SOLUTION: Prove that the set of all n-tuples of rational numbers {(q1, q2, . . . , qn)|qi ∈ Q} ⊂ R^n is NOT a subspace of R^n.

Question 1204483: Prove that the set of all n-tuples of rational numbers {(q1, q2, . . . , qn)|qi ∈ Q} ⊂ R^n is NOT a subspace of R^n.
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52781) (Show Source): You can put this solution on YOUR website!
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The simplest way is to start from 1d-case. 1d-case is simply the set of rational numbers as a part of all real numbers on the number line. If you multiply any rational number by irrational number, you will get irrational number, which is obvious. Thus, when you multiply rational number by irrational, you leave the set of rational numbers. Mathematicians say that "rational numbers cannot withstand multiplication by irrational numbers and become irrationals". It explains everything in 1d-case. In multidimensional space, everything is the same (everything is similar).

Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

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

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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