0E: Exercises

 

Exercise $\PageIndex{1}$

Define $h : \mathbb{Z} \rightarrow \mathbb{Z}$ by $h(x) = x^2+4$.  Determine (with reasons) whether or not $h$ is one−to−one and whether or not $h$ is onto.  

Exercise $\PageIndex{2}$

Suppose $f : A \rightarrow B$ and $g : B \rightarrow C$ are functions.  Prove or disprove the following statements: 

1. If $g \circ f$ is one-to-one then $g$ is one-to-one. 

2. If $g \circ f$ is one-to-one and $f$ is onto, then $g$ is one-to-one.

3.  If $g \circ f$ is onto and  $g$ is one-to-one, then  $f$ is onto.

Exercise $\PageIndex{3}$

Determine whether or not each of the following binary relations $R$ on the given set $A$ is reflexive, symmetric, antisymmetric, or transitive.  If a relation has a certain property, prove this is so; otherwise, provide a counterexample to show that it does not.  If $R$ is an equivalence relation, describe the equivalence classes of $A$. 

1. Let $S = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$. Define a relation $R$ on $A = S \times S$ by $(a, b) R (c, d)$ if and only if $10a + b \le 10c + d$. 

2. Let $A = \mathbb{Z} \backslash \{0\}$. Define a relation $R$ on $A$, by $a R b$ if and only if $ab > 0$.

3. Define a relation $R$ on $A = \mathbb{Z}$ by $a R b$ if and only if $4 | (3a + b)$.

4. Define a relation $R$ on $A = \mathbb{Z}$ by $a R b$ if and only if $3 | (a^2 - b^2 )$. 

5. Let $A = \mathbb{R}$, If $a,b \in \mathbb{R}$, define $a R b$ if and only if $a - b \in \mathbb{Z}$. 

6. Define a relation $R$ on the set $\mathbb{Z} \times \mathbb{Z}$ by $(a, b) R (c, d)$ if and only if $ac = bd$. 

7. Define a relation $R$ on $\mathbb{Z}$ by $a R b$ if and only if $2 \mid a^2+b$. 

8. Let $A = \mathbb{R}$, If $a,b \in \mathbb{R}$, define $a R b$ if and only if $a - b \in \mathbb{Q}$.

9. Let $A=\mathbb{R} \times \mathbb{R}$, If $(x,y),(x_1,y_1) \in \mathbb{R}\times \mathbb{R}$, define $(x,y) \, R \, (x_1,y_1)$ if and only if $x^2+y^2=x_1^2+y_1^2.$

10. Define a relation $R$ on $\mathbb{Z}$ by $a R b$ if and only if $5 | (2a + 3b)$.

Answer

 

1. $R$ is reflexive on $\mathbb{Z}$.

Proof:

Let $a \in \mathbb{Z}$.

We shall show that $a R a$, specifically $5|(2a+3a)$.

Consider that $2a+3a = 5a$ and $a \in \mathbb{Z}$.

Thus $5|(2a+3a)$,  and $aRa$.

Therefore $R$ is reflexive on $\mathbb{Z}$.◻

 

2. $R$ is symmetric on $\mathbb{Z}$.

Proof:

Let $a,b \in \mathbb{Z}$ s.t. $5|(2a+3b$.

Thus $2a+3b=5m$ for some $m\in \mathbb{Z}$.

We will show that $3a+2b=5k$ for some $k \in \mathbb{Z}$.

Consider that $-2a-3b=-5m$ for some $m\in \mathbb{Z}$.

Then $5(a+b) -2a-3b=5(a+b)-5m$.

Thus $3a+2b=5(a+b-m)$ where $a+b-m=k \in \mathbb{Z}$.

Hence $5|(3a+2b)$ and $bRa$.

Since $bRa$, $R$ is symmetric on $\mathbb{Z}$.

 

3. $R$ is not antisymmetric on $\mathbb{Z}$.

Counter Example:

Let $a=0$ and $b=5$.

