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[chriscase]

So you are using a relational DB and you want to associate multiple names to a given identity. What is the requirement to store multiple names coming from? Why not just pick one?

[edse]

My mother for example have had three different last names; a maiden name, a name from first marriage and a name from second marriage. It's useful to store all three, since if one want to find out more about her childhood and ancestry, her maiden name is needed. Some information can be find if one searches on her name from the first marriage. Her third name is what will be carried on and is also the name that most know her as.

I could pick one, that is true, but I feel like there is a lot of interesting data that goes missing. I don't know.

[chriscase]

So wait, is this for geneology?

[edse]

Yes, I did say it somewhere

[chriscase]

Ok, sorry for the confusion.

So the basic methodology would be first to look for a published standard information model. I believe GEDCOM is the de facto standard. Then you'd investigate whether there are any reference relational models that implement the standard, or, if you have a mind to try your hand at it, try to write your own. I usually split the difference by trying out a few reference implementations and then modify or roll my own.

[Finlander]

Does anyone here know how to solve this and put it in a simple form?

Edit. Is it just x * y^2? Just played a bit with the numbers.

[edse]

I believe you got it just a little bit wrong. I begin from the x in the middle.
(x^-8/3)³=x^(-8/3)*3=x^-8 so you have x^-8y⁸ under the root. The fourth root(?) of this is x^-8*(1/4)y^8*(1/4)=x^-2y². Now x²*x^-2*y²=x^2+(-2)*y²=y²

[Finlander]

Thanks edge! However, I am a bit confused right now here: shouldn't x^2+(-2) be x⁰= 1x?

[polymorphic]

Thanks edge! However, I am a bit confused right now here: shouldn't x^2+(-2) be x⁰= 1x?

No x⁰=1 even if x=o

[Finlander]

Alright I see it now, thanks!

[Nikitn]

Thanks edge! However, I am a bit confused right now here: shouldn't x^2+(-2) be x⁰= 1x?

No. The exponential of x defines the order of x. So:
1x = x = x¹ ≠ x⁰ = 1.

[Town Watch]

I have math problem I was mulling over.


a) calculate in long division 1/17 until you receive the repetend. Even though the repeating decimal doesn't appear immediately, why does long division eventually start to repeat itself?

b) why is the decimal form of every fraction m/n, either zero repetend (divided exactly), or at least repeating decimal. (repeating decimal, the "cycle" such as one third in decimal form. 0.33333, the repeating decimal is 33, or simply 3, or something like that... the decimal repeats itself forever)

Essentially construct a proof, which shows the b) part, to be true. Or, convince the author somehow else!


-probably it has to be assumed that, m and n belong to natural numbers(?). Something like this is implied, but it was not strictly given in this problem. It is implied, because of the earlier problem a), we were calculating in long division with pen and paper. Can you calculate in long division with negative numbers? You expand the range from natural numbers to integers? (all whole numbers, including negative, except zero obviously)

-I did calculate the long division of 1/17, and I am able to calculate in long division, but the question seems really challenging... Maybe my theoretical mastery of long division was always lacking!

-obviously in long division, when the remainder is zero, then in that case the whole procedure ends. The obvious conclusion of such cases where remainder is zero, then it means that there is zero repetend.

-it seems that when the range includes integers, from this directly follows, that m/n cannot be irrational number.

-When m/n cannot be irrational number, then it seems to logically clear, that it is impossible for the m/n, to have a decimal form which is unending, and random (without repeating decimal, without cycle). Maybe this is too circular for a proof?

-From there, can be concluded, that m/n must have some form of repeating decimal.

-obviously, n is unequal to 0, otherwise it's undefined.

- How should the proof be constructed in formal terms?

[chriscase]

I think the proposition needs to be stated in formal terms first. Perhaps a proof of the converse (every repeating decimal is rational) would be instructive -

http://math.stackexchange.com/questi...al-is-rational

It may also be simpler to treat "terminating" decimals simply as repeating decimals whose repeating place value is zero.

Looks like Wikipedia has a less formal proof - http://en.m.wikipedia.org/wiki/Repeating_decimal

[Town Watch]

I think the proposition needs to be stated in formal terms first. Perhaps a proof of the converse (every repeating decimal is rational) would be instructive -

http://math.stackexchange.com/questi...al-is-rational

It may also be simpler to treat "terminating" decimals simply as repeating decimals whose repeating place value is zero.

Looks like Wikipedia has a less formal proof - http://en.m.wikipedia.org/wiki/Repeating_decimal

Does Long Division algorithm (like the anglo-american long division algorithm, with the "division corner") only apply for, domain of integers? In other words, numerator and denominator belong to domain of integers? I've never even been taught: why long division even works, at all. Presumably the algorithm must satisfy some conditions to be valid, for dividing numbers.

The algorithms that I've used, are probably long division, and a particular algorithm for solving rubik's cube! I even memorized that rubik's cube algorithm once! I think it's really funny and also really cool, that some one invented that algorithm for rubik's cube for sure! It's not the most effective or fastest method, but it seems to work as long as you follow the algorithm.

