Bijection discrete math

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$f: BbbZ to BbbZ, f(x) = 3x + 6$



Is $f$ a bijection? If no, explain why it isn’t. If yes, find an expression computing $f^−1(y)$ for $y in BbbZ$.



How to approach that question?










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    Ask yourself whether or not the function is both injective and surjective. That is to say, will you always get different outputs if you give it different inputs or are there some inputs that give the same output? Is every element in the codomain actually an output that happens?
    – JMoravitz
    1 hour ago










  • Just try to solve $f(x)=1$.
    – Michael Hoppe
    57 mins ago














up vote
2
down vote

favorite
1












$f: BbbZ to BbbZ, f(x) = 3x + 6$



Is $f$ a bijection? If no, explain why it isn’t. If yes, find an expression computing $f^−1(y)$ for $y in BbbZ$.



How to approach that question?










share|cite|improve this question









New contributor




FreshmanUCSD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.















  • 2




    Ask yourself whether or not the function is both injective and surjective. That is to say, will you always get different outputs if you give it different inputs or are there some inputs that give the same output? Is every element in the codomain actually an output that happens?
    – JMoravitz
    1 hour ago










  • Just try to solve $f(x)=1$.
    – Michael Hoppe
    57 mins ago












up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1





$f: BbbZ to BbbZ, f(x) = 3x + 6$



Is $f$ a bijection? If no, explain why it isn’t. If yes, find an expression computing $f^−1(y)$ for $y in BbbZ$.



How to approach that question?










share|cite|improve this question









New contributor




FreshmanUCSD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$f: BbbZ to BbbZ, f(x) = 3x + 6$



Is $f$ a bijection? If no, explain why it isn’t. If yes, find an expression computing $f^−1(y)$ for $y in BbbZ$.



How to approach that question?







functions discrete-mathematics elementary-set-theory






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edited 58 mins ago









JMoravitz

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  • 2




    Ask yourself whether or not the function is both injective and surjective. That is to say, will you always get different outputs if you give it different inputs or are there some inputs that give the same output? Is every element in the codomain actually an output that happens?
    – JMoravitz
    1 hour ago










  • Just try to solve $f(x)=1$.
    – Michael Hoppe
    57 mins ago












  • 2




    Ask yourself whether or not the function is both injective and surjective. That is to say, will you always get different outputs if you give it different inputs or are there some inputs that give the same output? Is every element in the codomain actually an output that happens?
    – JMoravitz
    1 hour ago










  • Just try to solve $f(x)=1$.
    – Michael Hoppe
    57 mins ago







2




2




Ask yourself whether or not the function is both injective and surjective. That is to say, will you always get different outputs if you give it different inputs or are there some inputs that give the same output? Is every element in the codomain actually an output that happens?
– JMoravitz
1 hour ago




Ask yourself whether or not the function is both injective and surjective. That is to say, will you always get different outputs if you give it different inputs or are there some inputs that give the same output? Is every element in the codomain actually an output that happens?
– JMoravitz
1 hour ago












Just try to solve $f(x)=1$.
– Michael Hoppe
57 mins ago




Just try to solve $f(x)=1$.
– Michael Hoppe
57 mins ago










3 Answers
3






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3
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It's easy to see that the function's value is a multiple of 3. Is every integer a multiple of 3? No. So which property does that violate?






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    up vote
    1
    down vote













    Since $f(x)=3(x+2),$ the image of $f$ is not all of $mathbbZ$ but only the multiples of $3$. So there can be no inverse map from $mathbbZ$ to $mathbbZ.$ But $f$ does happen to be an injection.






    share|cite|improve this answer



























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      We can also use a different approach since it is easy to invert the function.



