Normally this would only be possible if the order of 4 was divisible by 3, but I am trying to figure out a way to do this if the order of 4 is not divisible by 3 but p-1 is. —Preceding unsigned comment added by 24.250.132.195 (talk) 00:26, 4 December 2007 (UTC)
 * I'm still not sure what you want. 4^x =2 mod 31 iff x=3 mod 5, and 4^x=4 mod 31 iff x=1 mod 5. Are those the kind of answers you want? Algebraist 00:40, 4 December 2007 (UTC)
 * ok 4^1 = 4 mod 31 should return nope;4^2 =16 mod 31 should return nope;4^3 =2 mod 31 should return yep;4^4 =8 mod 31 should return nope; 4^5 mod 31 =1 should return nope; 4^6 mod 31 would just mess everything up. I am just interested in residues generated by 4 during the first cycle (from exponent = 0 to exponent=4). It is assumed the the residue is always generated by the exponent being in this limited range.24.250.132.195 00:58, 4 December 2007 (UTC)ForgotMyLogin
 * After reading this 31 times I still don't understand what you are asking. It is a true statement that 41 ≡ 4 (mod 31), so why should this "return nope"? What does it mean that a statement returns nope? How, and isasmuch as what, would 46 mod 31 "mess everything up"? Is there a relationship between the example here with nope/nope/yep/nope/nope/mess and the initial powers mod 31 [4,16,2,8,1]? Generalizing the definition of quadratic residue, I could imagine the question: Given p, r, and n, does there exist a number x such that xp ≡ r (mod n). However, I don't recognize this in your question. Can you formulate the problem in the form: "Given a, b, c, ..., are there x, y, z, ... such that  holds?". --Lambiam 19:50, 4 December 2007 (UTC)

I don't know why I have to keep repeating myself. For an equation A^X mod P; I want to know the divisibility of the X, in this case by 3. Is X divisible by 3 when order of A is ((p-1)/3) for some residue generated by A? what is so hard to understand? To make to problem unambiguous I limit the exponent to be at most the order of A.128.227.194.167 (talk) 22:21, 4 December 2007 (UTC)ForgotMyLogin
 * I'm sorry to say this, but we are repeating ourselves in our requests for clarification because the way you express the problem is not abundantly clear. You want to know the divisibility of X by 3. I got that. But what is given? A^X mod P is not an equation. What do you mean in this context by the order of A? The order in the multiplicative group mod P? Is it given that P is a prime number? And why "for some residue"? What do you mean by "some residue generated by A"? Do you mean the remainder of A on division by P? Again, if you could take the time to reformulate the question, making explicit (1) which variables are given (and which are not); (2) for which variables a solution is asked; (3) what the constraints are relating the given (input) variables and the output variables; that will in the end save you more time than it costs, and it will save all of us frustration. --Lambiam 02:02, 5 December 2007 (UTC)


 * Have you taken a look at Quadratic reciprocity and Cubic reciprocity? I think they're along the right lines. Unfortunately, they're focused on solvability, not the solution itself, and would say "yes, x has a square root mod p" but not "the square root of x mod p is y". Black Carrot (talk) 04:02, 5 December 2007 (UTC)


 * There's also for the fourth-power case. The general case seems to be Artin reciprocity, but our article is both dense and a stub, so not much use.  looks a bit more detailed, but I haven't read it yet. Black Carrot (talk) 04:47, 5 December 2007 (UTC)