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answered Triangulation of matrices
Apr
21
comment For $n \ge 2$ , does every linear operator on $\mathbb R^n$ has an invariant subspace of dimension $2$ ?
@user ${\mathbb R}^n={\textsf{Ker}}((T-\lambda \textsf{id})^{s_1})$ by the Cayley-Hamilton theorem
Apr
21
revised For $n \ge 2$ , does every linear operator on $\mathbb R^n$ has an invariant subspace of dimension $2$ ?
edited body
Apr
15
accepted measure of a set invariant by rational translation
Apr
13
answered Prove $\{ (x,y) \in [0,1]^2: x-y\in \mathbb{Q}\}$ is measurable.
Apr
13
answered measure of a set invariant by rational translation
Apr
12
revised measure of a set invariant by rational translation
added 886 characters in body
Apr
12
comment measure of a set invariant by rational translation
@FardadPouran No, there's no mod 1. Only intersection with $[0,1]$. The set $A$ is free to do anything it wishes outside $[0,1]$.
Apr
12
asked measure of a set invariant by rational translation
Apr
8
comment Find a function $ f : R^2 → R $ such that $ f(λv) = λf(v) $, $\forall \lambda \in R, v ∈ R^2$, yet $f$ is not linear.
This is definitely a duplicate of some older question, whose location I cannot remember unfortunately
Apr
8
accepted Defining integer sum without using infinite sets
Apr
7
comment Defining integer sum without using infinite sets
@RobArthan $\phi$ is obvious in ZFC : in this case $\gamma$ is just $\alpha+\beta$. The axiom of infinity is used in the induction, which presuppose the existence of $\omega$
Apr
7
revised Integer induction without infinity
added 19 characters in body
Apr
7
asked Defining integer sum without using infinite sets
Apr
7
accepted Integer induction without infinity
Apr
7
comment Integer induction without infinity
Ah yes, that’s because we chose $M$ so that $\forall n \psi(n)$ is false ...
Apr
7
comment Integer induction without infinity
I don't quite follow your third paragraph. I agree that $A$ has been shown to be a class without a least element. But I fail to see how you deduce a contradiction from this. How do you find a $k$ such that $M\models k\in A$ ? $M$ satisfies the axiom of foundation for its sets, it does not necessarily satisfy it for its classes.
Apr
7
asked Integer induction without infinity
Apr
5
accepted Small integral representation as $x^2-2y^2$ in Pell's equation
Apr
5
comment Small integral representation as $x^2-2y^2$ in Pell's equation
The question has now become, can $|u|\leq \sqrt{2k}$ be strengthened to $|u|\leq \sqrt{k}$ ?