In the chorus of the song "Gone Country" by Alan Jackson, there is this sentence:     She/He's gone country, look at them boots   How come them is used instead of their? I am not a native English speaker so I am interested in knowing about this peculiar (at least to me) word choice.  This  would tend to indicate that them can be used for that, but look at that boots don't really make sense?
Background Nowadays, I see &quot;them&quot; used to mean &quot;those&quot; a lot. I don't know if it was as common in the past. For example, take &quot;one of &quot;. On researching about it, I found some people say it comes from a dialect of British English. Some others say it is a &quot;non-standard&quot; usage. I see this usage in Canadian English also, and it seems some people use it in a sarcastic way. Moreover, I have seen a song titled &quot;one of them days&quot;. I also read, in the book called &quot;A Broken Promise&quot;, &quot;&quot; Finally, Wikipedia says that it is a usage in  (a common name for the Southern Midland dialect of American English):  Pronouns and demonstratives &quot;Them&quot; is sometimes used in place of &quot;those&quot; as a demonstrative in both nominative and oblique constructions. Examples are &quot;Them are the pants I want&quot; and &quot;Give me some of them crackers.&quot;    Question(s): What would you say about the usage of this word? Is it correct? Could we use it in daily speech? Could this usage go beyond a specific dialect and be used in other dialects, regions, etc.? Does it really originate from Appalachian English? Why did this usage become popular among other English speakers? Note: I have already seen a similar question:  However, it only says, &quot;ungrammatical,&quot; there. This question is specific to this situation only, and there is more to it.
Show that $\textbf{ZF}$ shows $\phi$ is $\Delta_1$ $(S^1 \times [0,1])/\sim$ homeomorphic to unit disk $ D^2$ Let $\sim$ be the equivalence relation on $S_1 \times I$ given by $(x,t)\sim(y,s)$ if and only if $xt=ys$ where $I=[0,1]$  How do I prove that $(S_1 \times I)/\sim$ is homeomorphic to the unit disk $D^2$ with the induced topology? When should a supervisor be an author? How do electric currents make light? In an incandescent light bulb, electrons flow through wires to the bulb, where is passes through a filament, and the filament lights up.  Does this work because of several high energy electrons losing some of their energy to disturb the electromagnetic field, creating photons? The electrons just leave the filament afterward and are replaced by more high-energy electrons from the negative terminal. Or does it work differently?  As a secondary question, is there a type of light bulb that directly converts electron flow to photons? Command not found $ (dollar sign) Any command that starts with $ returns  command not found   How can I fix this?  I am  using 16.10 Ubuntu. How many Shutter Clicks on a New Camera considered normal? How many Shutter Clicks on a New Camera considered normal?    I purchased a quotes new Nikon D750 and checked the shutter count and it has 264 Shutter Clik count on it.    Is that suspect of a returned purchase or used camera? Converting expressions to polynomial form Operation not permitted turning off airplay mirroring How to render a pixelated, occluded wireframe with freestyle? I'd like to combine these features:    The pixelated look of a wire material with anti-aliasing disabled. Proper occlusion of hidden lines.   A wireframe material produces the first feature, freestyle produces the second. Can a line set be created that produces neither double pixels nor gaps? Line thickness alone does not seem to be the answer.  There is the trick with two materials and z-offset, but I have trouble getting a really clean solution at the outline of the object. So a freestyle solution would be best.    The main problem is that freestyle (to the right) produces double pixels: How can an 18 month old be taught that bullying is not ok? He seems to like causing the trouble it brings upon him, and bullying, mean acts also seem to please him. He likes to sit on his younger cousin (my son), push him down, and steal his toys. His parents seem to be at a loss about it, and I don't really have any good suggestions to give. Thanks. How do I simplify: $$\sum_{r = 0}^{\lfloor \frac{k}{2} \rfloor} \dbinom{k}{2r} \cdot \dbinom{n-k}{k-2r}?$$ Basically, the sum is: $\dbinom{k}{0} \cdot \dbinom{n-k}{k} + \dbinom{k}{2} \cdot \dbinom{n-k}{k-2} + \dbinom{k}{4} \cdot \dbinom{n-k}{k-4} \cdots.