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September 16[edit]

Constructible numbers[edit]

The Wikipedia article on constructible numbers states that :

If z is constructible, then it is algebraic, and its minimal irreducible polynomial has degree and power of 2, or equivalently, the field extension Q(z)/Q has dimension and power of 2. One should note that it is true, (but not obvious to show) that the converse is false — this is not a sufficient condition for constructibility. However, this defect can be remedied by considering the normal closure of Q(z)/Q.

Can anyone tell me how to construct such a number ? Even without an explicit construction, a Galois theoretic explanation of what type of polynomial this number should be a root of.

My understanding is that you need to find a number α such that [Q(α)/Q]=2^k but that, if E is the normal closure of Q(α), then [E/Q]!=2^k for any k. How is that possible ?

I thought that, as [Q(α)/Q]=2^k, there must be an irreducible polynomial P(x) in Q[x] of degree 2^k, of which α is a root. P(x) can't split over Q[x]/(P(x)), otherwise the extension would be normal. But all roots of P(x) must be of degree 2^k so I don't see how we could get a normal closure with degree over Q which isn't a power of two.

So, where does such a non-constructible number α with [Q(α)/Q]=2^k come from ?

Thanks. --Xedi^Talk 05:05, 16 September 2008 (UTC)[reply]

I believe the article is just saying that having the highest power of your polynomial a power of two isn't sufficient, for instance x in x⁴+x+1=0. However if you extend the rationals with the roots of this equation you'll get cube roots as well coming up. Dmcq (talk) 08:43, 16 September 2008 (UTC)[reply]

By the way if you like this then you might like Mathematics of paper folding where trisection of angles can be done. Dmcq (talk) 08:46, 16 September 2008 (UTC)[reply]

Well, the thing is, x⁴+x+1 is an irreducible polynomial of degree 4, so all its roots must be of degree 4 over Q, no ? And then, taking normal closure is not going to affect this degree, so I don't see how "the defect can be remedied" that way. I did have the impression something could go wrong if cube roots were involved somewhere, but I don't see how taking normal closure remedies this. --Xedi^Talk 16:10, 16 September 2008 (UTC)[reply]

In general, if p is an irreducible degree n polynomial over Q, then any root of p generates a n-dimensional extension. The normal closure of any of these extensions is the full splitting field of p. The degree of this splitting field is divisible by n and divides n!. In this case, the degree of the splitting field is 24. Algebraist 16:27, 16 September 2008 (UTC)[reply]

So, at which point does the factor of 3 in 24 come in then ? And how do you know that in this case the degree of the splitting field is 24 (Other than just giving the order of the Galois group of the extension) ? --Xedi^Talk 16:50, 16 September 2008 (UTC)[reply]

I calculated it using standard tricks for computing Galois groups using reduction mod m. The factor of 3 comes in because when we adjoin a single root of the polynomial, this splits the poly into a product of a linear poly and an irreducible cubic, and we have to adjoin a root of that cubic to make the poly split further. For a simpler example, consider x³-2. This is irreducible, so we adjoin a root a. The polynomial then splits as (x-a)(x²+ax+a²). The second factor is irreducible; to split it, we need to adjoin a square root of -3. Thus the splitting field has degree 6. Algebraist 17:15, 16 September 2008 (UTC)[reply]

Ah yes ok. Thanks a lot ! --Xedi^Talk 17:21, 16 September 2008 (UTC)[reply]

Baffling Proportions Problem[edit]

The other day I was helping my younger brother with a maths problem. His teacher had given him the problem in a quiz. "Working 8 hours a day, 24 men do a certain task in 15 days. Find how many days it would take 16 men working 9 hours a day to do the same task"

It is a inverse proportion problem but there are 5 pieces of information here. Usually these kind of problems involve 3 pieces of information. This problem has got me very confused and I am feeling pretty embarrassed. Would be glad if I could get some help here! Thank you! —Preceding unsigned comment added by 203.81.220.213 (talk) 16:33, 16 September 2008 (UTC)[reply]

Try breaking it all down into man-hours (the amount of work done by one man in one hour). Use the first scenario to work out how many man-hours the task requires, and then use that to work out the time required for the second scenario. --Tango (talk) 17:23, 16 September 2008 (UTC)[reply]

