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 Apr 28 asked Essentially bounded function which is continuously bounded Riemann integrable? Apr 28 comment If f*g is Riemann integrable, g continuous, nonzero and bounded, show that f is Riemann integrable @Arthur Upvoted, you beat me to the answer! Apr 28 answered If f*g is Riemann integrable, g continuous, nonzero and bounded, show that f is Riemann integrable Apr 28 revised Matrices Eqvialence Relation edited body Apr 28 comment Matrices Eqvialence Relation Yes, thanks. Will change it Apr 28 reviewed Approve For what values of $a$ will $y=ax$ be a tangent to $x^2+y^2+20x-10y+100=0$ Apr 28 reviewed Approve $\inf A = -\sup(-A)$ Apr 28 reviewed Approve From Cosine formula between two vectors to Schwarz inequality and Triangle inequality? Apr 28 answered Matrices Eqvialence Relation Apr 28 comment Logic - Binomial Theorem I believe there is an answer here: math.stackexchange.com/questions/11601/… Apr 28 revised Why do we study real numbers? added 485 characters in body Apr 28 answered Why do we study real numbers? Apr 28 comment Proof that there are infinitely many primes (Euclid) When you learn topology, try reading Furstenberg's topological proof of infinite primes! Apr 28 accepted Matrix that satisfies polynomial $x^{n}-1$ is diagonalizable Apr 27 comment Matrix that satisfies polynomial $x^{n}-1$ is diagonalizable @Omnomnomnom Shouldn't it be every eigenvalue is a root of the minimal polynomial and thus $p$, but not every root of $p$ is an eigenvalue? Apr 27 comment Matrix that satisfies polynomial $x^{n}-1$ is diagonalizable Update: $p$ isn't necessarily the characteristic polynomial Apr 27 revised Matrix that satisfies polynomial $x^{n}-1$ is diagonalizable added 56 characters in body Apr 27 comment Matrix that satisfies polynomial $x^{n}-1$ is diagonalizable Wait... How do we know the eigenvalues are precisely the roots of $p$? ($p$ is not necessarily the characteristic polynomial here, sorry for the confusing notation) Apr 27 comment Matrix that satisfies polynomial $x^{n}-1$ is diagonalizable Ok I think I understand. Apr 27 asked Matrix that satisfies polynomial $x^{n}-1$ is diagonalizable