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# Flashing question mark, low hard drive space

When I booted my Mac last night I got a flashing folder against a gray background. The computer started fine this morning, but when I closed the lid and opened it up again I had to reboot, and now it's persistently in this flashing folder state.

Booting while holding down ⌘ Cmd-R brings me to OS X Utilities. Running "Repair disk" from disk utilities indicates there's no problem, but I did notice that it also says I only have 16 MB free on a 110GB hard drive.

Might this be the problem? If so, is there some way to delete data from my computer while in this state, to clear enough space to boot correctly?

• How long ago is your last backup and can you afford to delete some files or applications to then reinstall them again? (For instance, you could reinstall Word or iMovie or GarageBand easily and not lose your data or settings) – bmike Feb 21 '15 at 1:18

Though the following should be valid for non-encrypted volumes and unlocked encrypted volumes, it doesn't help much because the main volume here is encrypted and locked. So skip to FileVault2-encrypted volumes

Non-encrypted volumes

Almost no free space on your start volume is a problem. At least 10 % free space are recommended.

It's indeed possible to remove files & folder in Recovery Mode but if you have an external disk/thumb drive (hfs+ or fat32-formatted) you may move them instead:

• attach your external device (if you have one).
• If FileVault2 is enabled on your main volume start Disk Utility, right click the volume and unlock it
• Quit Disk Utility and open /Utilities/Terminal in the menubar
• enter cd /Volumes/NameOfYourMainVolume/Users/accountname to change to your user folder (example: cd "/Volumes/Macintosh HD/Users/ausername"). To get the names/mountpoints of your main volume/external disk enter diskutil list or df.
• enter ls -la to show all files and folders

Example:

drwxr-xr-x+ 56 ausername  staff   1904 20 Feb 23:45 .
drwxr-xr-x+  7 root       admin    238 15 Jun  2012 ..
-rw-------   1 ausername  staff      3 21 Nov 16:11 .CFUserTextEncoding
-rw-r--r--@  1 ausername  staff  43012 20 Feb 23:35 .DS_Store
drwxr-xr-x+  3 ausername  staff    102 15 Feb 17:41 .TemporaryItems
drwxr-xr-x   4 ausername  staff    136  2 Nov 23:22 SomeFolder

• now move further with cd SomeFolder
• enter ls -la to show all files and folders

Example:

drwxr-xr-x+  7 ausername  staff        238 15 Jun  2012 ..
-rw-------   1 ausername  staff 4736527884 16 Okt 22:33 BigFile.mov
drwxr-xr-x   4 ausername  staff        136  2 Nov 23:22 UnwantedFolder

• to move BigFile.mov to your external device enter
mv BigFile.mov /Volumes/ExternalDevice/
or to move UnwantedFolder to your external device enter
mv UnwantedFolder /Volumes/ExternalDevice/.

• if you don't have an external drive remove folders and files with rm -dr UnwantedFolder to remove UnwantedFolder and all its content or rm BigFile.mov to remove BigFile.mov (4.7 GB in the example above). rm *.mov removes all .mov files in your working directory.

rm is a powerful command to delete a file or directory without confirmation by default. Because of this behavior, users should really be sure before deleting files or folders. So double check where you are - enter pwd to print your working directory - and what you want to delete!

FileVault2-encrypted volumes

• started to Recovery Mode open Disk Utility

If you had one partition spanning your whole hard drive and encrypted it with FileVault2 it's completely normal that you have only some MB available space left. That's unallocated free space on your hard drive. In the example above 12,7 MB are left.

• highlight your FileVault2 volume and unlock it.

• if you succeed and the encrypted volume isn't corrupted your FileVault2 volume will be mounted. Check the available free space. In the example below the availbale free space on the volume is 52,8 GB

If the unlocking fails - using the correct password - the FileVault2 volume probably is corrupted. The only possible solution then is completely erase/repartition the hard drive and restore a backup.

