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The NSA uses the square roots of small primes in the SHA1 and SHA2 algorithms. https://en.wikipedia.org/wiki/Nothing_up_my_sleeve_number These are referred to as nothing up my sleeve numbers. In cryptography, nothing up my sleeve numbers are any numbers which, by their construction, are above suspicion of hidden properties. They are used in creating ...

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Let's zoom out a bit. In math, there are these things called algebraic structures. Their role is similar to the role of interfaces in Java programming, or typeclasses in Haskell programming. If you ask a Java programmer what a list is, they might reply, In Java, the formal definition of a list is given by the List interface. For an object to qualify as ...

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Other people alluded to the formal definition about being closed under linear combinations. Perhaps what you're looking for is a slightly more intuitive, less rigorous answer. A finite dimensional vector space is isomorphic to some $\mathbf{R}^n$ for some $n$. So for instance, a $1$-dimensional vector space "looks like" the real number line, and a ...

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Linear alludes to the fact (rather, definition) that the space is closed under addition and scalar multiplication. If $x,y \in \mathbb{R}^n$ and $\alpha, \beta \in \mathbb{R}$, then $\alpha x + \beta y \in \mathbb{R}^n$, i.e. any linear combination of vectors in $\mathbb{R}^n$ belongs to the space as well.

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a semi famous example from number theory, Erdos Proof of Bertrands postulate (paper by Galvin). In 1845 Bertrand postulated that there is always a prime between n and 2n, and he verified this for n < 3 × 106 . Tchebychev gave an analytic proof of the postulate in 1850. In 1932, in his first paper, Erd˝os gave a beautiful elementary proof using ...

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Assume $\gcd(n,2i)=1$. Given any $x$, $y$, let $x^n+y^{n-i}=w$. Then for any $r$ we have $$x^nw^{rn(n-i)}+y^{n-i}w^{rn(n-i)}=w^{1+rn(n-i)}$$ which is to say $$(xw^{r(n-i)})^n+(yw^{rn})^{n-i}=w^{1+rn(n-i)}$$ Since $n$ (and, thus, $n-i$) is prime to $n-2i$, we can choose $r$ such that $1+rn(n-i)=(n-2i)s$ for some integer $s$, and so we have a solution of ...

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Say there are $k$ blocks in the box. Say Kim tries to build a cube of side $n$, so: $n^3 = k+6$ Say Takashi tries to build a cube of side $m$, so: $m^3= k-85$ Where $k, m, n$ are all integers. Oh, and we can see: $n^3-m^3 = 91$

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Let Kim be building a cube of side $x$ and Takashi be building a cube of side $y$ where $y<x$. So $x^3 -6 = y^3 + 85$ $x^3 - y^3 = 91$ $(x-y)(x^2 + xy +y^2) = (7)(13) = (1)(91)$ where the RHS has been factorised in the only two unique ways among natural numbers. Now try $x-y = 7 \implies x = y + 7$ and then $x-y = 1 \implies x = y+1$, substitute ...

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Dvir's solution to Kakeya Conjecture over finite fields is probably a good example of this. See http://arxiv.org/abs/0803.2336 and https://terrytao.wordpress.com/2008/03/24/dvirs-proof-of-the-finite-field-kakeya-conjecture/.

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You can see : André Weil, Number Theory : An approach through history (1984), Preface, page ix : Fermat, Euler, Lagrange, Legendre. These are the founders of modern number theory. The greatness of Gauss lies in his having brought to completion what his predecessors had initiated, no less than in his inaugurating a new era in the history of the ...

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Fermat contributed a lot of results (not so many proofs, unfortunately) to the community, and did lots of work in number theory. For example, both Fermat's Little Theorem and Fermat's Last Theorem are named after him, and they are clearly number theoretic. Fermat is particularly notable in that he worked mostly in isolation (if I remember correctly; if ...

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Well, the simplest one I could find in a few minutes was $100^4 + 500^3 = 15000^2$.

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Informally, but intuitively: Let's say $a = 3$. Then, we're looking at the sequence $\{1, 3, 9, 27, 81\}$. The numbers are clearly too far apart for adding two of them to reach the next one. This leads us to a slightly more formal proof: Let's say that $x \leq y$. Then, $$a^x + a^y \leq a^y + a^y = 2a^y < a(a^y) = a^{y+1}$$ So $z$ must be between $y$ ...

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Here's a less algebraic way to interpret your question: Look at things in base $a$. Then the equation $a^x+a^y=a^z$ looks like $1 \ldots 0 + 1 \ldots 0 = 1\ldots0$. (where the number of zeros is $x$, $y$, or $z$) This can only work if we're in base $2$ and the addition carries to produce a $1$ in the next digit.

