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Apr 11 |
answered | Find an orthogonal vector to 2 vector |
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Apr 11 |
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Show that $f^{-1}(A \cup B) = f^{-1}(A) \cup f^{-1}(B)$ I think of $f^{-1}(S)$ informally as "stuff that lands inside $S$ (when I hit it with $f$)". You're being asked to show that "x lands inside $A\cup B$" iff "x lands inside $A$ or $B$", which is obviously true. You're also being asked to disprove the claim that "x lands inside $A\cap B$" iff "x lands inside both $A$ and $B$", but this claim is obviously true too, as Cameron Buie says. |
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Apr 11 |
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Show that $f^{-1}(A \cup B) = f^{-1}(A) \cup f^{-1}(B)$ I think there's a typo (twice) on the line *, where x should say f(x). |
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Apr 10 |
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Proving that an element in an algebra presentation is nonzero Forget about A for a second. The element xy-1 (in F[x,y]) is invertible if and only if there is some polynomial u = u(x,y) in F[x,y] such that (xy-1)u = 1. Now take deg (the usual degree map on F[x,y]) of both sides, to show that deg u = -2, which is impossible as u is a polynomial. So xy-1 is not invertible, and so the ideal (xy-1) is not all of F[x,y], so B = F[x,y]/(xy-1) is a non-zero ring. But if yx = 0, it's easy to see that x = y = 0, and so the ideal (xy-1) = (1) = F[x,y], and B is the zero ring. This is a contradiction. |
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Apr 10 |
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Proving that an element in an algebra presentation is nonzero At the risk of spamming, I'll make the final comment that these are not just very similar: they're actually the same argument. (The map $\varphi: F\langle x,y \rangle\to F[x,y]$ sending x to x and y to y is a ring homomorphism, and the grading I imposed on the non-commutative ring is just induced from the degree map on the commutative ring. So, in fact, you can push everything through this homomorphism, and say that if $xy-1$ could be somehow inverted, then $\varphi(xy-1)$ could too.) |
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Apr 10 |
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Proving that an element in an algebra presentation is nonzero If you prefer, think first of the commutative polynomial ring F[x,y], where this is obvious. Suppose u is an invertible polynomial, say uv = 1; then deg(uv) = deg(u) + deg(v) = deg(1) = 0, but all degrees are at least 0, so deg(u) = deg(v) = 0. In particular, deg(xy-1) = 2, so this can't be a unit. I'm just doing the same thing in the non-commutative case, a little more carefully. |
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Apr 10 |
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Proving that an element in an algebra presentation is nonzero The grading is on F<x,y>. I had a specific one in mind: the natural one whose nth part is the homogeneous polynomials of degree n. (So the 0th part is F, the 1st part is generated by x and y, the 2nd part is generated by x^2, xy, yx and y^2, ...) It's easy to check that this is a grading. Make sense of the statement "if $abf \neq 0$, then $\deg(f) \leq \deg(afb)$". |
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Apr 10 |
answered | Relationship between two variables that does not seem correlated |
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Apr 10 |
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Solving for x in a mod relation? 7 and 10 are very small numbers. Work out $7, 7^2, 7^3, 7^4, 7^5, \dots$ mod 10, and see if you can spot a pattern. Or even just notice that $7^2 = -1$, so $7^4 = 1$. |
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Apr 10 |
answered | complex maths puzzle problems |
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Apr 10 |
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Proving that an element in an algebra presentation is nonzero John's given a great answer, but here's a very direct method for 1: assume that yx = 0. Then, by right-multiplying by y, we see that y = 0; and by left-multiplying by x, we see that x = 0. Now 0 = xy = 1. So in fact A is just the zero ring: that is, the ideal (xy-1) is all of F<x,y>, so xy-1 was a unit in F<x,y>. That's obviously nonsense - pick your favourite reason, e.g. because of the natural $\mathbb{Z}_{\geq 0}$-grading on F<x,y>. (Informally, $\deg(f) \leq deg(afb)$ for all $a, b, f$: multiplying by stuff never reduces the degree, modulo some silly discussion about what deg(0) means.) |
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Apr 10 |
answered | How to find nearest point on line of rectangle from anywhere? |
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Apr 10 |
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Infinite sets don't exist!? @Kaz Precision is a way of thinking. Defining Q from Z like that looks a little embarrassing, sure. But here are its advantages: it's precise and clear; basically the same construction generalises to localisations of arbitrary rings (it's far less obvious that these exist); understanding how R and C came from Q gave us the fields Q_p and C_p; and so on. This isn't just nonsense. Precision is a way of thinking - without these tools, some genuine mathematical problems would still be unsolved. |
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Apr 10 |
answered | Distance between point and line in the complex plane |
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Mar 21 |
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Why are these two equivalent? (Modular multiplicative inverse) I never said that your answer was wrong. I only said it was unhelpful. The OP asked "why?", and you said "that's just what it means". That's not mathematics. In any case, I wouldn't have learnt anything from your comment when I was a beginner at modular arithmetic, so I downvoted it. Isn't that how this site works? |
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Mar 21 |
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Why are these two equivalent? (Modular multiplicative inverse) The downvotes were mine. They were answers that I would have found very unhelpful when I was a complete beginner at modular arithmetic. "That's what $e^{-1}$ means" does not explain why a number with this property exists, or why you can do things like multiply both sides of a congruence by $e$. If the OP already knew these facts, I'm sure (s)he would have had no problem multiplying one equation by $e$ or $e^{-1}$ to get the other. |
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Mar 21 |
answered | Why are these two equivalent? (Modular multiplicative inverse) |
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Mar 21 |
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For a given metric space, how to show that $d(f(x),x) \ge \epsilon$? First of all, prove the hint (you'll need that X is compact - is this in the question?). Now, suppose there is not some e > 0 such that g(x) >= e for all x. Then, for all e' > 0, there is some x such that g(x) < e'. As g is continuous, it will attain both its infimum and its supremum... |
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Mar 21 |
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Is the free group on an empty set defined? I can't work out from the comments and postscripts whether there's still an issue here, but while the free product and the direct product match up in certain small cases, it's still far more honest to call it a free product, since it's a free product by definition and a direct product by accident. |
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Mar 21 |
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Linearly independent set can be completed to a basis $\{e_2\}$ is a linearly independent set, yes. Extensions to a basis, as you correctly guess, are usually very non-unique. Given a linearly independent set, look at the vector space W spanned by it; if this space is equal to V, then your linearly independent set is a basis; if it's not equal to V, then pick a vector in V but not in W. Notice that it's linearly independent to all of the rest. Throw it into your linearly independent set, and repeat. Can you prove that you'll eventually span all of V? (This argument only works for finite-dimensional vector spaces.) |