Then $5|(2(0)+3(5))$ and $5|(2(5)+3(0)$.

However, since $0 \ne 5$ $R$ is not antisymmetric on $\mathbb{Z}$.◻

 

4. $R$ is transitive on $\mathbb{Z}$.

 

Let $a,b,c \in \mathbb{Z}$ s.t. $aRb$, $5|(2a+3b)$ and $bRa$, $5|(2b+3c)$.

We will show that $aRc$, $5|(2a+3c)$.

Since $5|(2a+3b)$, $2a+3b=5(k)$ for some $k \in \mathbb{Z}$.

Since $5|(2b+3c)$, $2b+3c=5(m)$ for some $m \in \mathbb{Z}$.

Consider $2a+3c=(2a+3b)+(2b+3c)-5b$

       $=5(k)+5(m)-5(b)$

       $=5(k+m-b)$, where $k+m-b \in \mathbb{Z}$.

Hence $5|2a+3c$ and $aRc$.

Hence $R$ is transitive on $\mathbb{Z}$.◻

Since $R$ is reflexive, symmetric and transitive on $\mathbb{Z}$, $R$ is an equivalence relation.

 

Te equivalence classes of $aRb$ iff $5 | (2a + 3b)$ are $[0], [1], [2], [3]$ and $[4]$.

Let $a \in \mathbb{Z}$, then $[a]=\{a\in \mathbb{Z}:x \sim a\}$.

$[0]=\{x \in \mathbb{Z}: x\sim 0\}$

      $=\{x \in \mathbb{Z}: 5|(2x+3(0)) \}$

      $=\{x \in \mathbb{Z}: 5|2x \}$

      $=\{x \in \mathbb{Z}: 2x=5m, m\in \mathbb{Z} \}$

      $=\{\ldots, -10,-5,0,5,10,\ldots\}$.

  

$[1]=\{x \in \mathbb{Z}: x\sim 1\}$

     $=\{x \in \mathbb{Z}: 5|(2x+3(1)) \}$

      $=\{x \in \mathbb{Z}: 5|(2x+3) \}$

      $=\{x \in \mathbb{Z}: 2x+3=5m, m\in \mathbb{Z} \}$

      $=\{x \in \mathbb{Z}: 2x=5m-3, m\in \mathbb{Z} \}$

      $=\{\ldots, -9,-4,1,6,11,\ldots\}$.

  

$[2]=\{x \in \mathbb{Z}: x\sim 2\}$

     $=\{x \in \mathbb{Z}: 5|(2x+3(2)) \}$

      $=\{x \in \mathbb{Z}: 5|(2x+6) \}$

      $=\{x \in \mathbb{Z}: 2x+6=5m, m\in \mathbb{Z} \}$

      $=\{x \in \mathbb{Z}: 2x=5m-6, m\in \mathbb{Z} \}$

      $=\{\ldots, -8,-3,2,7,12,\ldots\}$.

 

$[3]=\{x \in \mathbb{Z}: x\sim 3\}$

     $=\{x \in \mathbb{Z}: 5|(2x+3(3)) \}$

      $=\{x \in \mathbb{Z}: 5|(2x+9) \}$

      $=\{x \in \mathbb{Z}: 2x+9=5m, m\in \mathbb{Z} \}$

      $=\{x \in \mathbb{Z}: 2x=5m-9, m\in \mathbb{Z} \}$

      $=\{\ldots, -7,-2,3,8,13,\ldots\}$.    

  

$[4]=\{x \in \mathbb{Z}: x\sim 4\}$

     $=\{x \in \mathbb{Z}: 5|(2x+3(4)) \}$

      $=\{x \in \mathbb{Z}: 5|(2x+12) \}$

      $=\{x \in \mathbb{Z}: 2x+12=5m, m\in \mathbb{Z} \}$

      $=\{x \in \mathbb{Z}: 2x=5m-12, m\in \mathbb{Z} \}$

      $=\{\ldots, -6,-1,4,9,14,\ldots\}$