• I would delete something in /Applications unless you have some large files you can download again (torrents?) or media you can rip from CD again. Also, if the OP has a backup, then anything can be removed safely... – bmike Feb 21 '15 at 1:20
• Thanks, that's greatly appreciated (and nice touch incorporating my username!). Though I am a Mac novice (as is probably obvious), I'm actually very comfortable with the command line. But I'm still having problems. I used Internet Recovery, and when I open the terminal as you instruct, the only folder in /Volumes is 'Mac OS X Base System', which includes 'Install OS X Mountain Lion.app' and 'usr' but not 'Users' (and the find command shows no 'Users' directory exists), so it seems like my actual hard drive volume isn't accessible from here. – ausername Feb 21 '15 at 1:29
• The output of diskutil list shows /dev/disk0s2 is the Apple_HFS Mac OS X Base System (1.2GB), and includes the output 0: FDisk_partition_scheme *120.0 GB disk12 1: 0xEE 120.0 GB disk12s1 But the mount command seems to fail: running diskutil mount -mountPoint /Volumes/myhd /dev/disk12s1 gives the error: Volume on disk12s1 failed to mount If it's relevant, my hard drive is encrypted. Am I doing something obviously wrong? – ausername Feb 21 '15 at 7:45
• @ausername But now i really doubt that low disk space causes the question mark: Enabling FileVault on your main volume always leaves some unallocated space in the MByte range on your main drive. Maybe your FileVault got corrupted. I think we have to start from scratch. – klanomath Feb 21 '15 at 8:39


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# Best carbon dating formula derivation

Radiocarbon dating (also referred to as carbon dating or carbon-14 dating) is a method for determining the age of an object containing organic material by using the properties of radiocarbon, a radioactive isotope of carbon. The method was developed in the late 1940s by Willard Libby, who received the Nobel Prize in Chemistry for his work in 1960. It is based on the fact that radiocarbon (14C) is constantly being created in the atmosphere by the interaction of cosmic rays with atmospheric nitrogen .

Carbon radiometric dating is based on measuring the amount of carbon-14 isotope in a sample of carbon of organic origin (from a living thing that is now dead). In a living thing the amount of carbon-14 in its carbon remains at equilibrium with its environment (~1.5 PPB) which remains roughly constant due to the production of carbon-14 by cosmic ray impacts with atoms in the upper atmosphere.

This carbon-14 is continuously decaying by beta decay to nitrogen-14, but by interconnection with its environment the lost carbon-14 is replaced. Once the living thing dies however it is no longer interconnected to its environment and the decayed carbon-14 is no longer replaced.

Carbon-14 decays with a halflife of 5730 years. By measuring the amount of carbon-14 in the sample and calculating how long on the exponential curve it took to drop from the equilibrium level it had when last alive to its current measured level, you get the age of the sample. The technique is not perfect: • for a variety of reasons the carbon-14 level in the environment is not actually constant over time, this requires checking carbon dating ages against other dating methods ages in many cases • materials older than about 40000 to 50000 years old are usually not datable with carbon dating as the level of carbon-14 in the sample has decayed to low to be reliably measured with any accuracy • of course it will not work on samples that do not contain carbon or which are contaminated with either carbon of nonorganic origin or ancient organic carbon (carbon in which all the carbon-14 has already decayed) • etc.

Burning can mean two different things in chemistry: Usually it refers to combustion, but especially with hydrocarbons, it could also mean pyrolysis or "cracking". Combustion of straight-chain alkanes (normal carbon-hydrogen structures) follows the formula: C n H 2n+2 + (2n)O 2 ----> (n)CO 2 … + (2n)H 2 O + heat where n is a positive integer. If your question was what is the formula of carbonate then the answer would be CO 3.

or a compound containing one Carbon atom and Three Oxygen atoms. Some common places carbonate can be found are in your home. Baking Soda is made of sodium bicarbonate It also can be found in the form of carbonic a … cid. H 2 CO 3 Which can be found in acid rain and is formed by the combination of CO 2 and H 2 O or Carbon Dioxide and Water.

This is also the main mode of transportation of Carbon Dioxide in the blood.

