Question
If $g(x)$ is 1:1, and $g=\{(3,2), (a,0), (-1,b), (1,3)\}$, what is $g^{-1}(g(b))$?
I thought that it would just be $b$ but the answer key said that it does not exist.
I was wondering why it doesn't exist?
If $g(x)$ is 1:1, and $g=\{(3,2), (a,0), (-1,b), (1,3)\}$, what is $g^{-1}(g(b))$
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algebra-precalculus
functions
inverse
1 Answers
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The thing is that you do not map $b$ to anything since no tuple has $b$ as its first coordinate (only one as it second). So $g(b)$ does not exist. Hence also $g^{-1}(g(b))$. To be precise: A function $f: X \to Y$ is a binary relation $f \subseteq X \times Y$ (with some uniquenessproperty of course), but $b \notin X$.
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0But what guarantees here that $b\notin\{3,a,-1,1\}$? I think that here special cases must be discerned. If e.g. $b=1$ then the answer is $1$. – 2017-01-23
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0@drhab Yes, you are right. But nothing is stated about the nature of $a$ or $b$. So it is like a riddle. – 2017-01-23
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0I think the problem to be solved is not well-posed. No blame for the OP but for the one who composed this problem. – 2017-01-23
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0@drhab Yes also right. I just wanted to give an idea, why it might be wrong. – 2017-01-23
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0@drhab $b$ cannot be $3$ as $g(-1)=b$ and $g(1)=3$. But I would have thought that you can deduce $a\not \in \{1,3\}$, and that $b=-1$ could be possible if $a\not=-1$, that $b=1$ could be possible if $a\not=-1$, and that $b=a \not \in \{-1,0,1,2,3\}$ could be possible. I think all of these possibilities would lead to $g^{-1}(g(b))=b$ – 2017-01-23