Question 1204483
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Review these rules to see what makes a subspace
Scroll down to "Definition  2.6.2: Subspace"
<a href = "https://math.libretexts.org/Bookshelves/Linear_Algebra/Interactive_Linear_Algebra_(Margalit_and_Rabinoff)/02%3A_Systems_of_Linear_Equations-_Geometry/2.06%3A_Subspaces">https://math.libretexts.org/Bookshelves/Linear_Algebra/Interactive_Linear_Algebra_(Margalit_and_Rabinoff)/02%3A_Systems_of_Linear_Equations-_Geometry/2.06%3A_Subspaces</a>



The zero vector is easy to prove it belongs in the set, since we can make all of the coordinates 0 and 0 is indeed rational. 
We have satisfied condition 1.


Furthermore, condition 2 works as well because any two n-tuples of the same size added together gets some other n-tuple in the set.
Adding any two rational numbers yields a rational number.


Condition 3 is where things break down. If c and q were rational, then c*q would be rational as well. 
But if c is irrational while q is rational, then c*q would be irrational. 
We have escaped the set of rational numbers and therefore we don't have closure with multiplication. 
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