Asked by yengiang
Let A={0,1,2,3,4}. Define a function f from A to A by f(n)=2n mod 5.
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.
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.
Answers
Answered by
Mgraph
f(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
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
Answered by
MathMate
The 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
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
Answered by
yengiang
Thank 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}
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}
Answered by
MathMate
Looks all good to me!
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