Irrational numbers forming dense subset

In summary, the conversation is about proving that the set A, defined as the set of all numbers in the form of n+ma where n and m are integers, is dense in the real numbers. The conversation discusses using contradiction to show that the smallest positive linear combination of any two real numbers divides both numbers, and deducing that the set of positive linear combinations of irrational numbers and 1 does not have a smallest element. It also explores how the greatest lower bound of the set of positive linear combinations is 0 and how this relates to A containing all its multiples.
  • #1
bw0young0math
27
0
Hello. I have some problems with proving this. It is difficult for me. Please help me.:confused:

"For arbitrary irrational number a>0, let A={n+ma|n,m are integer.}
Show that set A is dense in R(real number)
 
Physics news on Phys.org
  • #2
Let's say that $x$ divides $y$ if there exists an integer $k$ such that $y=kx$. Also, let's call any number of the form $am+bn$ where $a,b\in\mathbb{R}$ and $m,n\in\mathbb{Z}$ a linear combination of $a$ and $b$.

Prove by contradiction that the smallest positive linear combination of any two real numbers divides both numbers. Deduce that the set of positive linear combinations of $a\in\mathbb{R}\setminus\mathbb{Q}$ and 1 does not have the smallest element (otherwise, $a$ and 1 would be commensurate). Next show that the greatest lower bound of the set of positive linear combinations is 0. Now that you have a positive linear combination as small as you'd like, note that $A$ contains all its multiples.
 

FAQ: Irrational numbers forming dense subset

What is the definition of "Irrational numbers forming dense subset"?

The term "irrational numbers forming dense subset" refers to a set of irrational numbers that are infinitely close to each other, with no gaps or empty spaces between them. This means that any real number can be approximated by an irrational number in the subset.

How are irrational numbers formed into a dense subset?

Irrational numbers are formed into a dense subset by taking a set of irrational numbers and arranging them in a way that there are no gaps between them. This requires the numbers to be infinitely close to each other, with no empty spaces in between.

What is the significance of irrational numbers forming a dense subset?

The significance of irrational numbers forming a dense subset is that it shows the completeness of the real number system. It means that there are no "holes" or gaps in the number line and that any real number can be approximated by an irrational number. This has important implications in mathematics, particularly in the study of limits and continuity.

Can rational numbers form a dense subset?

No, rational numbers cannot form a dense subset. This is because there will always be gaps or empty spaces between rational numbers, no matter how close they are to each other. This is because rational numbers can be expressed as fractions, and there will always be infinitely many fractions between any two rational numbers.

What is an example of a set of irrational numbers forming a dense subset?

An example of a set of irrational numbers forming a dense subset is the set of all square root of prime numbers. This set includes numbers such as √2, √3, √5, √7, and so on. These numbers are all irrational and are infinitely close to each other, forming a dense subset of the real number line.

Back
Top