## best carbon dating formula derivation - Derivation Of Lens Maker Formula

Derivative Formula: Actual derivative formula: If y = f(x), then $$\LARGE f’(x)=\lim_{\triangle x \rightarrow 0}\frac{f(x+ \triangle x)-f(x)}{\triangle x}$$ Basic Derivative formula: • $$\frac{d}{dx}$$( c) = 0, where c is constant. • $$\frac{d}{dx}$$( x) = 1 • $$\frac{d}{dx}$$( x n) = n x n-1 • $$\frac{d}{dx}$$[ f(x)] n = n [f(x)] n-1 $$\frac{d}{dx}$$ f(x) • $$\frac{d}{{dx}}\sqrt x= \frac{1}{{2\sqrt x }}$$ • $$\frac{d}{dx}$$ C∙f(x) = C ∙ $$\frac{d}{dx}$$ f(x) = C ∙f’(x) • $$\frac{d}{{dx}}[f(x) \pm g(x)] = \frac{d}{{dx}}f(x) \pm \frac{d}{{dx}}g(x) = f'(x) \pm g'(x)$$ • $$\frac{d}{dx}$$ [f(x) ∙ g(x)] = f(x) $$\frac{d}{dx}$$ g(x) + g(x) $$\frac{d}{dx}$$ f(x) This is called product rule of derivative.

• $$\frac{d}{{dx}}[\frac{{f(x)}}{{g(x)}}] = \frac{{g(x)\frac{d}{{dx}}f(x) – f(x)\frac{d}{{dx}}g(x)}}{{{{[g(x)]}^2}}}$$ This is quotient rule of derivative. Chain Rule: • [f(g(x))]’= f’(g(x))g’(x) • $$\frac{du}{dx}$$ = $$\frac{du}{dv}$$ ∙$$\frac{dv}{dx}$$ • $$\large \frac{du}{dx}=\frac{\frac{du}{dv}}{\frac{dx}{dv}}$$ • $$\large \frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}$$ Logarithm Derivative Formula: • $$\frac{d}{dx}$$ln x = $$\frac{1}{x}$$ • $$\frac{d}{dx}$$log ax = $$\frac{1}{x\:ln\:a}$$ • $$\frac{d}{{dx}}\ln f(x) = \frac{1}{{f(x)}}\frac{d}{{dx}}f(x)$$ • $$\frac{d}{{dx}}{\log _a}f(x) = \frac{1}{{f(x)\ln a}}\frac{d}{{dx}}f(x)$$ Derivative formula for exponential functions: • $$\frac{d}{dx}$$e x = e x • $$\frac{d}{{dx}}{a^x} = {a^x}\ln a$$ • $$\frac{d}{dx}$$e f(x) = e f(x) f’(x) • $$\frac{d}{dx}$$ a f(x) = a f(x)ln a f’(x) • $$\frac{d}{dx}$$ x x = x x(1 + ln x) Derivative Formula for Trigonometric Formula: • $$\frac{d}{dx}$$ sin x = cos x • $$\frac{d}{dx}$$ cos x = – sinx • $$\frac{d}{dx}$$ tan x = sec 2 x • $$\frac{d}{dx}$$ cot x = – cosec 2x • $$\frac{d}{dx}$$ sec x = sec x ∙tan x • $$\frac{d}{dx}$$ cosec x = – cosec x ∙cot x Derivative formula for Inverse Trigonometric functions: • $$\frac{d}{{dx}}si{n^{ – 1}}x = \frac{1}{{\sqrt {1 – {x^2}} }},{\text{ }} – 1 1$$ • $$\frac{d}{{dx}}co{\sec ^{ – 1}}x = \frac{{ – 1}}{{x\sqrt {{x^2} – 1} }},{\text{ }}\left| x \right| > 1$$ Derivative formula for hyperbolic functions: • $$\frac{d}{dx}$$ sinh x = cosh x • $$\frac{d}{dx}$$ cosh x = sinh x • $$\frac{d}{dx}$$ tanh x = sech 2x • $$\frac{d}{dx}$$ coth x = – cosech 2x • $$\frac{d}{dx}$$ sech x = -sech x ∙tanh x • $$\frac{d}{dx}$$ cosech x = – cosech x ∙coth x Derivative formula for Inverse Hyperbolic functions: • $$\frac{d}{{dx}}Sin{h^{ – 1}}x = \frac{1}{{\sqrt {1 + {x^2}} }}$$ • $$\frac{d}{{dx}}Cos{h^{ – 1}}x = \frac{1}{{\sqrt {{x^2} – 1} }}$$ • $$\frac{d}{{dx}}Tan{h^{ – 1}}x = \frac{1}{{1 – {x^2}}},{\text{ }}\left| x \right| 1$$ • $$\frac{d}{{dx}}Sec{h^{ – 1}}x = \frac{{ – 1}}{{x\sqrt {1 – {x^2}} }},{\text{ }}0 0$$ Derivative formula examples: • Find the derivative of the function given by f(x) = sin (x) 2.

