Saturday

April 19, 2014

April 19, 2014

Posted by **yengiang** on Saturday, April 16, 2011 at 8:58pm.

a/ Is f one-to-one?

b/ Is f onto?

Could you show me how to solve this problem, please? I have no idea what this function is.

Your help is greatly appreciated.

- Discrete Math -
**Mgraph**, Saturday, April 16, 2011 at 9:38pmf(0)=2x0=0=0 mod 5

f(1)=2x1=2=2 mod 5

f(2)=2x2=4=4 mod 5

f(3)=2x3=6=1 mod 5

f(4)=2x4=8=3 mod 5,

so f(A)={0,1,2,3,4}

a/- yes

b/- yes

- Discrete Math -
**MathMate**, Saturday, April 16, 2011 at 9:39pmThe function has been defined already:

f : A -> A

where

f(n) = 2n mod 5

To determine if the function is one-to-one and onto, we can enumerate the values of x and f(x) and compare with the elements of A = {0,1,2,3,4}.

x f(x)=2n mod 5

0 0 [2*0 mod 5 = 0 mod 5 = 0]

1 2 [2*1 mod 5 = 2 mod 5 = 2]

2 4

3 1 [2*3 mod 5 = 6 mod 5 = 1]

4 3

Recall the definition of "one-to-one" is

"A function f from A to B is called one-to-one (or 1-1) if whenever

f (a) = f (b) then a = b. No element of B is the image of more than one element in A."

Since no element in B (=f(x)) can be mapped from more than one element in A, the function is one-to-one.

Recall the definition of "onto".

"A function f from A to B is called onto if for all b in B there is an a in A such that f (a) = b. All elements in B are used."

Since every element of set B (=f(x)) maps to an element in A, the function is onto.

Many links are available for the definition and explanation of the two terms, you can try the following as a start, as it explains with clear examples:

http://www.regentsprep.org/Regents/math/algtrig/ATP5/OntoFunctions.htm

- Discrete Math -
**yengiang**, Saturday, April 16, 2011 at 10:13pmThank you so much for your replies.

Could you also verify what I did on this problem please?

The set of course numbers for a collection of math courses is:

M = {0099,1111,1113,2200,2210,2220,2300,2450,2500,3500,4500,4900}

Define a relation R on M by (x,y) in R if course numbers x and y start with the same number.

a/ Verify that R is an equivalence relation

Reflexive: yes, because course number x start with the number itself.

Symmetric: yes, because if x starts with the same number as y, then y also starts with the same number as x.

Transitive: Yes, because for all courses x,y, and z, if x starts with the same number as y and y starts with

the same number as x, then x starts with the same number as z.

R is an equivalence because it is reflexive, symmetric, and transitive.

b/ Describe the distinct equivalence classes of R.

There are 5 equivalence classes:

[1] = {0099}

[2] = {1111, 1113}

[3] = {2200,2210,2220,2300,2450,2500}

[4] = {3500}

[5] = {4500,4900}

- Discrete Math -
**MathMate**, Sunday, April 17, 2011 at 8:06amLooks all good to me!

**Related Questions**

maths - The mapping is f:N →N, given by F(n)= 2n^3+1,n belongs to N. A. ...

maths - The mapping is f:N →N, given by F(n)= 2n^3=1,n belongs to N. A. ...

sant singh sukha singh school - The mapping is f:N →N, given by F(n)= 2n^3...

algebra - So... I know the answer to the question because it's something we ...

Mathematics - I have a question about onto and one-to-one.? Determine(with an ...

Math - Which Statements of congruence are true and which are false and why? 177 ...

math - Which Statements of congruence are true and which are false and why? 177 ...

Math - Which Statements of congruence are true and which are false and why? 177 ...

discrete math - 5= (1+4) (1+4)^n= 1^n + 2*1*4 + ...+4^n which is 1 mod 4 Prove ...

Discrete Math - Let a, b, c, and d be integers, and let n be a positive integer...