Gavran's latest activity

The answer to the post #8: Yes, it does.

The Axiom of Choice and the existence of a Hamel basis imply each other.

A ⇔ B The exclusion of A does not produce B is false.

With Venus having a shorter period around the sun than Earth, with Earth having a much shorter period than Jupiter, Venus has now...

It can be shown that if the Axiom of Choice is false, then there are vector spaces with no Hamel basis. The exclusion of the Axiom of...

You already have the proof of the theorem on page 29 of the book. By the way, the more common statement of the Schroeder-Bernstein...

I'm working on the Schroether-Bernstein theorem, this is, the next step in the textbook "Introduction to topology and modern analysis"...

##zz^*## is always real.

Thks @Gavran ,I will check on your approach...

Although you have not used the standard step-by-step approach, your solution is also correct.

This is the question, I understand the concept, in ##\mathbb{Z_n}## an element is a is a unit if and only if gcd( a,n) =1. My...

The back-substitution in the context of the Euclidean algorithm is: 1=9-4⋅2, 2=29-3⋅9, 1=9-4⋅(29-3⋅9)=-4⋅29+13⋅9, 9=96-3⋅29...

To be on the set means occupying the whole set in a particular way and the word “onto” follows the definition of a surjective function...

Thank you PeroK. I think I got it. But to be sure, let me try to reiterate what you said in the context of my original question...

This is not a simple continued infinite fraction. $$ a_0+\frac{1}{a_1+\frac{1}{a_2+\frac{1}{a_3+\frac{1}{...}}}} $$ All integers in the...