Answer to Question #110879 in Discrete Mathematics for Justin Lee

Question

Let Q+ be the set of positive rational numbers. Prove that if x is in q+ there is some y in q+ such that y < x. Provide two proofs of this fact, one using a direct proof and one using a proof by contradiction.

Answer 1

$Direct\;proof.\\Let\;y\;=\;\frac x2\in Q_+\\Since\;x>0\Rightarrow\frac x2<x\\y<x\\Proof\;by\;contradiction.\\Let\;x=\frac ab\;be\;the\;smallest\;positive\;rational\\number.\\Consider\;\;y\;=\frac a{b+1}.\\Since\;both\;a,\;b\;are\;natural,\;we\;have\;\\\frac a{b+1}<\frac ab\Rightarrow y<x\\So,\;contradiction.\\Thus, \,for\;every\;x\in Q_{+\;}there\;is\;always\;some\\y\in Q_+\;such \,that\;\;y<x.$