Recent content by burak100

making discontinues function, continues.! Homework Statement Given a function g, which is not continuous everywhere and g is increasing. The problem is how to approach to this function to make it continuous. Homework Equations The Attempt at a Solution I am not sure but one way...

Actually I have a function of curvature for some transcendental curve. and this curve considered as a coxeter group generated by the reflections about the normal lines through two adjacent extrema of the curvature function. and it says that since this function of curvature is always non...

what does it mean "non integral"

I try to calculate possibilities, I, A, B, AB, BA, AAB, AAA, BAA and there are 8 elements , is it the answer?

So , should I try to find all possible products of A and B , or is there some trick to find it?

I am confused, because \langle A, B \rangle = \Big\langle \left( \begin{matrix} i & 0 \\ 0 & -i \end{matrix} \right) \left( \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right)\Big\rangle = \left( \begin{matrix} 0 & i \\ -i & 0 \end{matrix} \right) right? and this is just an element of...

\langle A, B \rangle = \left( \begin{matrix} 0 & i \\ -i & 0 \end{matrix} \right) and det(<A,B>)=-1, hence det(<A,B>) in GL_2(\mathbb{C}). right? on the other hand, if we want to show \langle A, B \rangle generated by A and B, we need to show that A and B are linear independent ?

Homework Statement A= \left( \begin{matrix} i & 0 \\ 0 &-i \end{matrix} \right) , B= \left( \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right) \\ Show that \langle A, B \rangle is subgroup of GL_2(\mathbb{C}). And Show that \langle A, B \rangle generated by A and B, and order of...

Homework Statement F(\mathbb{Q},\mathbb{R}) is the set of maps from \mathbb{Q} to \mathbb{R}. Then show that F(\mathbb{Q},\mathbb{R}) and \mathbb{R} have same potency (cardinal number?).. Homework Equations The Attempt at a Solution I am no tsure but I think I need to...

Ok. I think; we have \mathbb{R}= \bigcup\limits_{q \in \mathbb{Q}} f^{-1}(q), and we suppose that for all q in Q , f^{-1}(q) finite, then union would be countable but R is uncountable so contradiction... right?

I think that is right because just now we said " every element of R is in one of the set f^{-1}(q) "

I didn't understand clearly but ; for all q \in \mathbb{Q}, we suppose all the sets, f^{-1}(q) are finite. That means, every element of \mathbb{R} is one of these sets, f^{-1}(q) ...right?

yeah R is uncountable but I can't find a relation to this question..

infinite set means ; we can't find 1-1 correspondence between the set {1,...,n} and f^{-1}(q) , here f^{-1}(q) has n elements. So, can we say f^{-1}(q) is uncountable then we can't find 1-1 correspondence between the set {1,...,n}...?

[b] infinite set [b] Homework Statement f: \mathbb{R} \rightarrow \mathbb{Q} , show that there is a q \in \mathbb{Q} st. f^{-1}(q) is infinite set in \mathbb{R}. Homework Equations The Attempt at a Solution how can we show that is true?