3,346 reputation
412
bio website
location Struga
age
visits member for 2 years, 1 month
seen Aug 14 at 14:03

Aug
8
answered If $A,B,C\in L(X)$ commutes, then $A\leq B,$ and $C\geq 0$ follows $AC\leq BC.$
Jul
30
accepted Prove the equation $\vert d(x,z)-d(y,t)\vert\leq d(x,y)+d(z,t)$
Jul
30
comment Prove the equation $\vert d(x,z)-d(y,t)\vert\leq d(x,y)+d(z,t)$
thanks sir, now I understood very well after your solution
Jul
30
comment Prove the equation $\vert d(x,z)-d(y,t)\vert\leq d(x,y)+d(z,t)$
thanku very much sir,
Jul
30
comment Prove the equation $\vert d(x,z)-d(y,t)\vert\leq d(x,y)+d(z,t)$
I now this inequlity, but but please help me with the necessary steps proving, thanks
Jul
30
asked Prove the equation $\vert d(x,z)-d(y,t)\vert\leq d(x,y)+d(z,t)$
Jul
27
comment If there is the inverse operator of the operator A, then $(A^{-1})^{-1}=A$?
can you please explain this to add to your solution to be more clear
Jul
27
comment If there is the inverse operator of the operator A, then $(A^{-1})^{-1}=A$?
@ Jonas Meyer: i.e., The above solution is not correct
Jul
27
comment If there is the inverse operator of the operator A, then $(A^{-1})^{-1}=A$?
thanks sir, please help me for the last part, it is easy to see that $AB≠I$.
Jul
27
comment If there is the inverse operator of the operator A, then $(A^{-1})^{-1}=A$?
$A$ is a linear operator
Jul
27
asked If there is the inverse operator of the operator A, then $(A^{-1})^{-1}=A$?
Jul
1
awarded  Yearling
May
14
awarded  limits
May
1
answered Prove: $\frac{1}{2}\cdot\frac{2}{3}\cdots\frac{2n-1}{2n}<\frac{1}{\sqrt{2n+1}}$
May
1
answered Prove the inequality $\frac{a}{b}<\frac{a+k}{b+k},(a<b, a,b,k>0)$
Apr
21
answered Finding roots of cubing equation
Apr
21
revised Finding the limit of $\lim_{x\to 1} (x^2-\sqrt x)/(1-\sqrt x)$
added 9 characters in body; edited title
Apr
21
comment Finding the limit of $\lim_{x\to 1} (x^2-\sqrt x)/(1-\sqrt x)$
Now is correct,
Apr
21
comment Finding the limit of $\lim_{x\to 1} (x^2-\sqrt x)/(1-\sqrt x)$
$$\frac{x^2-\sqrt x}{1-\sqrt x}=\frac{x^2-\sqrt x}{1-\sqrt x}\cdot \frac{{1+\sqrt x}}{{1+\sqrt x}}=\frac{x^2+x^2\sqrt x- \sqrt x - x}{1-x}$$
Apr
21
answered Finding the limit of $\lim_{x\to 1} (x^2-\sqrt x)/(1-\sqrt x)$