Solution: f(x) = sin(x 2) f’(x) = $$\frac{d}{dx}$$( sin x 2) x $$\frac{d}{dx}$$ x 2 = (cos x 2) (2x) = 2x cos x 2 • Find $$\frac{dy}{dx}$$ if x – y = π. Solution: We can write the equation as y = x – π $$\frac{dy}{dx}$$ = 1 • Find $$\frac{dy}{dx}$$, if y + sin y = cos x. Solution: We differentiate the relationship directly with respect to x, $$\frac{dy}{dx}$$ + $$\frac{d}{dx}$$(sin y) = $$\frac{d}{dx}$$(cos x) which implies using chain rule $$\frac{dy}{dx}$$ + cos y $$\frac{dy}{dx}$$ = -sin x This allows $$\frac{dy}{dx}$$ = – $$\frac{sin x}{1 + cos y}$$ where y ≠ (2n + 1) π More from Calculus

After 5 years the amount of carbon 14 left in body is half original dating formula derivation rated 3 this is known as the isochron equation the plotted here is from g w wetherill ann rev nucl sci 25 283 1975 and involves dating of a meteorite cosmological geological and archaeological dating derivation of radioactivity equation 02

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# zbMATH — the first resource for mathematics

Stochastic averaging principle for dynamical systems with fractional Brownian motion. (English) Zbl 1314.60122
Summary: Stochastic averaging for a class of stochastic differential equations (SDEs) with fractional Brownian motion, of the Hurst parameter $$H$$ in the interval $$(\frac{1}{2},1)$$, is investigated. An averaged SDE for the original SDE is proposed, and their solutions are quantitatively compared. It is shown that the solution of the averaged SDE converges to that of the original SDE in the sense of mean square and also in probability. It is further demonstrated that a similar averaging principle holds for SDEs under stochastic integrals of pathwise backward and forward types. Two examples are presented and numerical simulations are carried out to illustrate the averaging principle.

