Irrational numbers forming dense subset

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SUMMARY

The discussion focuses on proving that the set A, defined as A={n+ma | n,m are integers and a is an arbitrary irrational number greater than 0}, is dense in the real numbers R. The proof involves demonstrating by contradiction that the smallest positive linear combination of any two real numbers divides both numbers. It concludes that the set of positive linear combinations of an irrational number a and 1 lacks a smallest element, leading to the conclusion that the greatest lower bound of this set is 0, thereby confirming the density of set A in R.

PREREQUISITES
  • Understanding of irrational numbers and their properties
  • Familiarity with linear combinations in real analysis
  • Knowledge of the concept of density in real numbers
  • Basic proof techniques, particularly proof by contradiction
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  • Study the properties of irrational numbers and their implications in real analysis
  • Learn about linear combinations and their applications in proving density
  • Explore the concept of greatest lower bounds in real number sets
  • Review proof by contradiction techniques in mathematical proofs
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Mathematicians, students studying real analysis, and anyone interested in the properties of irrational numbers and their role in forming dense subsets of real numbers.

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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)
 
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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.
 

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