Loading [MathJax]/jax/output/HTML-CSS/jax.js
Skip to main content
Library homepage
 

Text Color

Text Size

 

Margin Size

 

Font Type

Enable Dyslexic Font
Mathematics LibreTexts

0E: Exercises

( \newcommand{\kernel}{\mathrm{null}\,}\)

 

Exercise 0E.1

Define h:ZZ by h(x)=x2+4.  Determine (with reasons) whether or not h is one−to−one and whether or not h is onto.  

Exercise 0E.2

Suppose f:AB and g:BC are functions.  Prove or disprove the following statements: 

  1. If gf is one-to-one then g is one-to-one. 

  2. If gf is one-to-one and f is onto, then g is one-to-one.

  3.  If gf is onto and  g is one-to-one, then  f is onto.

Exercise 0E.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×S by (a,b)R(c,d) if and only if 10a+b10c+d

  2. Let A=Z{0}. Define a relation R on A, by aRb if and only if ab>0.

  3. Define a relation R on A=Z by aRb if and only if 4|(3a+b).

  4. Define a relation R on A=Z by aRb if and only if 3|(a2b2)

  5. Let A=R, If a,bR, define aRb if and only if abZ

  6. Define a relation R on the set Z×Z by (a,b)R(c,d) if and only if ac=bd

  7. Define a relation R on Z by aRb if and only if 2a2+b

  8. Let A=R, If a,bR, define aRb if and only if abQ.

  9. Let A=R×R, If (x,y),(x1,y1)R×R, define (x,y)R(x1,y1) if and only if x2+y2=x21+y21.
  10. Define a relation R on Z by aRb if and only if 5|(2a+3b).

Answer

 

  1. R is reflexive on Z.

Proof:

Let aZ.

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

Consider that 2a+3a=5a and aZ.

Thus 5|(2a+3a),  and aRa.

Therefore R is reflexive on Z.◻

 

2. R is symmetric on Z.

Proof:

Let a,bZ s.t. 5|(2a+3b.

Thus 2a+3b=5m for some mZ.

We will show that 3a+2b=5k for some kZ.

Consider that 2a3b=5m for some mZ.

Then 5(a+b)2a3b=5(a+b)5m.

Thus 3a+2b=5(a+bm) where a+bm=kZ.

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

Since bRa, R is symmetric on Z.

 

3. R is not antisymmetric on Z.

Counter Example:

Let a=0 and b=5.

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

However, since 05 R is not antisymmetric on Z.◻

 

4. R is transitive on Z.

 

Let a,b,cZ 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 kZ.

Since 5|(2b+3c), 2b+3c=5(m) for some mZ.

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

       =5(k)+5(m)5(b)

       =5(k+mb), where k+mbZ.

Hence 5|2a+3c and aRc.

Hence R is transitive on Z.◻

Since R is reflexive, symmetric and transitive on Z, R is an equivalence relation.

 

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

Let aZ, then [a]={aZ:xa}.

[0]={xZ:x0}

      ={xZ:5|(2x+3(0))}

      ={xZ:5|2x}

      ={xZ:2x=5m,mZ}

      ={,10,5,0,5,10,}.

  

[1]={xZ:x1}

     ={xZ:5|(2x+3(1))}

      ={xZ:5|(2x+3)}

      ={xZ:2x+3=5m,mZ}

      ={xZ:2x=5m3,mZ}

      ={,9,4,1,6,11,}.

  

[2]={xZ:x2}

     ={xZ:5|(2x+3(2))}

      ={xZ:5|(2x+6)}

      ={xZ:2x+6=5m,mZ}

      ={xZ:2x=5m6,mZ}

      ={,8,3,2,7,12,}.

 

[3]={xZ:x3}

     ={xZ:5|(2x+3(3))}

      ={xZ:5|(2x+9)}

      ={xZ:2x+9=5m,mZ}

      ={xZ:2x=5m9,mZ}

      ={,7,2,3,8,13,}.    

  

[4]={xZ:x4}

     ={xZ:5|(2x+3(4))}

      ={xZ:5|(2x+12)}

      ={xZ:2x+12=5m,mZ}

      ={xZ:2x=5m12,mZ}

      ={,6,1,4,9,14,}

 

 

 

 


This page titled 0E: Exercises is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Pamini Thangarajah.

Support Center

How can we help?