##### MSC:
 60H10 Stochastic ordinary differential equations (aspects of stochastic analysis) 60G22 Fractional processes, including fractional Brownian motion 60H05 Stochastic integrals 34F05 Ordinary differential equations and systems with randomness 37H10 Generation, random and stochastic difference and differential equations 93E03 Stochastic systems in control theory (general)
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Duncan, Stochastic calculus for fractional Brownian motion I: Theory,, SIAM J. Control Optim., 38, 582, (2000) · Zbl 0947.60061 [9] D. Feyel, Fractional integrals and Brownian processes,, Potential Analysis., 10, 273, (1996) [10] H. Holden, Stochastic Partial Differential Equations,, Birkhäuser, (1996) [11] Y. Hu, Fractional white noise calculus and application to finance,, Infin. Dimens. Anal. Quantum Probab. Relat. Topics, 6, 1, (2003) · Zbl 1045.60072 [12] W. T. Jia, Stochastic averaging of quasi-non-integrable Hamiltonian systems under combined Gaussian and Poisson white noise excitations,, Int. J. Nonlin. Mech., 51, 45, (2013) [13] G. Jumarie, Stochastic differential equations with fractional Brownian motion input,, Int. J. Syst. Sci., 24, 1113, (1993) · Zbl 0771.60043 [14] G. Jumarie, On the solution of the stochastic differential equation of exponential growth driven by fractional Brownian motion,, Appl. Math. Lett., 18, 817, (2005) · Zbl 1075.60068 [15] R. Z. Khasminskii, A limit theorem for the solution of differential equations with random right-hand sides,, Theory Probab. Appl., 11, 390, (1963) [16] R. Z. Khasminskii, Principle of averaging of parabolic and elliptic differential equations for Markov process with small diffusion,, Theory Probab. Appl., 8, 1, (1963) [17] A. N. Kolmogorov, Wienersche Spiralen und einige andere interessante Kurven im Hilbertschen,, Raum, 26, 115, (1940) · JFM 66.0552.03 [18] S. C. Kou, Generalized Langevin Equation with Fractional Gaussian Noise: Subdiffusion within a Single Protein Molecule,, Phys. Rev. Lett., 93, (2004) [19] W. E. Leland, On the self-similar nature of ethernet traffic,, IEEE/ACM Trans. Networking, 2, 1, (1994) [20] R. Liptser, On estimating a dynamic function of a stochastic system with averaging,, Statistical Inference for Stochastic Processes, 3, 225, (2000) · Zbl 0982.62070 [21] T. Lyons, Differential equations driven by rough signals,, Rev. Mat. 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# Find the PDE for a function that makes it a martingale

Given the SDE, find the PDE for the function $V(t,x)$ such that $V(t,S_t)$ is a martingale.

$dS_t = \kappa(m - S_t)dt + \sigma\sqrt{S_t}dB_t$ where $\kappa$,$m$, and $\sigma$ are constants.

Attempted solution: The stochastic differential equation or SDE can be written in differential form such that $$dS_t = \mu(t,S_t)dt + \sigma(t,S_t)dB_t$$ thus from c.) we have $$\mu(t,S_t) = \kappa(m-S_t) \ \ \ \text{and} \ \ \ \sigma(t,S_t) = \sigma\sqrt{S_t}$$ $V(t,S_t)$ is martingale if and only if $V(t,x)$ satisfies $$\partial_t V(t,x) + \partial_x V(t,x)\mu(t,x) + \frac{1}{2}\partial_{xx}V(t,x)\sigma^2(t,x) = 0$$ therefore the PDE required to satisfy this condition is $$\partial_t V(t,x) + \partial_x V(t,x)\kappa(m - S_t) + \frac{1}{2}\partial_{xx}V(t,x)\sigma^2 S_t = 0$$

Not sure if this is correct any suggestions is greatly appreciated.

• Not sure why this got voted down, I will gladly edit my question if there was something I did to make me get marked down – Wolfy Apr 1 '16 at 17:35
• I do not see anything particularly wrong. However, you need to use $x$ consistently, that is, the PDE should be write as $$\partial_t V(t,x) + \partial_x V(t,x)\kappa(m - x) + \frac{1}{2}\partial_{xx}V(t,x)\sigma^2 x = 0.$$ – Gordon Apr 1 '16 at 17:52


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# All Science Fair Projects

## Science Fair Project Encyclopedia for Schools!

 Search    Browse    Forum  Coach    Links    Editor    Help    Tell-a-Friend    Encyclopedia    Dictionary

# Science Fair Project Encyclopedia

For information on any area of science that interests you,
enter a keyword (eg. scientific method, molecule, cloud, carbohydrate etc.).
Or else, you can start by choosing any of the categories below.

# Lens space

A lens space is an example of a topological space, considered in mathematics. The term often refers to a specific class of 3-manifolds, but in general can be defined for higher dimensions.

