MHB Irrational numbers forming dense subset

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The discussion centers on proving that the set A, defined as A={n+ma | n,m are integers and a is an arbitrary positive irrational number, is dense in the real numbers R. It introduces the concept of linear combinations of real numbers and explores the implications of the smallest positive linear combination of two real numbers. A contradiction is proposed to show that this smallest positive linear combination cannot exist, leading to the conclusion that the set of positive linear combinations of an irrational number and 1 lacks a smallest element. Ultimately, it is established that the greatest lower bound of these combinations is 0, confirming the density of set A in R. The discussion highlights the intricate relationship between irrational numbers and their linear combinations in the context of real number density.
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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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