(Duplicate) I'm solving some questions in discrete mathematics, and I stuck with this problem.  Prove that there are uncountably many non-isomorphic linear orderings in N.
I would like to find a way to construct uncountably many non-isomorphic linear (total) orderings of natural numbers (as stated in the title).  I've already constructed two non-isomorphic total orderings. They are $$ 1 &lt; 2 &lt; 3 &lt;\ldots\\ 1 &gt; 2 &gt; 3 &gt;\ldots $$ where "$&lt;$" and "$&gt;$" are  strict total ordering relations.  Thanks in advance for your help.
Given a file like this:  1,768,12,46576457,7898 1,123,435,134,146 2,345,6756856,12312,1311 5,234,567465,12341,1341 1,3245,4356345,2442,13 9,423,2342,121,463 9,989,342,121,1212   I would like to list all rows (in bash terminal) such that the value in column 1 appears at least twice (in column 1). The result should be  1,768,12,46576457,7898 1,123,435,134,146 1,3245,4356345,2442,13 9,423,2342,121,463 9,989,342,121,1212 How to remove snap completely without losing the Chromium browser? Is there anything special with complex fraction $\left|\frac{z-a}{1-\bar{a}{z}}\right|$? Prove that the square root of a positive integer is either an integer or irrational Is my proof that the square root of a positive integer is either an integer or an irrational number correct?  The proof goes like this:  Suppose an arbitrary number n, where n is non-negative. If $\sqrt{n}$ is an integer, then $\sqrt{n}$ must be rational. Since $\sqrt{n}$ is an integer, we can conclude that n is a square number, that is  for some integer a.  Therefore, if n is a square number, then $\sqrt{n}$ is rational.  Suppose now that n is not a square number, we want to show that the square root of any non-square number is irrational.  We prove by contradiction. That is, we suppose that the square root of any non-square number is rational. So $\sqrt{n} = \frac{a}{b}$, where $a,b \in Z^+, b \neq 0$. We also suppose that $a \neq 0$, otherwise $\frac ab = 0$ , and n will be a square number, which is rational.   Hence $n = \frac {a^2}{b^2}$, so $nb^2 = a^2$.    Suppose $b=1$. Then $\sqrt n = a$ , which shows that n is a square number. So $b \neq 1$. Since $\sqrt n &gt; 1$, then $a&gt;b&gt;1$.  By the unique factorization of integers theorem, every positive integer greater than $1$ can be expressed as the product of its primes.  Therefore, we can write $a$ as a product of primes and for every prime number that exists in $a$, there will be an even number of primes in $a^2$. Similarly, we can express $b$ as a product of primes and for every prime number that exists in $b$, there will be an even number of primes in $b^2$.  However, we can also express $n$ as a product of primes. Since $n$ is not a square number, then there exist at least one prime number that has an odd number of primes.  Therefore, there exists at least one prime in the product of $nb^2$  that has an odd number of primes. Since $nb^2=a^2$ , then this contradicts the fact that there is an even number of primes in $a^2$ since a number can neither be even and odd.  Therefore, this contradicts the fact that $\sqrt n$ is rational. Therefore, $\sqrt n$ must be irrational.  Is this sufficient? Or is there any parts I did not explain well? Take for example  both at SO and SU. 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That function returns a coefficient $\beta$ such that $$y = \beta x.$$  Today I learned about total least squares and that princomp() function (principal component analysis, PCA) can be used to perform it. It should be good for me (more accurate). I have done some tests using princomp(), like:  r &lt;- princomp( ~ x + y)   My problem is: how to interpret its results? How can I get the regression coefficient? By "coefficient" I mean the number $\beta$ that I have to use to multiply the $x$ value to give a number close to $y$. . While the reasons for his doing this are not yet public, this is a real loss of a valuable service. Does anyone know of similar services available to the general public?  Edit:  is definitely relevant, but that approach is more appropriate for finding top journals rather than identifying bottom ones. I.e., following that method would probably exclude lots of valid, lower-tier journals. Are there any approaches to easily identifying a predatory publisher? 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I've been watching  and in one lecture, he discusses a very simple gauge theory.  We have a complex scalar field $\phi(x)$ with Lagrangian  $$\mathscr{L} = \partial_\mu \phi^* \partial^\mu \phi - V(\phi^* \phi)$$  and we want to do the gauge transformation $\phi(x) \to e^{i\theta(x)} \phi(x)$.  But the derivative term in the Lagrangian is not invariant if $\theta$ varies from place to place, so we add a new vector field $A_\mu$, and define it to transform like $$A_\mu \to A_\mu + \partial_\mu \theta$$ under the gauge transformation.  Then we change the Lagrangian to use covariant derivatives $D_\mu = \partial_\mu + iA_\mu$ instead of ordinary ones; now when the gauge transform is done all the derivatives of $\theta$ cancel out, and the result is invariant.  My question is: was there any freedom in choosing to add a vector field, with that specific gauge transformation law?  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