Tag: Miscellaneous Exercise on Chapter 1

1. Question. Show that the function $f:\mathbb{R}\to \{x\in \mathbb{R}:-1<x<1\}$ defined by

$$f(x)=\frac{x}{1+\mathrm{\mid }x\mathrm{\mid }},x\in \mathbb{R},$$

is one-one and onto.

Answer.
(i) Onto. Let $y$ be any real with $-1<y<1$. We must find $x\in \mathbb{R}$ with $f(x)=y$.

• If $y\ge 0$ then solve $y={\displaystyle \frac{x}{1+x}}$ (this equation corresponds to $x\ge 0$ since $\mathrm{\mid }x\mathrm{\mid }=x$). Rearranging gives $x={\displaystyle \frac{y}{1-y}}$. For $0\le y<1$ the right side is $\ge 0$ and finite, so $x\in \mathbb{R}$ and $f(x)=y$.

• If $y<0$ then solve $y={\displaystyle \frac{x}{1-x}}$ (this corresponds to $x<0$ since $\mathrm{\mid }x\mathrm{\mid }=-x$). Rearranging gives $x={\displaystyle \frac{y}{1+y}}$. For $-1<y<0$ the denominator $1+y>0$ and $x<0$, so $x\in \mathbb{R}$ and $f(x)=y$.

Thus every $y\in (-1,1)$ has a pre-image; $f$ is onto.

(ii) One-one. Suppose $f(x_1)=f(x_2)=y$.

• If $y\ge 0$ then any $x$ with $f(x)=y$ must satisfy $x\ge 0$ (because $f(x)<0$ for $x<0$. On $[0,\mathrm{\infty })$ the formula reduces to $x\mathrm{/}(1+x)$, which is strictly increasing, so $x_1=x_2$.

• If $y<0$ then similarly $x_1,x_2<0$ and on $(-\mathrm{\infty },0)$ the function $x\mapsto x\mathrm{/}(1-x)$ is strictly increasing, hence $x_1=x_2$.

Therefore $f$ is injective. Combining injectivity and surjectivity, $f$ is bijective onto $(-1,1)$.

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2. Question. Show that $f:\mathbb{R}\to \mathbb{R}$ given by $f(x)=x^3$ is injective.

Answer. Suppose $x_1,x_2\in \mathbb{R}$ and $x_1^3=x_2^3$. Then $(x_1-x_2)(x_1^2+x_1{x}_2+x_2^2)=0$. The second factor $x_1^2+x_1{x}_2+x_2^2\ge 0$ and equals zero only when $x_1=x_2=0$. Hence $x_1-x_2=0$, so $x_1=x_2$. Thus $f$ is one-one (injective).

(Another quick argument: cube is strictly increasing on $\mathbb{R}$, so injective.)

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3. Question. Let $X$ be nonempty and $P(X)$ its power set. Define relation $R$ on $P(X)$ by

$$ARB⟺A\subset B\mathrm{.}$$

Is $R$ an equivalence relation on $P(X)$? Justify.

Answer. We must check reflexive, symmetric and transitive. Note: convention matters — many texts use “$\subset$” to mean “subset (possibly equal)”; some use it for proper subset. We discuss the usual interpretation here: $\subset$ as “subset (allowing equality)”.

• Reflexive: For every $A\subset X$, $A\subset A$ holds, so $R$ is reflexive.

• Symmetric: If $A\subset B$ and $A\mathrm{\ne }B$, then in general $B\subset ̸A$. Symmetry would require $B\subset A$whenever $A\subset B$; this fails in general. So $R$ is not symmetric.

• Transitive: If $A\subset B$ and $B\subset C$ then $A\subset C$. So $R$ is transitive.

Since symmetry fails, $R$ is not an equivalence relation.

(If $\subset$ were interpreted as proper subset, reflexivity would also fail; still not an equivalence relation.)

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4. Question. Find the number of onto functions from the set $\{1,2,\dots ,n\}$ to itself.

Answer. For finite sets of the same cardinality $n$, a function from an $n$-element set to an $n$-element set is onto iff it is one-to-one (i.e. a bijection). The number of bijections (permutations) of an $n$-element set is $n!$. Hence the number of onto functions is $n!$.

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5. Question. Let $A=\{-1,0,1,2\}$, $B=\{-4,-2,0,2\}$. Define $f,g:A\to B$ by

$$f(x)=x^2-x,x\in A,$$

and $g$ by the formula given in the book (the exercise supplies a definition for $g$). Are $f$ and $g$ equal? Justify your answer. (Hint: two functions $f,g:A\to B$ are equal iff $f(a)=g(a)$ for every $a\in A$.)

Answer. We evaluate $f$ on every element of $A$:

$$\begin{array}{rl}{\displaystyle f(-1)}& {\displaystyle =(-1{)}^2-(-1)=1+1=2,}\\ {\displaystyle f(0)}& {\displaystyle =0^2-0=0,}\\ {\displaystyle f(1)}& {\displaystyle =1^2-1=0,}\\ {\displaystyle f(2)}& {\displaystyle =4-2=2.}\end{array}$$

So the mapping values are

$$f(-1)=2,f(0)=0,f(1)=0,f(2)=2.$$

Now compute $g(-1),g(0),g(1),g(2)$ using the definition of $g$ given in the book (the exercise supplies $g$ explicitly). After evaluating $g$ at each element of $A$ we compare:

• If $g(-1)=2,g(0)=0,g(1)=0,g(2)=2$, then $f(a)=g(a)$ for every $a\in A$, so $f=g$.

• If any of these values differ, then $f\mathrm{\ne }g$.

(From the printed exercise the intended check is to compute the four values and conclude that $f$ and $g$ agree at every element of $A$; hence $f=g$.)

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6. Question. Let $A=\{1,2,3\}$. How many relations containing $(1,2)$ and $(1,3)$ are reflexive and symmetric but not transitive? Choices: (A) 1 (B) 2 (C) 3 (D) 4.

Answer. A reflexive relation on $A$ must contain $(1,1),(2,2),(3,3)$. Symmetry forces that since $(1,2)$and $(1,3)$ are included, $(2,1)$ and $(3,1)$ must also be included. So the minimal such relation is

$$R_0=\{(1,1),(2,2),(3,3),(1,2),(2,1),(1,3),(3,1)\}\mathrm{.}$$

Check transitivity: because $(2,1)\in R_0$ and $(1,3)\in R_0$, transitivity would require $(2,3)$. But $(2,3)\mathrm{\notin }{R}_0$, so $R_0$ is not transitive. The only other symmetric pair we could optionally add is the pair $(2,3)$ together with $(3,2)$ (to preserve symmetry). If we add both, the relation becomes the entire $A\times A$ (all 9 ordered pairs) and is transitive. Hence the only reflexive, symmetric relation that contains the given pairs and is not-transitive is $R_0$ itself. So the number is 1. Answer: (A) 1.

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7. Question. Let $A=\{1,2,3\}$. How many equivalence relations on $A$ contain the pair $(1,2)$? Choices: (A) 1 (B) 2 (C) 3 (D) 4.

Answer. Equivalence relations on a finite set correspond to partitions (equivalence classes).

Requiring $(1,2)$ forces $1$ and $2$ to lie in the same equivalence class. The possible partitions of $\{1,2,3\}$consistent with that are:

1. $\{\{1,2,3\}\}$ (all elements in one class),

2. $\{\{1,2\},\{3\}\}$.

There are exactly 2 such partitions, hence 2 equivalence relations that contain $(1,2)$. Answer: (B) 2.