$ This seems very similar to Vandermonde's Identity but because we skip two everytime instead of one, I don't see how I can apply it to this problem. Was Harry Potter a direct descendant of Godric Gryffindor? If light is an electric and (magnetic field), how can it be absorbed? I was wondering how light or any electromagnetic radiation can be "absorbed" if it consists of electric and magnetic fields. For example if there is a charge at point A, and the  light reaches point A, it will cause that charge to move depending on the value of the electric field. However, why wouldn't this electric field continue on and influence others particles everywhere else? For example, if you have a static charge creating an electric field, then if you put a charge at point A it will feel its effects and then if you bring a charge at another location B it will also feel its effects, namely it will not have been absorbed. Please shed some light on this. Thanks  edit: Does this mean the electron that absorbs this light moves in such a way that it creates an electric field of its own canceling the initial one? Or am I way off? Are Living Beings Turned Into Horcruxes Suceptible to Age, Disease, or Injury? From the HP books I know that Horcruxes grant a form of immortality. So, it is not possible for someone who created a Horcrux to die from old age or succumb to diseases. I guess the same is true for living beings turned into a Horcrux. But do living Horcruxes age normally? Is it possible to cripple them or sever their limbs, head or something like that, or do they instantly "repair" themselves like inanimate Horcruxes? Does the equivalence $$\neg\neg (P \rightarrow Q) \leftrightarrow (\neg\neg P \rightarrow \neg\neg Q)$$ hold in propositional intuitionistic logic? In propositional classical logic the equivalence holds obviously since $P \leftrightarrow \neg\neg P$. However, the similar-looking implication $\neg\neg (P \vee Q) \rightarrow (\neg\neg P \vee \neg\neg Q)$ is . The validity of the backward implication comes up in  as well. Is it possible to mount on a PC an SD card that has been encrypted by an Android Marshmallow device? I have a Samsung Galaxy S5 duos with stock Android Marshmallow, and I'm considering encrypting my SD card to prevent access to its contents in case I lose my phone. The main reason of not doing this is that I'm not sure if there's a way to mount an encrypted SD card on Windows (or other desktop OS's)   When I encrypt the SD card on my Samsung Galaxy S5, is it using a known implementation of encryption, or is it something that only works on Android (or perhaps Samsung) phones?   I do not want to use USB connection to my phone to access my SD card, it's terrible. Mounting the SD card is much easier and faster and less annoying.   EDIT: just to clarify, I am NOT talking about adoptable storage. I want a portable encrypted SD card that can both be read on PCs and Android phones (i.e. I want to be able to remove the SD card from my phone, plug it into my PC, provide a password to decrypt and read/write it like any other SD card) How to prove that $\quad\displaystyle\frac{4^{n}}{\sqrt{4n}}&lt;\binom{2n}{n}&lt;\frac{4^{n}}{\sqrt{3n+1}}\quad$ for all $n$ &gt; 1 ?     Does anyone know any better elementary estimates?   Attempt.  We have $$\frac1{2^n}\binom{2n}{n}=\prod_{k=0}^{n-1}\frac{2n-k}{2(n-k)}=\prod_{k=0}^{n-1}\left(1+\frac{k}{2(n-k)}\right).$$ Then we have $$\left(1+\frac{k}{2(n-k)}\right)&gt;\sqrt{1+\frac{k}{n-k}}=\frac{\sqrt{n}}{\sqrt{n-k}}.$$ So maybe, for the lower bound, we have $$\frac{n^{\frac{n}{2}}}{\sqrt{n!}}=\prod_{k=0}^{n-1}\frac{\sqrt{n}}{\sqrt{n-k}}&gt;\frac{2^n}{\sqrt{4n}}.$$ By Stirling, $n!\approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$, so the lhs becomes $$\frac{e^{\frac{n}{2}}}{(2\pi n)^{\frac14}},$$ but this isn't $&gt;\frac{2^n}{\sqrt{4n}}$. Extracting tree crown areas from remote sensing data (visual images and LiDAR) I've seen many posts regarding a combinatorial proof of the following question. But for a non-combinatorial proof would the following method work? Also... is this the easiest way to arrive at a proof? It seems to be rather verbose.  Show the formula $k {n\choose k} = n {n-1 \choose k-1}$ is true for all integers $n,k$ with $0\le k \le n$.  My answer:  Observe that  $$\begin{align*} {n\choose k}&amp;= \frac {n}{k} {n-1 \choose k-1}\\ &amp;= \frac {n}{k}\left[ {n-2 \choose k-2} + {n-2 \choose k-1}\right]\\ &amp;= \frac {n}{k}\left[ \frac{(n-2)!}{(k-2)!(n-2-(k-2))!} + \frac {(n-2)!}{(k-1)!(n-2-(k-1))!}\right]\\ &amp;= \frac {n}{k}\left[ \frac{(n-2)!}{(k-2)!(n-k)!} + \frac {(n-2)!}{(k-1)!(n-k-1)!}\right]\\ &amp;= \frac{(n-1)!}{(k-1)!(n-k)!} + \frac {(n-1)!}{(k)!(n-k-1)!}\\ &amp;= \frac{(n-1)!}{(k-1)!(n-1-(k-1))!} + \frac {(n-1)!}{(k)!(n-1-k)!}\\ &amp;={n-1 \choose k-1} + {n-1 \choose k}={n \choose k}. \end{align*}$$ How do I prove that in every commuting family there is a common eigenvector?