 Contents

## Definition

Sit the 2n - 1-sphere S2n - 1 inside $\mathbb C^n$ as the set of all n-tuples of unit absolute value. Let ω be a primitive pth root of unity and let $q_1,\ldots,q_n$ be integers coprime to p. Let the set of powers $\mathbb Z_p=\{1,\omega,\ldots,\omega^{p-1}\}$ act on the sphere by

$\omega\cdot(z_1,\ldots,z_n)=(\omega^{q_1}z_1,\ldots,\omega^{q_n}z_n).$

The resulting orbit space is a lens space, written as $L(p;q_1,\ldots,q_n)$.

We can also define the infinite-dimensional lens spaces as follows. These are the spaces $L(p;q_1,q_2,\ldots)$ formed from the union of the increasing sequence of spaces $L(p;q_1,\ldots,q_n)$ for $n=1,2,\ldots$. As before, the $q_1,q_2,\ldots$ must be coprime to p.

## In three dimensions

By specializing the above definition to n = 2, we get 3-manifolds. In this case, a more picturesque description of a lens space is that of a space resulting from gluing two solid torii together by a homeomorphism of their boundaries. Of course, to be consistent, we should exclude the 3-sphere and $S^2 \times S^1$, both of which can be obtained as just described; some mathematicians include these two manifolds in the class of lens spaces.

Three-dimensional lens spaces were the first known examples of 3-manifolds which were not determined by their homology and fundamental group alone. J.W. Alexander in 1919 showed that the lens spaces L(1;5) and L(2;5) were not homeomorphic even though they have isomorphic fundamental groups and the same homology.

There is a complete classification of three-dimensional lens spaces.

## References

• G. Bredon, Topology and Geometry, Springer Graduate Texts in Mathematics 139, 1993.
• A. Hatcher, Algebraic Topology, Cambridge University Press, 2002.


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# Which Brownian motion property is the most important? [closed]

Which Brownian motion property is the most important?

A standard Brownian motion is a stochastic process $(W_t, t\geqslant 0)$ indexed by nonnegative real numbers t with the following properties:

1. $W_0=0$;
2. With probability 1, the function $t \to W_t$ is continuous in t;
3. The process $(W_t, t\geqslant 0)$ has stationary, independent increments;
4. The increment $W_{t+s}-W_s$ has the $\mathrm{NORMAL}(0, t)$ distribution.

## closed as primarily opinion-based by Nate Eldredge, Did, Davide Giraudo, Alexander Gruber♦May 10 '14 at 17:03

Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise. If this question can be reworded to fit the rules in the help center, please edit the question.

• What do you mean by "most important"? Most important with regard to what...? Actually, all these properties are very imporant. – saz May 10 '14 at 11:53
• Actually I think the same - that all are important. However, during the exam I got the question: "Which Brownian motion property is the most important?" and got confused... So now I am curious if I can exclude some property as the most important one. – kevi May 10 '14 at 11:58

## 2 Answers

I agree with @FooBar that the first property, i.e. that the initial point equals $0$ a.s., is not an important one - in the sense that this property is not needed in order to characterize the Brownian motion. In fact, the shifted process $(x+W_t)_{t \geq 0}$ is called Brownian motion started at $x \in \mathbb{R}$, and there are further generalizations which basically allow to consider arbitrary initial distributions.

Concerning the continuity of the sample paths: Any process $(W_t)_{t \geq 0}$ which satisfies the properties 3+4 has a modification which has (a.s) continuous sample paths. This follows from the Kolmogorov-Chentsov theorem. (Obviously, this does not mean that the continuity is not important.)

Concerning the stationary independent increments: If a process $(L_t)_{t \geq 0}$ has stationary independent increments (and is stochastically continuous), then it is called a Lévy process. This is a rather important class of stochastic processes and, in particular the sample path properties differ from these of the Brownian motion. Most importantly, the sample paths can have jumps. In fact, the Brownian motion is the only Lévy process with continuous sample paths. This means that the properties 2+3 imply property 4.

This does not answer the question which property is the most important one; but if I would have to strip down the definition to two properties, I guess I would choose property 3 and 4.

