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September 13

Adding up numbers with margins of error

How would I solve this expression? Sorry, not sure how to work with LaTeX yet.

(b ± Δ b) + (k ± Δ k) - (r ± Δ r)= b + k - r ± ????

Thanks alot
Señor Purple 01:00, 13 September 2007 (UTC)[reply]

± (Δ b + Δ k + Δ r). Geesh, I sure hope this isn't homework. - hydnjo talk 01:09, 13 September 2007 (UTC)[reply]

I believe this should be ± (Δ b + Δ k - Δ r). Tesseran 02:16, 13 September 2007 (UTC)[reply]

It depends on what is meant by ±. If it means that they are either all pluses or all minuses, Tesseran is correct. If (b ± Δ b) means any number between b - Δ b and b + Δ b, hydnjo is correct. If it means there is a certain probability that the latter is correct i.e. Δ b is the margin of error (which is what you said in the title), they're both incorrect. It's less than what hydnjo said, but the amount depends on what kind of distribution it is. I'd tell you what it is with normal distribution, but I don't know. — Daniel 02:25, 13 September 2007 (UTC)[reply]

For independent normal distributions it's

\pm {\sqrt {\Delta b^{2}+\Delta k^{2}+\Delta r^{2}}}

. -- BenRG 02:35, 13 September 2007 (UTC)[reply]

Yes, that's the answer I was going to give. If they're measurement uncertainties (and you have no particular reason to assume the distribution is "weird", i.e., not Gaussian), then they "add in quadrature". The square of the total uncertainty is equal to the sum of the squares of the individual uncertainties. —Keenan Pepper 20:39, 13 September 2007 (UTC)[reply]

Thanks Daniel -- I wasn't thinking. My interpretation is almost certainly wrong, given the mention of "margins of error". Tesseran 00:27, 14 September 2007 (UTC)[reply]

sudoku help

Could someone help me w/this sudoku I can't get any more numbers and I've spent over 15 mins on it!

File:Stuckonsudoku.PNG

Portal111 01:10, 13 September 2007 (UTC)[reply]

Start guessing. 202.168.50.40 01:11, 13 September 2007 (UTC)[reply]

I shouldn't have to guess on a sudoku, every move should be logical.--Portal111 01:13, 13 September 2007 (UTC)[reply]

A case could be made that that assertion is equivalent to P=NP. —Blotwell 23:44, 14 September 2007 (UTC)[reply]

The questioner means: a properly constructed sudoku problem is such that a competent solver can, at each stage of solution, using deductive reasoning, find the assignment for some empty cell, so that the whole puzzle can be solved by repeating this, without need to resort to backtracking. This requirement is then part of the definition of "properly constructed". As a criterium it is not sharp: one solver's guess-and-check is another solver's deductive step. After all, from a (purely!) theoretical point of view, sudoku puzzles can be solved by precomputing a table indexed by the solvable puzzles and giving the complete solution by an O(1) lookup step. --Lambiam 00:37, 15 September 2007 (UTC)[reply]

Unbelievable, 15 whole minutes!!! It's perfectly fine to guess and then to eliminate that guess as a result of your subsequent options. A guy named Thomas Edison spoke of this strategy. - hydnjo talk 01:15, 13 September 2007 (UTC)[reply]

I usually do them in 6 mins.--Portal111 01:17, 13 September 2007 (UTC)[reply]

For a starter, provided the other numbers are correct, you must have a 4 in the top left corner of the top-middle box. i.e. a 4 in the 4th column and top row. Look at the other boxes in the 4th column to see why you can't have a 4 in any of these. You'll probably find the rest of the numbers will slot into place now. Richard B 01:41, 13 September 2007 (UTC)[reply]

That's logical. - hydnjo talk 01:47, 13 September 2007 (UTC)[reply]

Thanks Richard B that one # let me solve it. Odd that we have the same first name and first letter in our last name too.. —Preceding unsigned comment added by Portal111 (talk • contribs) 02:14, 13 September 2007 (UTC)[reply]

Prove identity law

If $ax=a\,$ for some number where $a\,$ is nonzero then $x=1\,$

How would I go about proving this? It's *so* fundamental I can't think where to start.

Depends on how you start. If you're given a mathematical system that contains the identity, such as the real numbers or integers, it's usually already an axiom, and there's no need to prove it because it's already true. If you're trying to show that some mathematical system contains an identity, it's slightly harder. What is this for? I can help you better if you explain what you're doing this for. Gscshoyru 15:46, 13 September 2007 (UTC)[reply]

I don't think this particular statement is commonly an axiom (though it's quite close). More likely axioms are that 1 is a multiplicative identity, that if

a\neq 0

then there is a number

a^{-1}

such that

a^{-1}a=1

, and that multiplication is associative. If you have all of those, then

1=a^{-1}a=a^{-1}(ax)=(a^{-1}a)x=1x=x\;\!

