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) $R=\{(a,b)\in\mathbb{R}^2 : a-b\in\mathbb{Z}\}$. (b) $f(x)=\sin(x)+x$ on $\mathbb{R}$.

Find. (a) verification of reflexivity/symmetry/transitivity and the equivalence classes; (b) injectivity, surjectivity, and existence of an inverse for f.

Approach. (a) check the three equivalence-relation axioms directly from the integer difference condition. (b) examine the sign of $f'(x)=\cos x+1$ to determine monotonicity (hence injectivity), and the limiting behaviour of f as $x\to\pm\infty$ to determine surjectivity onto $\mathbb{R}$.

1. Part (a) — reflexive, symmetric, transitive. Reflexive: $a-a=0\in\mathbb{Z}$ for every real a, so $(a,a)\in R$. Symmetric: if $a-b\in\mathbb{Z}$ then $b-a=-(a-b)\in\mathbb{Z}$ too, so $(a,b)\in R\Rightarrow(b,a)\in R$. Transitive: if $a-b\in\mathbb{Z}$ and $b-c\in\mathbb{Z}$, their sum $(a-b)+(b-c)=a-c$ is a sum of two integers, hence an integer, so $(a,c)\in R$. All three hold, so $\boxed{R\text{ is an equivalence relation}}$. Equivalence classes: $a\sim b$ exactly when a and b share the same fractional part, so each class is $$[a] = \{a+n : n\in\mathbb{Z}\},$$ i.e. one class per real number $r\in[0,1)$ — so there are uncountably many classes, each of which is a countable shifted copy $r+\mathbb{Z}$ of the integers, and together they partition $\mathbb{R}$.
2. Part (b) — monotonicity via the derivative. $$f'(x)=\cos x + 1$$ Since $\cos x\in[-1,1]$, $f'(x)\in[0,2]$ — never negative, so f is non-decreasing everywhere. $f'(x)=0$ only at the isolated points $x=\pi+2k\pi$ (where $\cos x=-1$), which have measure zero and do not form any interval, so f cannot be flat on any sub-interval; f is therefore strictly increasing on all of $\mathbb{R}$, hence injective (one-to-one).
3. Surjectivity. f is continuous, and since $|\sin x|\le1$ while $x\to\pm\infty$, $$\lim_{x\to+\infty}f(x)=+\infty,\qquad \lim_{x\to-\infty}f(x)=-\infty.$$ By the Intermediate Value Theorem a continuous function unbounded in both directions on all of $\mathbb{R}$ takes every real value, so f is onto (surjective) $\mathbb{R}$.
4. Conclusion. f is already a bijection $\mathbb{R}\to\mathbb{R}$ (strictly increasing + unbounded both ways), so $$\boxed{f\text{ has an inverse on }X=\mathbb{R},\ Y=\mathbb{R}\text{ --- no domain/range restriction needed}}$$ (the inverse has no elementary closed form, since $y=\sin x+x$ cannot be solved for x in terms of standard functions, but it exists because f is a bijection; $f^{-1}$ is itself continuous and strictly increasing, and is differentiable everywhere except at the images $y=f(\pi+2k\pi)$ of the isolated points where $f'(x)=0$).