The book answer hints at using the long division, for asserting the statement to be true. Well, the proof, or convincing argument, should somehow answer the question: why does the division result in recurring decimal, or divide evenly, in other words, either repeating decimal, or zero repetend.

We can use the terms division, and fraction interchangably, right? They are the same operation. Assuming that m/n in which both m and n, belong to domain of integers (or rational numbers?)

I suppose, one part of the proof would have to state formally, that you cannot have irrational numbers in this long division thing. The statement about recurring decimal or zero repetend seems to hold true for rational numbers, though.

Assuming m/n, where both m, n belong to integers (or rational numbers). It follows that irrational numbers cannot result from the division, according to some axiom, I forget which one. This accounts, for the purposes of "proof", dismissing all cases, where the repeating decimal has no recurring cycle, only an infinitely continuing decimal, but not a recurring one...(?)

The question of, why repeating decimals are rational numbers, seems to be more complicated. According to some math professor (on mathforum), it's about geometric series being used to prove this fact.

[John Doe]

Assuming m/n, where both m, n belong to integers.

When you divide integers n by m, n=q*m+r where r is the leftover: you know its value is 0 or any integer strictly below m, basically r belong to {0,1,2,...,m-1} a finite set of numbers. If the leftover isn't 0, then you multiply the leftover by 10 and try to divide it again by m (to get the next decimal of you result), and you repeat the process until either the leftover is 0 or of the same value of a previous leftover (start of the cycle). So at the worst case scenario, if you havent got a leftover equal to zero or equal to a previous leftover at the (m-1)^th division, you will at the m^th division because you have exhausted all other possible result.

I did not demonstrate that m/n is either zero repetend or repeating decimal, I demonstrated it can't be otherwise (it's obviously not very rigourous but it will do).

[chriscase]

Informally, as soon as the remainder in a given long division step is the same as the remainder in a previous step, the long division algorithm determines that the sequence of remainders will repeat. Given that there are a finite number of possible remainders, it is inevitable that the remainders will have repeated at a certain point.

[Town Watch]

When you divide integers n by m, n=q*m+r where r is the leftover: you know its value is 0 or any integer strictly below m, basically r belong to {0,1,2,...,m-1} a finite set of numbers. If the leftover isn't 0, then you multiply the leftover by 10 and try to divide it again by m (to get the next decimal of you result), and you repeat the process until either the leftover is 0 or of the same value of a previous leftover (start of the cycle). So at the worst case scenario, if you havent got a leftover equal to zero or equal to a previous leftover at the (m-1)^th division, you will at the m^th division because you have exhausted all other possible result.

I did not demonstrate that m/n is either zero repetend or repeating decimal, I demonstrated it can't be otherwise (it's obviously not very rigourous but it will do).

You mean the value of the remainder, in the underlined, you seem to be indicating that way?

assuming n/m, where n and m are integers, m is unequal to 0 (you accidentally switched the numbers around, compared to original example, but let's continue with n/m)

Why must the remainder be smaller than the denominator?

Assuming you calculate in long division with pen and paper. Correct me if I'm wrong, but the remainder is actually always, this specific number, which is the result of subtraction procedure?

When you calculate in long division, 5/4

You calculate how many times does 4 fit into 5. Result is one times. Then you calculate one times four. Then you subtract 4 from 5. The result of the subtraction is the remainder of division, one remains left over?

[Town Watch]

When you divide integers n by m, n=q*m+r where r is the leftover: you know its value is 0 or any integer strictly below m, basically r belong to {0,1,2,...,m-1} a finite set of numbers. If the leftover isn't 0, then you multiply the leftover by 10 and try to divide it again by m (to get the next decimal of you result), and you repeat the process until either the leftover is 0 or of the same value of a previous leftover (start of the cycle). So at the worst case scenario, if you havent got a leftover equal to zero or equal to a previous leftover at the (m-1)^th division, you will at the m^th division because you have exhausted all other possible result.

I did not demonstrate that m/n is either zero repetend or repeating decimal, I demonstrated it can't be otherwise (it's obviously not very rigourous but it will do).

What if, n/m in which, n, m both belong to rational numbers? m is unequal to 0. (this means that either n, or m, can be fractions themselves, such as 1/2 divided by 1/3 for example)

How could you prove, or convince the reader, of repeating decimals or zero repetend?

The question never claimed that n,m both belong to integers. (This had to be assumed by the reader, as such this was a bit of a move by the author)

[John Doe]

5/4 has a left over of one, but any integer divide by four will give a leftover of 0,1,2 and 3. I'm french, so we might call leftover what the anglosaxon call the carry.

if n an m are rational, then themself could be turn into a fraction of integers.

[Town Watch]

5/4 has a left over of one, but any integer divide by four will give a leftover of 0,1,2 and 3. I'm french, so we might call leftover what the anglosaxon call the carry.

if n an m are rational, then themself could be turn into a fraction of integers.

In Finnish, remainder is called "jakojäännös", I think. "Leftover /after/of the/ division" or something like that. The word itself doesn't mean anything else in the Finnish language, I think. In wikipedia it is called modulo. I think the Finnish word was invented for that specific purpose.