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.

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$Direct\;proof.\\Let\;y\;=\;\frac x2\in Q_+\\Since\;x>0\Rightarrow\frac x2<x\\y<x\\Proof\;by\;contradiction.\\Let\;x=\frac ab\;be\;the\;smallest\;positive\;rational\\number.\\Consider\;\;y\;=\frac a{b+1}.\\Since\;both\;a,\;b\;are\;natural,\;we\;have\;\\\frac a{b+1}<\frac ab\Rightarrow y<x\\So,\;contradiction.\\Thus, \,for\;every\;x\in Q_{+\;}there\;is\;always\;some\\y\in Q_+\;such \,that\;\;y<x.$ $D i rec t p roo f . L e t y = \frac{x}{2} \in Q_{+} S in ce x > 0 \Rightarrow \frac{x}{2} < x y < x P roo f b y co n t r a d i c t i o n . L e t x = \frac{a}{b} b e t h e s ma ll es t p os i t i v e r a t i o na l n u mb er . C o n s i d er y = \frac{a}{b + 1} . S in ce b o t h a, b a re na t u r a l, w e ha v e \frac{a}{b + 1} < \frac{a}{b} \Rightarrow y < x S o, co n t r a d i c t i o n . T h u s, f or e v ery x \in Q_{+} t h ere i s a lw a ys so m e y \in Q_{+} s u c h t ha t y < x .$

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.

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Question ID: mtid-5-stid-8-sqid-3975-qpid-2674