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It is clear that $1+a^{y-x}=a^{z-x}$ where $a=1$,$x=y=z$ is an obvious solution. On the other hand the minimun difference of powers of integers is of the forme $2k+1$ which corresponds to squares.So, (making abstraction of the previous division by zero!) this would gives $a=0$.

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I will just provide short answers and a link to wikipedia. 1.Antiderivative and Integral: $\int$ is used to denote integration. Usually written as $$\int f(x)\, \text{d}x$$ the so called antiderivative. Or as: $$\int_a^b f(x)\, \text{d}x,$$ this is the so called definite case. $\int\int\int$ is used to denote multiple(3) applications of this. see above ...

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Clearly we must have $x, y<z$. Suppose $x\not=y$; WLOG, suppose $x<y$. Then this can be rewritten as $$a^x(1+a^{y-x})=a^xa^{z-x},$$ with $z-x, y-x>0$. This gives $$1+ a^{y-x}=a^{z-x};$$ but the right hand side is divisible by $a$, while the left hand side is not. So we must have $x=y$. But then this yields $$2a^x=a^z,$$ which in turn yields ...

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The equation $a^x+a^y=a^z$ has no solutions in positive integers with $a\gt 2$. For suppose the relation holds. Without loss of generality we may assume $x\le y$. Then $$1+a^{y-x}=a^{z-x}.$$ This is only possible if $y=x$, $a=2$, and $z=x+1$.

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The proof of the Calabi conjecture. This is much beyond the other examples here in level, and is also the most illustrious one.

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Advanced "analytics" are progressing slower in (American) football than in basketball and especially baseball, perhaps because the interactions are more complex. In baseball, practically all tangible interactions are mediated by the baseball itself. The sorts of ancillary actions represented by picks in basketball and by blocks in football are much reduced ...

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In 2012, Javier Lopez Pena and Hugo Touchette published a paper (http://arxiv.org/pdf/1206.6904v1.pdf) describing the application of network theory to the analysis of football team strategies. Using passing data recorded during the 2010 FIFA World Cup, they constructed networks and calculated centralities to "identify play pattern, determine hot-spots on the ...

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"Does a tangent exist at x=0 to y=sgn(x)?." It all depends on how you are defining the tangent. If you are calling the line with the slope equal to the derivative at a point as a tangent, then the answer would be no, because now sgn(x) is discontinous. If you are calling the limiting chord a tangent then the Y-axis is tangent at x=0.

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Begin with dimension $1$, as we can use the motivation behind homotopy groups. So, we have $1$-dimensional loops, and we count the number of 'distinct' loops which wind round the holes, if any. The circle has one loop going round itself $n$ times, so the homology group of dimension one is $\Bbb Z$. With the disc, all loops can be shrunk to a point, so the ...

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In some contexts, when direct products are defined (e.g. vector spaces, groups, modules...) given two maps $$f: A \to B \\ g : C \to D$$ one defines $$f \times g : A \times C \to B \times D$$ as $(a,c) \mapsto (f(a), g(c))$. Sometimes (when you work with modules or vector spaces) products are denoted with the symbol $\oplus$ and are called direct sums. This ...

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It's just a question of terminology and usefulness. As far as terminology, there's the standard definition of linear independence and there's your definition. We need to pick one to be called "linearly independent" and the other needs to be called something else. It just so happens that the standard definition is the one we've picked. Of course this ...

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We still have very little progress towards the Riemann hypothesis (RH), which may well be one of the most important open problems in mathematics. This link to MathOverflow (the professional math site on StackExchange) may impart just how far we are from solving this problem. This difficulty is further obscured by the attractiveness of this problem to ...

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The most complex topics in mathematics are undoubtedly Cauchy-Gorsaut, Chauchy's Integral formula,De Moivre's Formula, Picard's Theorem and Louville's Thoerem among other other results which can be obtained after complex thinking and careful analysis .

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Let $C$ be a category containing a final objects, products and coproducts either finite or arbitrary. Let $1$ denote the final object. For $A$ an object of $C$, the space $Hom(1, A)$ are often called the elements of $A$. The reason for this can be explained as follows, if there is an adjuction $C\to Set$ that is a forgetful funtor that preserves products ...

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They knew most of the cutting-edge mathematics of their times, by studying with the people who invented it, or by voraciously reading all they could put their hands on. They maintained extensive contact with colleagues, be it by meeting in some Academy or other or just by mail. Not that far from how today's prominent mathematicians came to know what they ...