The 5 pieces of information fall into one base piece (the given time to do the task, i.e. 15 days), and two pairs of adjustment pieces (the number of men, and how many hours they work each day), where each pair forms a fraction to multiply the base piece. So the fraction for men is either 24/16 or 16/24, and for hours either 8/9 or 9/8. Which fraction to use in each case will come by considering the direction of change to the base figure resulting from (a) fewer men (b) more hours per day. When that has been decided, the answer will come by multiplying the original time by the two fractions in turn. This approach lends itself to there being more than two factors in the problem, say a rate of efficiency as well. If the 24 men worked at 80% and the 16 men at 90%, there would just be a third fraction (80/90 or 90/80, to be decided) in the multiplicative chain.—86.148.186.156 (talk) 22:15, 16 September 2008 (UTC)[reply]

Random trivia: hapless schoolboys once solved these problems using the "double rule of three," immortalized in Lewis Carroll's Mad Gardener's Song:

He thought he saw a Garden-Door that opened with a key:

He looked again, and found it was a Double Rule of Three:

‘And all its mystery,’ he said, ‘Is clear as day to me!’

The "mystery" was a reference to the fact that this rule was famously difficult to teach. I don't recommend teaching it to your brother—far better to teach general problem-solving techniques and real understanding of the problem. But here's an explanation of the rule from an 1830 textbook. It's interesting to note that the word problem they chose as their example can't correctly be solved this way—if $100 gains $6 in one year, it's not likely to gain $6n in n years. I wonder if any of the students ever noticed that. For some real soul-killing awfulness, check out this page. "Three sailors having been aboard 9 months 1/4, received 40ℓ 3/15; I demand how much 100 sailors must receive for 28 months 3/7 service? Answer 4118ℓ 6s 0d 1qrs 1305/11655." -- BenRG (talk) 23:13, 16 September 2008 (UTC)[reply]

Is there really any 'problem' in this?

Input data: "Working 8 hours a day, 24 men do a certain task in 15 days."

Twenty-four men do the work, so each of them does a 24-th part of the work in 15 days. He works 8 hours per day, so he completes his 1/24 of work in 15×8=120 hours — that means he does a 1/120 part of his work per hour, so one worker does 1/120×1/24 = 1/2880 of the task per hour.

Question: "Find how many days it would take 16 men working 9 hours a day to do the same task"

Each man, working 9 hr/day, does 9×1/2880 = 1/320 of the task per day. Sixteen workes do 16×1/320 = 1/20 of the taks a day. So they need 20 days to complete the work. Is anything obscure here? --CiaPan (talk) 07:19, 17 September 2008 (UTC)[reply]

This is a problem which requires more logical thinking than calculations. The answer from my calculations is: 23 days, 7 hours and 30 minutes. Try to see whether you get the same answer. Topology Expert (talk) 12:56, 17 September 2008 (UTC)[reply]

Nope. But this way is non-baffling (as outlined by Tango above):

8 hours/day $\times$ 15 days $\times$ 24 men = 2880 man-hours of work to be done (calculated by doing it the original way).
9 hours/day $\times$ 16 men = 144 man-hours/day available (by doing it the new way)
2880 man-hours (of work to be done) $\div$ 144 man-hours/day (by doing it the new way) = 20 days. -hydnjo talk 13:00, 17 September 2008 (UTC)[reply]

Huh? 'but this is non-baffling'...??? I apologize then, must have missed something important. I understand 'problems' are sometimes given in a complex, sophisticated form, but thought the solution should always be made as simple as possible, not baffling. At least that's what I remember from schools. --CiaPan (talk) 14:03, 17 September 2008 (UTC)[reply]

Yes, and Hydnjo's solution (which is the one I recommended at the top of this section, although I didn't actually give the answer) is simpler than yours. At least, it is to me. --Tango (talk) 17:27, 17 September 2008 (UTC)[reply]

Yeah, I spelt it out because no one seemed to be picking up on your simple explanation. -hydnjo talk 23:17, 17 September 2008 (UTC)[reply]