Question 3.43: Let Q+ = {x ∈ Q | x ≥ 0}. If a/b, c/d ∈ Q+ prove that a ⁄ b ......
Let \mathbb{Q}_{+}=\lbrace x\in\mathbb{Q}\mid x\geqslant\mathbf{0}\rbrace. If a/b, c/d ∈ \mathbb{Q}_{+} prove that
{\frac{a}{b}}={\frac{c}{d}}\Rightarrow\left|{\frac{a+b}{{hcf}(a,b)}}\right|=\left|{\frac{c+d}{\operatorname{hcf}(c,d)}}\right|.Deduce that the prescription
f{\biggl(}{\frac{a}{b}}{\biggr)}=\left|{\frac{a+b}{\operatorname{hcf}(a,b)}}\right|defines a mapping f:\mathbb{Q}_{+}\to\mathbb{Q}_{+}. Is f a bijection?
Let \alpha=\operatorname{hcf}(a,\,b). Then we have a=a^{\prime}{\alpha},\,b=b^{\prime}{\alpha}~\mathrm{and}~a/b=a^{\prime}/b^{\prime}, the latter quotient being ‘in its lowest terms’ in the sense that \operatorname{hcf}(a^{\prime},b^{\prime})=1. Similarly, if \beta=\mathrm{hcf}\left(c,\,d\right)\,\mathrm{then}\ c=c^{\prime}\beta,\,d=d^{\prime}\beta\ \mathrm{and}\ c/d=c^{\prime}/d^{\prime} the latter being in its lowest terms. Thus, if a/b = c/d we have a^{\prime}/b^{\prime}=c^{\prime}/d^{\prime} so that either a^{\prime}=c^{\prime}, b^{\prime}=d^{\prime}\operatorname{or}a^{\prime}=-c^{\prime},b^{\prime}=-d^{\prime}. It follows that
\left|\frac{a+b}{\alpha}\right|=|a^{\prime}+b^{\prime}|=|c^{\prime}+d^{\prime}|=\left|\frac{c+d}{\beta}\right|.The first part of the question is precisely the condition that is necessary to ensure that the given prescription defines a mapping from \mathbb{O}_{+} to itself. This mapping is not a bijection. For example, it fails to be injective: we havef(2/3) = 5/1=f(3/2).