      Suppose for a contradiction that $f$ is a bijection, which implies $f$ is invertible. Then inverse function $f^-1$ of $f$ is
      $$f^-1(x) = fracx-63, xinmathbbZ, f^-1:mathbbZtomathbbZ$$
      But this is not even a function since $x$ values that are not multiple of $3$ are not even mapped to an element in codomain. So we have a contradiction as required.






      share|cite|improve this answer




















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        3 Answers
        3






        active

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        3 Answers
        3






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        active

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        active

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        up vote
        3
        down vote













        It's easy to see that the function's value is a multiple of 3. Is every integer a multiple of 3? No. So which property does that violate?






        share|cite|improve this answer
























          up vote
          3
          down vote













          It's easy to see that the function's value is a multiple of 3. Is every integer a multiple of 3? No. So which property does that violate?






          share|cite|improve this answer






















            up vote
            3
            down vote










            up vote
            3
            down vote









            It's easy to see that the function's value is a multiple of 3. Is every integer a multiple of 3? No. So which property does that violate?






            share|cite|improve this answer












            It's easy to see that the function's value is a multiple of 3. Is every integer a multiple of 3? No. So which property does that violate?







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 53 mins ago









            Matt Samuel

            35k43060




            35k43060




















                up vote
                1
                down vote













                Since $f(x)=3(x+2),$ the image of $f$ is not all of $mathbbZ$ but only the multiples of $3$. So there can be no inverse map from $mathbbZ$ to $mathbbZ.$ But $f$ does happen to be an injection.






                share|cite|improve this answer
























                  up vote
                  1
                  down vote













                  Since $f(x)=3(x+2),$ the image of $f$ is not all of $mathbbZ$ but only the multiples of $3$. So there can be no inverse map from $mathbbZ$ to $mathbbZ.$ But $f$ does happen to be an injection.






                  share|cite|improve this answer






















                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    Since $f(x)=3(x+2),$ the image of $f$ is not all of $mathbbZ$ but only the multiples of $3$. So there can be no inverse map from $mathbbZ$ to $mathbbZ.$ But $f$ does happen to be an injection.






                    share|cite|improve this answer












                    Since $f(x)=3(x+2),$ the image of $f$ is not all of $mathbbZ$ but only the multiples of $3$. So there can be no inverse map from $mathbbZ$ to $mathbbZ.$ But $f$ does happen to be an injection.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 50 mins ago









                    coffeemath

                    1,6841313




                    1,6841313




















                        up vote
                        0
                        down vote













                        We can also use a different approach since it is easy to invert the function.



                        Suppose for a contradiction that $f$ is a bijection, which implies $f$ is invertible. Then inverse function $f^-1$ of $f$ is
                        $$f^-1(x) = fracx-63, xinmathbbZ, f^-1:mathbbZtomathbbZ$$
                        But this is not even a function since $x$ values that are not multiple of $3$ are not even mapped to an element in codomain. So we have a contradiction as required.






                        share|cite|improve this answer
























                          up vote
                          0
                          down vote













                          We can also use a different approach since it is easy to invert the function.



                          Suppose for a contradiction that $f$ is a bijection, which implies $f$ is invertible. Then inverse function $f^-1$ of $f$ is
                          $$f^-1(x) = fracx-63, xinmathbbZ, f^-1:mathbbZtomathbbZ$$
                          But this is not even a function since $x$ values that are not multiple of $3$ are not even mapped to an element in codomain. So we have a contradiction as required.






                          share|cite|improve this answer






















                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            We can also use a different approach since it is easy to invert the function.



                            Suppose for a contradiction that $f$ is a bijection, which implies $f$ is invertible. Then inverse function $f^-1$ of $f$ is
                            $$f^-1(x) = fracx-63, xinmathbbZ, f^-1:mathbbZtomathbbZ$$
                            But this is not even a function since $x$ values that are not multiple of $3$ are not even mapped to an element in codomain. So we have a contradiction as required.






                            share|cite|improve this answer












                            We can also use a different approach since it is easy to invert the function.



                            Suppose for a contradiction that $f$ is a bijection, which implies $f$ is invertible. Then inverse function $f^-1$ of $f$ is
                            $$f^-1(x) = fracx-63, xinmathbbZ, f^-1:mathbbZtomathbbZ$$
                            But this is not even a function since $x$ values that are not multiple of $3$ are not even mapped to an element in codomain. So we have a contradiction as required.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 48 mins ago









                            ArsenBerk

                            7,35021234




                            7,35021234




















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