# Donkey_2009

Cambridge, United Kingdom

 10h comment Korean Math Olympiad (Construct rectangle)I think it simplifies things just to point out that the number of white squares covered by each tile is odd; since there are an even number of such squares, there must be an even number of tiles. Nice method, though: +1. 10h revised Korean Math Olympiad (Construct rectangle)deleted 2 characters in body 17h reviewed Leave Closed Finding pattern (5,8,2,1) 17h reviewed Reopen Induction proof; help needed. 17h reviewed No Action Needed Percentage Change of (A+B) different than percentage change of A + percentage change of B 17h reviewed No Action Needed Distributing partially known data between n parties 17h reviewed Leave Open function inequality $f(x+y)+y \leq f(f(f(x)))$ 17h reviewed Close Exponent and powers - Find the value of x 2d revised compact set and proveFixe MathJAX formatting Dec 4 comment Is there any other name for limits?@aspiring No - that's a programming question, and you should post it on Stack Overflow instead. Dec 3 revised Determine if equation will generate perfect squaresRetagged Dec 3 reviewed Close computer graphic question about Polygon Clipping? Dec 3 answered Is $f(x)>g(x) \iff \frac {\operatorname d}{\operatorname d x }f(x)>\frac {\operatorname d}{\operatorname d x }g(x)$? Dec 3 comment Difference of squaresOr, as I prefer to think about it, the difference of squares formula is just a congruence written in a non-fancy way. Dec 3 comment Is there any other name for limits?No problem! I'm glad I was able to help. Dec 1 awarded Fanatic Nov 29 accepted Is every injective rational function $f:\mathbb Q\to\mathbb Q$ a polynomial? Nov 29 comment Is every injective rational function $f:\mathbb Q\to\mathbb Q$ a polynomial?Yes it is. I think I might have to reformulate the question, but thanks. Nov 29 asked Is every injective rational function $f:\mathbb Q\to\mathbb Q$ a polynomial? Nov 29 comment Polynomials irreducible over $\mathbb{Q}$ but reducible over $\mathbb{F}_p$ for every prime $p$OK - got the answer myself. If $f\mod p$ is inseparable, then it has some irreducible factor of the form $g(X^p)$ for some irreducible $g$. But every element of $\mathbb F_p$ is a $p$-th power so, applying the Frobenius automorphism, we can write $g(X^p)$ as $(h(x))^p$, where the coefficients of $h$ are the $p$-th roots of the coefficients of $g$. But this is clearly reducible.