. -- Meni Rosenfeld (talk) 15:51, 13 September 2007 (UTC)[reply]

Oh -- I read the title and didn't read the equation. Oops. Meni's proof works as long as your system has those axioms, and most do. Gscshoyru 15:54, 13 September 2007 (UTC)[reply]

It needs to be proved because it is not always true - it depends on certain properties of the algebraic structure in which you are working. In the ring of integers modulo 4, 2x3=2, but 3≠1. This is because 2 has no multiplicative inverse in this ring. Gandalf61 16:04, 13 September 2007 (UTC)[reply]

(edit conflict) The nice thing about this question is that it connects with some serious mathematics. The difficult thing is … the same. :-)

We must study the details of the request. Two important questions are: (1) what is "some number", and (2) is x = 1 a unique solution?

Multiplication is evidently required, but that means we may be talking about natural numbers (0,1,2,…), integers (…,−2,−1,0,1,2,…), rational numbers (fractions), real numbers (like √2 and π), or something else. For natural numbers and integers the equation 2x = 2 cannot be solved by multiplying both sides by ¹⁄₂, because it is not available among our numbers; yet we can consult a definition for multiplication and prove that x = 1 is a solution. Do we know that some other value for x is not lurking undiscovered; can we prove it? This is more awkward, but we can do it.

If we have multiplicative inverses (fractions, reciprocals), we can follow the proposed strategy of multiplying both sides by ¹⁄_a. Again we can prove uniqueness, but our method will probably be different.

Number theory and abstract algebra study examples where our arithmetic often requires careful attention to detail. We encounter kinds of "multiplication" where xy = 0 can occur even when neither x nor y is 0. Or we might have xy = −yx, so that the order of multiplication matters. And so on. This teaches us to be very careful about the details of our assumptions, of what properties we use for our proofs.

Uniqueness is something we cannot take for granted. Consider x²+1 = 0. Using real numbers we have no solutions; using complex numbers we have two solutions; and using quaternions we have an infinite number of solutions.

So mathematicians have learned to ask questions that may at first seem peculiar.

Does a solution exist?
Is the solution unique?
Can we find a solution?

The last question is perhaps the most unsettling, because sometimes we can prove that solutions exists, yet be forever denied any way to exhibit one. --KSmrq^T 17:03, 13 September 2007 (UTC)[reply]

This looks like a chapter one section one abstract algebra question to me. Meni's approach is the way to go in that case. We're probably assuming a group. Donald Hosek 18:40, 13 September 2007 (UTC)[reply]

I doubt the OP had abstract algebra in mind. He is probably just trying to understand better things he has long known about the real numbers. -- Meni Rosenfeld (talk) 18:49, 13 September 2007 (UTC)[reply]

Sorry for the late reply. I forgot to mention the universe of discourse for all variables is

\mathbb {R}

. This isn't an intentional study into abstract algebra but obviously questions of this sort demand it. I was just starting to think about proofs on a purely personal, recreational basis and I thought something this simple might be easy to prove. Thanks Meni.

At this point, I think you should be asking yourself less "how do I prove this property of the real numbers?" and more "what are the real numbers, anyway?". There are many approaches for defining the real numbers, and articles like Real number and Construction of real numbers highlight some of them. Each approach comes with its own set of proofs for their most basic properties. One you have those basic properties, you can essentially "forget" what the real numbers are and how did you construct them, and continue to explore based solely on those basic properties. The 3 "axioms" I mentioned are an example; once you have established those for your favored construction, you can move on to using them, in an argument like I presented above, to prove more "advanced" statements like the one in your question.

Of course, the ordered field of real numbers is just one algebraic structure, and you will, no doubt, also want to explore others, like those suggested by KSmrq and Gandalf. -- Meni Rosenfeld (talk) 21:22, 13 September 2007 (UTC)[reply]

Working with real numbers, the most important facts we need are that 1 is an identity (two-sided, in fact) for multiplication, and that every nonzero real has a unique multiplicative inverse (also two-sided). Then we deduce

{\begin{aligned}ax&=a&\qquad &({\text{given}})\\a^{-1}(ax)&=a^{-1}a&&(a^{-1}{\text{ exists, multiplication preserves equality}})\\(a^{-1}a)x&=a^{-1}a&&({\text{multiplication is associative}})\\1x&=1&&({\text{inverse property}})\\x&=1&&({\text{identity property}})\end{aligned}}

By carefully observing the properties we used, we find that the same proof applies without change to other algebras. For example, it works with rational numbers, and with complex numbers, and with integers modulo 17 (and other Galois fields), and with invertible matrices, and so on.

For example, let's use a specific 2×2 matrix for a.

