Notice that f^+=\max\{f,0\} and f^-=\max\{-f,0\}. If g and h are continuous functions, so is \max\{g,h\}=\frac{g+h+|g-h|}{2}.

Here is one simple way to show that (9 - 5 \sqrt{3}) ( 2- \sqrt{2}) is not a square in \mathbb{Q}(\sqrt{2}, \sqrt{3}). Assume that we have an equality (9-5 \sqrt{3})(2-\sqrt{2}) = \alpha ^2 ...

I think the example with G=GL_2(\mathbb R) and X=\mathbb R^2 you are trying to make is not correct. If you want to use two balls with different centers, then there is no transformation in G ...

As you've discovered, evaluation no longer works like you want it to. Another illustration is that ix=xi, but it isn't valid if j is substituted for x. The problem is that x is assumed to be ...

For brevity let x^6 +x^5+ x^4 +x^3+ x^2+x+1=f(x).... For x\in \Bbb Z we have: (Ia).\quad |x|\geq 4\implies f(x)=(x^6+x^4+x^2)(1+1/x)+1\geq \geq (x^6+x^4+x^2)(1-1/|x|)+1> > x^6(1-1/|x|)\geq ...

\begin{align*} 3\cos^2 x & = \sin^2x\\ 3 & = \frac{\sin^2x}{\cos^2x}\\ \tan^2x & = 3\\ \tan x & = \sqrt{3}\\ \tan x & = \tan 60 \end{align*} From there, you can use the tangent equation: x= \pm 60+180k