Question 7 of 12

Question 7

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A relation on $\{(m,n): m,n\in\mathbb{Z}, n\neq0\}$ defined by cross-ratio equality, and the function $f(x)=\lfloor x^3+0.5\rfloor$ from $\mathbb{R}$ to $\mathbb{Z}$.

Find. (a) whether the relation is an equivalence relation and, if so, its equivalence classes; (b) whether $f$ is injective, surjective, and invertible.

Approach. Check reflexivity, symmetry, and transitivity of the fraction-equality relation directly; analyze $f$ by exhibiting a collision (non-injective witness) and by using the intermediate-value/continuity of $x^3$ to argue surjectivity.

1. 7a) Reflexive. $(m,n)\sim(m,n)$ since $m/n=m/n$ trivially. Symmetric. If $m/n=p/q$ then $p/q=m/n$, so $(p,q)\sim(m,n)$. Transitive. If $m/n=p/q$ and $p/q=r/s$ then $m/n=r/s$ (equality of real numbers is transitive), so $(m,n)\sim(r,s)$. All three properties hold, so $\sim$ is an equivalence relation.
2. 7a) Equivalence classes. Two pairs are related exactly when they represent the same ratio $m/n$; so each equivalence class is the set of all integer pairs $(m,n)$, $n\neq0$, that reduce to one particular rational number. The classes are in exact one-to-one correspondence with $\mathbb{Q}$ (each class has a unique lowest-terms representative $(a,b)$ with $\gcd(a,b)=1$, $b>0$). $\boxed{\text{equivalence classes} \leftrightarrow \mathbb{Q} \text{ (the rational numbers)}}$
3. 7b) Not one-to-one — exhibit a collision. Compute $f$ at two nearby points: $f(0)=\lfloor 0+0.5\rfloor=\lfloor0.5\rfloor=0$ and $f(0.7)=\lfloor0.343+0.5\rfloor=\lfloor0.843\rfloor=0$. Two different inputs ($x=0$ and $x=0.7$) give the same output ($0$), so $f$ is not injective — in fact every $x$ with $x^3\in[-0.5,0.5)$ maps to $0$.
4. 7b) Onto. $g(x)=x^3$ is continuous and strictly increasing on $\mathbb{R}$, ranging over all of $\mathbb{R}$ as $x$ ranges over $\mathbb{R}$. For any target integer $k$, the interval $[k-0.5,\,k+0.5)$ is nonempty, and by the intermediate value theorem some $x$ satisfies $x^3\in[k-0.5,k+0.5)$ (e.g. $x=\sqrt[3]{k}$ gives $x^3=k$ exactly), so $f(x)=\lfloor x^3+0.5\rfloor=k$. Every integer $k$ is hit, so $f$ is onto.
5. 7b) Inverse. A function has an inverse iff it is a bijection (both one-to-one and onto). $f$ is onto but not one-to-one, so $f$ does not have an inverse. $\boxed{f \text{ is not 1-1, is onto, and has no inverse}}$