{\begin{aligned}{\begin{bmatrix}4&5\\7&9\end{bmatrix}}X&={\begin{bmatrix}4&5\\7&9\end{bmatrix}}\\{\begin{bmatrix}9&-5\\-7&4\end{bmatrix}}\left({\begin{bmatrix}4&5\\7&9\end{bmatrix}}X\right)&={\begin{bmatrix}9&-5\\-7&4\end{bmatrix}}{\begin{bmatrix}4&5\\7&9\end{bmatrix}}\\\left({\begin{bmatrix}9&-5\\-7&4\end{bmatrix}}{\begin{bmatrix}4&5\\7&9\end{bmatrix}}\right)X&={\begin{bmatrix}9&-5\\-7&4\end{bmatrix}}{\begin{bmatrix}4&5\\7&9\end{bmatrix}}\\{\begin{bmatrix}1&0\\0&1\end{bmatrix}}X&={\begin{bmatrix}1&0\\0&1\end{bmatrix}}\\X&={\begin{bmatrix}1&0\\0&1\end{bmatrix}}\end{aligned}}

This works in spite of the fact that, for matrices, AB ≠ BA. In fact, we need merely demand that the algebra be a group under multiplication for the proof to go through. Thus we have a powerful motivation to study abstract algebra. --KSmrq^T 16:30, 14 September 2007 (UTC)[reply]

You don't need a group, just the cancellation property. Thus the usual generality in which to state this result would be an integral domain. —Blotwell 23:42, 14 September 2007 (UTC)[reply]

The proof just above (essentially that proposed by Meni Rosenfeld) requires a multiplicative identity, a guarantee that a multiplicative inverse exists, and associativity (but not commutativity); hence, it requires a multiplicative group.

Perhaps you are thinking of the different method of proof I referred to earlier, in my first post, where we might want to prove this for, say, the natural numbers. But, as I said at the time, if a multiplicative identity, 1, exists, then it clearly satisfies the equation without needing any other properties. We don't need inverses, nor cancellation, nor even associativity. That leaves the question of uniqueness, as I said; but it is not clear that uniqueness is being asked.

Permit me a fanciful example. Suppose a is a guy, x is a gal, ax is a kiss, and the equation ax = a says the kiss leaves the guy unmoved. Certainly a gal whose kisses have no effect on any guy would satisfy the equation (if not the guys). But it could also happen that some guy other than a will respond strongly to x even though a does not; she's not his type, apparently. In fact, a might only respond to a unique "soul mate", so most kissing partners would satisfy the equation. And in these modern times, we need not assume that a is always a guy nor that x is always a gal (so we don't need a two-sorted algebra). Well, I did say it would be a fanciful example; but I could have illustrated the same mathematical points with, say, matrices (albeit less memorably). --KSmrq^T 06:11, 15 September 2007 (UTC)[reply]

Solve

solve 25+56+67+89100/678*345-1000 —Preceding unsigned comment added by 59.180.11.224 (talk) 15:40, 13 September 2007 (UTC)[reply]

That's what calculators are for. If you don't have one, theree's usually one built into your operating system -- start-programs-accessories on windows. But see the box at the top of this page, we're not here to do your homework for you. Gscshoyru 15:48, 13 September 2007 (UTC)[reply]

If you can edit this page, you can access your friend Google. -- Meni Rosenfeld (talk) 15:53, 13 September 2007 (UTC)[reply]

Although I imagine that this is a homework problem and you need to show your work. You should take a look at the article on the order of operations as well as your class notes. The key thing to keep in mind for this problem is that you do multiplication and division together, from left to right (so you'll do the division before the multiplication in this case), then, once you have all the multiplications and divisions done, you do the additions and subtractions, again left to right. One other nit: You can solve an equation (no equals sign, then no equation), but an expression like this you can only evaluate. Donald Hosek 16:41, 13 September 2007 (UTC)[reply]

Approximately 44,486.496 I think... YeoungBraxx 17:52, 13 September 2007 (UTC)YeoungBraxx[reply]

And as an aside for our original poster, if you were my student and gave as your (entire) answer "Approximately 44,486.496", I would give you no points. Or maybe 1/3 if I were feeling generous, but I seldom am. Donald Hosek 18:28, 13 September 2007 (UTC)[reply]

Almost [1] - 44 486.4956. Unless of course your teacher wants it rounded down to three decimal places.martianlostinspace ^{email me} 20:15, 13 September 2007 (UTC)[reply]

If you like, I will even give you permission to cite my link (to Google) as your working out (that is, how you arrived at your answer. Something I worked very hard for :-)martianlostinspace ^{email me} 20:18, 13 September 2007 (UTC)[reply]

You realise, of course, that this value is also rounded? -- Meni Rosenfeld (talk) 20:20, 13 September 2007 (UTC)[reply]

So it would appear. Although maybe less so than the other above answer. Maybe Google can only handle nine digits. [2].martianlostinspace ^{email me} 20:58, 13 September 2007 (UTC)[reply]

Assuming you are on a Windows machine. You can built your own calculator.

http://windowstipoftheday.blogspot.com/2005/03/ms-dos-calculator.html

Of course, if you are on a windows machine why don't you use the windows calculator. Start - All Programs - Accessories - Calculator

Make sure you select the scientific calculator by selecting : View - Scientific

202.168.50.40 00:47, 14 September 2007 (UTC)[reply]