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You could horizontally flip the $5$, making a $2$. $$2+13+15=30$$ You could also vertically flip $15$, making $12$. $$7+11+12=30$$

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I would have liked to put this as a comment but I do not have enough points to do it. I think there is some magic in this formula because it tells us that the Lie derivative $\mathscr{L}_X$ is homotopic to zero with the homotopy $i_X$ going from top-right $\Omega^{p+1}(M)$ to bottom-left $\Omega^p(M)$ diagonally in the following diagram: $$\require{AMScd} ... 1 I would argue neither is harder in general, or at least there's no objective way of making such a judgement. Finding a counterexample doesn't necessarily require an ingenious guess though, you can find one in a relatively systematic fashion. When proving a theorem, the main question you are trying to answer is "why do these hypotheses imply the conclusion?" ... 0 Going to research talks while being an undergraduate can be very challenging (frustrating I would say). This is a subjective answer, from my own experience and some friends'. Empirically (very subjective too) it appears that taking exhaustive notes of what is happening in a talk is usually not so efficient. The pace is not really the same as in a classical ... 0 You are asking an incredibly broad question here. My best answer is to question your question. What exactly do you mean by a counter-example? We usually mean something that goes against our intuition. Formally though, a counterexample is just something that goes against a claim e.g. "All integers are greater than 0" has the obvious counter example of -1. ... 0 I think the opposite...disproving is easier than proving...reason is that for proving something you need to show that all cases under the given condition is going to be true...but for disproving something you need to provide just one example which is not in accordance with the given mathematical statement. Also before proving anything you will have to deal ... 2 I completely disagree with your statement that it is easier to prove things than it is to find counterexamples. There are statements that are easy to prove For every real valued function f, there exists a function g such that \forall x\in \mathbb R: f(x)<(g) and statements hard to prove. There are no integer solutions to the equation ... 0 I would call such an encyclopedia a series of textbooks. I think the closest thing should be all the Elements of Mathematics books by Bourbaki. They start with simple topics like set theory and algebra that do not have any prerequisites and try to cover all those topics completely. Anyone who read them will however tell you that it is nigh impossible to ... 0 You can try with Encyclopedia of Mathematical Sciences. It is a multi-volume work of surveys, and some volumes start at an elementary level, like the one by Shafarevich on algebra. The whole work is not as elementary as calculus, though. 2 I would say yes in general. Of course this depends on what kind of course you take. The huge difference between ODEs and PDEs is that for ODEs one is usualy interested in finding a solution. For PDEs this proves to be way to complicated very fast. There are some simple cases where explicit solutions can be constructed, sometimes by reducing to an ODE. But ... 3 Short answer: no. Long answer: noooooooooooooooooooooo. Long serious answer: PDEs are tough. Very tough. And there are two types of PDEs in general: The type we can generalize to ODEs and the type we cannot. When solving PDEs, you consider your work finished when you reduced them to a set of ODEs, because solving the ODEs then should be "trivial". A ... 1 The deeper we dive into the ocean of mathematics, the stronger is my conviction that we actually do not know anything. Let me give some examples: 1. Every polynomial p(z) has a root in \mathbb{C} 2. For a Hilbert space H, H=W\oplus W^{\perp}, where W is a closed subspace of H and W^{\perp} is its orthogonal complement. 3. Every subgroup ... 2 Module 2 syllabus can be found in almost all books on Algebra but I recommend Dummit and Foote. For Module 1, I would recommend "Linear Algebra done right" by Axler for abstract approach avoiding Matrices, and "Linear algebra done wrong " by Sergie Treil (Google it for e-copy). These two books covers all topics espically Sergie's book, but Axler is ... 1 Take a peek at some no-cost alternatives around the 'net, like Treil's "Linear Algebra done Wrong" (very nice, but might not be all the abstract you'd want). Often lecture notes are available, and add a different explanation that helps you over some rough spot. -4 Basic Mathematics by Serge Lang for a start in Algebra, his Linear Algebra book was okay, look up other math books by Springer (publisher). 0 Using the integration by parts, one has \begin{eqnarray} \int_0^1\ln x\ln(1-x)dx&=&x\ln x\ln(1-x)|_0^1-\int_0^1x(\frac{\ln(1-x)}{x}-\frac{\ln x}{1-x})dx\\ &=&-\int_0^1(\ln x+\ln(1-x))dx+\int_0^1\frac{\ln x}{1-x}dx\\ &=&2-\frac{\pi^2}{6}. \end{eqnarray} Here$$ \int_0^1\frac{\ln x}{1-x}dx=-\frac{\pi^2}{6}  is well-known.

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We say "$A$ injects into $B$", and may write $f: A \hookrightarrow B$.

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Probably the most standard term of the sort you're looking for is "equipotent", though it isn't used particularly often (more often, people will just say two sets "have the same cardinality" or "are in bijection"). People who think about categories a lot sometimes say "isomorphic as sets".

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You can call them equinumerous.

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It seems you think your only choices are between pure math and software engineering. I doubt that is true even at an engineering university. If it is true, it seems you have already made your choice. And I have to agree there is no joy in the prospect of spending your one and only life doing something you really don't like. However, I'd like you to ponder ...

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http://www.purplemath.com/modules/index.htm http://www.sosmath.com/algebra/algebra.html Here you are as requested, hope it helps.

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