Recent content by Gavran

For more complicated problems than this one in the original post the formula is $$ N=\frac1n\sum_{d|gcd(n_1,n_2,...,n_k)}^{}\phi(d)\binom{n/d}{n_1/d,n_2/d,...,n_k/d} $$ where: ## n ## - the number of balls, ## k ## - the number of colors, ## n_i ## - the number of balls of the i-th color, ##...

You do not need any formula to solve this problem. With two blue balls and two red balls you have two cases BBRR and BRBR. How many different options do you have when you insert the yellow ball into BBRR? How many different options do you have when you insert the yellow ball into BRBR?

The first statement of the theorem was published in 1887 by Georg Cantor. If M and N are two such sets that components M' and N' can be separated from them, of which it can be shown that |M|=|N'| and |M'|=|N|, then M and N are equivalent sets.

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.

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 Choice means the exclusion of the proof based on the Axiom of Choice and this does not mean there are vector spaces with no Hamel basis. Every vector space has a...

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 theorem is this: if there are injections f : A → B and g : B → A, then there is a bijection h : A → B.

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

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, 1=-4⋅29+13⋅(96-3⋅29)=13⋅96-43⋅29, 29=125-1⋅96, 1=13⋅96-43⋅(125-1⋅96)=-43⋅125+56⋅96, 96=471-3⋅125, 1=-43⋅125+56⋅(471-3⋅125)=56⋅471-211⋅125=260⋅125.

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 at all. To be in the set means occupying the part of the set but it does not mean that every element of the part of the set is occupied only once and this is the...

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 sequence ## \{a_i\} ##, other than the first, must be positive.

It can be proved that the sequence of convergents of any simple continued infinite fraction converges.

ODE | Slope fields and isoclines example

The question is more general than it looks and there is a difference between the coded term and the name of the coded term. For example, The coded term is 1. The name of the coded term is 1. In the first case 1 presents the coded term while in the second case 1 presents the symbol which is used...