I would just exclude $W_0$. The initial value is quite irrelevant and is typically chosen to be zero only for normalization purposes. If you compare it to discrete-time, the Brownian motion is the equivalent of an iid $N(0,\sigma)$ process.

Then, it depends on the context, whether you care most about it being continuous, independent or normal.


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# First order language in logic

1. May 26, 2010

### rainwyz0706

1. The problem statement, all variables and given/known data

Let L = {P } be a ﬁrst-order language with a binary relation symbol
P as only non-logical symbol. By exhibiting three suitable L-structures prove
(informally) that no two of the following sentences logically implies the other
(i) ∀x∀y∀z(P (x, y) → (P (y, z) → P (x, z))),
(ii) ∀x∀y(P (x, y) → (P (y, x) → x = y)),
(iii) (∀x∃yP (x, y) → ∃y∀xP (x, y)).

I really don't have a clue how to handle this problem. Could anyone please give me some hints of the L-structures? Any help is appreciated!

2. May 26, 2010

### Martin Rattigan

An L-structure is a formalisation of an interpretation of a language. If you think first of an interpretation, then constructing the corresponding L-structure is just turning the handle of the sausage machine.

So the first thing to do is to think of interpretations for which each of (i),(ii) and (iii) (in turn) are false while the other two are true. This would then mean that no two of the sentences logically implies the third. You can then construct corresponding L-structures.

You should check exactly how the L-structures are constructed in the exposition you're following, and exactly how the phrase "logically implies" is defined (this may be in terms of the L-structures). If for some reason you can't find this information there's a reasonably straightforward explanation http://www.maths.ed.ac.uk/~s0571100/Logic.pdf [Broken], but try not to mix and match definitions because there can be slight differences which don't affect the meaning in general but can affect whether your proof is considered correct.

Last edited by a moderator: May 4, 2017
3. May 26, 2010

### rainwyz0706

Thanks a lot for your help. I can only think of <Q, >>, which would make 1,2 true and 3 false. And I'm not sure that I've interpreted 3 correctly. Could you explain it a little bit more please?

4. May 26, 2010

### Martin Rattigan

OK that example is correct.

Conditions (i) and (ii) are almost the requirement for P to be a weak partial order. (By weak I mean "$\leq$" type rather than "$<$" type.) But there's something missing - what?

You've chosen a weak partial (actually total) order with no maximum and that makes the third condition false, because it says, "if everything $\leq$ something, then something $\geq$ everything". That is, "if the domain of P is its whole field (that is the whole of the domain of interpretation in this case) then P must have a maximum", which $\mathbb{Q}$ of course doesn't.

If you answer the question in the first paragraph that should give you possible other interpretations to use.

5. May 26, 2010

### Martin Rattigan

Correction; I didn't read your answer properly. You chose a strong partial order ("$>$" type, rather than "$\leq$" type).

If you'd said $(\mathbb{Q},\leq)$ what I said would make sense. The changes to make it correspond with what you said are straightforward.

That should give you a further clue. If you're still struggling ask again.

Last edited: May 26, 2010
6. May 26, 2010

### rainwyz0706

Thanks. What if I change Q into all non-positive rational numbers, then it has a maximum. Would that work?
Also, that's only one l-structure. Could you give me some hints about the other two possible l-structure?

7. May 26, 2010

### Martin Rattigan

With your suggested change it would still be correct. But notice I was talking about a maximum in connection with $(\mathbb{Q},\leq)$, because I'd misread the answer. If I'd talked about $(\mathbb{Q},\geq)$ instead (which is still not quite what you wrote), I'd have talked about a minimum. In fact with a strict order the definitions of maximum and minimum in terms of the ordering relation have to change slightly.
The domain of interpretation will be a pair of distinct cabbages and $P(x,y)$ will be interpreted as, "$x$ is a cabbage or $y$ is a cabbage". It should be apparent that with this interpretation (i) and (iii) are true whereas values of $x$ and $y$ can be found so that (ii) is not satisfied, which should also mean that (i) and (iii) don't logically imply (ii) (when you have checked what that means).