Welcome to our community

Be a part of something great, join today!

Irrational numbers forming dense subset


New member
Jun 14, 2013
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)


Well-known member
MHB Math Scholar
Jan 30, 2012
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.