Prove that a specific function is injective…

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... In the context of the question:



Say we have a set $S$, that contains all the strings with the alphabet $a$ and $b$, i.e. $$S = lefta, b, aa, bb, ab, ba, aaa, bab, bba, dotsright$$



Now, say we have a function $R(t)$ that replaces the leftmost occurrence of $a$ in $t$ with $b$, i.e. $R(aaabb) = baabb$, and $R(bbaaaab) = bbbaaab$, and so on.



How do we prove that $R$ is injective, or otherwise?






functions elementary-set-theory







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  • 5




    I think it isn't injective. Because $R(a)=b$ and $R(b)=b$...
    – Doyun Nam
    3 hours ago






  • 1




    @DoyunNam, thank you; that was short and sweet.
    – SRSR333
    3 hours ago














up vote
1
down vote

favorite












... In the context of the question:



Say we have a set $S$, that contains all the strings with the alphabet $a$ and $b$, i.e. $$S = lefta, b, aa, bb, ab, ba, aaa, bab, bba, dotsright$$



Now, say we have a function $R(t)$ that replaces the leftmost occurrence of $a$ in $t$ with $b$, i.e. $R(aaabb) = baabb$, and $R(bbaaaab) = bbbaaab$, and so on.



How do we prove that $R$ is injective, or otherwise?






functions elementary-set-theory







share|cite|improve this question

















  • 5




    I think it isn't injective. Because $R(a)=b$ and $R(b)=b$...
    – Doyun Nam
    3 hours ago






  • 1




    @DoyunNam, thank you; that was short and sweet.
    – SRSR333
    3 hours ago












up vote
1
down vote

favorite









up vote
1
down vote

favorite











... In the context of the question:



Say we have a set $S$, that contains all the strings with the alphabet $a$ and $b$, i.e. $$S = lefta, b, aa, bb, ab, ba, aaa, bab, bba, dotsright$$



Now, say we have a function $R(t)$ that replaces the leftmost occurrence of $a$ in $t$ with $b$, i.e. $R(aaabb) = baabb$, and $R(bbaaaab) = bbbaaab$, and so on.



How do we prove that $R$ is injective, or otherwise?






functions elementary-set-theory







share|cite|improve this question













... In the context of the question:



Say we have a set $S$, that contains all the strings with the alphabet $a$ and $b$, i.e. $$S = lefta, b, aa, bb, ab, ba, aaa, bab, bba, dotsright$$



Now, say we have a function $R(t)$ that replaces the leftmost occurrence of $a$ in $t$ with $b$, i.e. $R(aaabb) = baabb$, and $R(bbaaaab) = bbbaaab$, and so on.



How do we prove that $R$ is injective, or otherwise?






functions elementary-set-theory




functions elementary-set-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 3 hours ago









SRSR333

304




304







  • 5




    I think it isn't injective. Because $R(a)=b$ and $R(b)=b$...
    – Doyun Nam
    3 hours ago






  • 1




    @DoyunNam, thank you; that was short and sweet.
    – SRSR333
    3 hours ago












  • 5




    I think it isn't injective. Because $R(a)=b$ and $R(b)=b$...
    – Doyun Nam
    3 hours ago






  • 1




    @DoyunNam, thank you; that was short and sweet.
    – SRSR333
    3 hours ago







5




5




I think it isn't injective. Because $R(a)=b$ and $R(b)=b$...
– Doyun Nam
3 hours ago




I think it isn't injective. Because $R(a)=b$ and $R(b)=b$...
– Doyun Nam
3 hours ago




1




1




@DoyunNam, thank you; that was short and sweet.
– SRSR333
3 hours ago




@DoyunNam, thank you; that was short and sweet.
– SRSR333
3 hours ago










1 Answer
1






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It is not injective. Take $s_1 = ab$ and $s_2 = bb$. Then, $R(s_1) = bb = R(s_2) = bb$. But, $s_1 ne s_2$. So, we have one output mapping to two inputs. This is just a more exhaustive explanation of Doyun's answer.






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  • Thanks very much for the explanation.
    – SRSR333
    3 hours ago










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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote



accepted










It is not injective. Take $s_1 = ab$ and $s_2 = bb$. Then, $R(s_1) = bb = R(s_2) = bb$. But, $s_1 ne s_2$. So, we have one output mapping to two inputs. This is just a more exhaustive explanation of Doyun's answer.






share|cite|improve this answer




















  • Thanks very much for the explanation.
    – SRSR333
    3 hours ago














up vote
3
down vote



accepted










It is not injective. Take $s_1 = ab$ and $s_2 = bb$. Then, $R(s_1) = bb = R(s_2) = bb$. But, $s_1 ne s_2$. So, we have one output mapping to two inputs. This is just a more exhaustive explanation of Doyun's answer.






share|cite|improve this answer




















  • Thanks very much for the explanation.
    – SRSR333
    3 hours ago












up vote
3
down vote



accepted







up vote
3
down vote



accepted






It is not injective. Take $s_1 = ab$ and $s_2 = bb$. Then, $R(s_1) = bb = R(s_2) = bb$. But, $s_1 ne s_2$. So, we have one output mapping to two inputs. This is just a more exhaustive explanation of Doyun's answer.






share|cite|improve this answer












It is not injective. Take $s_1 = ab$ and $s_2 = bb$. Then, $R(s_1) = bb = R(s_2) = bb$. But, $s_1 ne s_2$. So, we have one output mapping to two inputs. This is just a more exhaustive explanation of Doyun's answer.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 3 hours ago









D.B.

6337




6337











  • Thanks very much for the explanation.
    – SRSR333
    3 hours ago
















  • Thanks very much for the explanation.
    – SRSR333
    3 hours ago















Thanks very much for the explanation.
– SRSR333
3 hours ago




Thanks very much for the explanation.
– SRSR333
3 hours ago

















 

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