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Math Help - identity and inverse

  1. #1
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    identity and inverse

    Here is another part of the question:


    For part (c), this is what I've done so far:

    suppose both e and are identities of \mathcal{P}(M) and suppose x \in \mathcal{P}(M). Therefore

    xe=x .....(1)
    x=x .....(2)

    Since e, \in \mathcal{P}(M), we can set x= in equation (1) and set x=e in equation (2)

    e=
    e=e

    Thus e and are equal and the identity is uniqe.

    Is this correct? I think the problem here is that I've shown that the identity of (\mathcal{P}(M), \star) is unique but I have not shown that it must have an identity. Can anyone help?

    For part (d), I did this:

    Let a,b,c \in \mathcal{P}(M_n).
    Suppose b and c are both inverses of a. Then ab=e also ac=e. Hence ab=bc \iff b=c.

    Again, I think my approach just shows the uniqueness, and not the existence of an inverse... I appreciate any help with this.
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  2. #2
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    You need to exhibit the identity. What does the identity do, in this case? Well, given any set x \in \mathcal{P}(M), you have that x\star e=e\star x=x, right? Now, x and e must be the same kind of object; that is, members of \mathcal{P}(M). Which member of \mathcal{P}(M) will do what you need?
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  3. #3
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    Quote Originally Posted by Ackbeet View Post
    You need to exhibit the identity. What does the identity do, in this case? Well, given any set x \in \mathcal{P}(M), you have that x\star e=e\star x=x, right? Now, x and e must be the same kind of object; that is, members of \mathcal{P}(M). Which member of \mathcal{P}(M) will do what you need?
    I'm still confused... I think \mathcal{P}(M) = \{ \emptyset , \{1 \} \}, so

     1 \star \emptyset = (\emptyset \cup 1) \backslash (\emptyset  \cap 1). Then what?
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  4. #4
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    Quote Originally Posted by demode View Post
    I'm still confused... I think \mathcal{P}(M) = \{ \emptyset , \{1 \} \}, so

     1 \star \emptyset = (\emptyset \cup 1) \backslash (\emptyset  \cap 1). Then what?
    A \star \emptyset  = \left( {A\backslash \emptyset } \right) \cup \left( {\emptyset \backslash A} \right) = A \cup \emptyset  = A
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  5. #5
    MHF Contributor Swlabr's Avatar
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    So, Plato has told you that the identity is the empty set. You now need to prove this this identity is unique, and you need to find the inverse of an element.

    To prove the identity is unique you do not need to use this example - any identity of a set is unique. This is because if you have two identities e_1 and e_2 then clearly e_1e_2 = e_1 but also e_1e_2=e_2. Thus they are equal.

    To find the inverse, you want to find a set B such that A \cup B = A \cap B. Note that |A \cup B| \geq \max\{|A|, |B|\} and |A \cap B| \leq \min\{|A|, |B|\}. What does this tell you? (Remember you want A \cap B = A \cup B).
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  6. #6
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    Thanks...

    Quote Originally Posted by Swlabr View Post
    To find the inverse, you want to find a set B such that A \cup B = A \cap B. Note that |A \cup B| \geq \max\{|A|, |B|\} and |A \cap B| \leq \min\{|A|, |B|\}. What does this tell you? (Remember you want A \cap B = A \cup B).
    I think A \cup B = A \cap B iff A=B, which means every element is its own inverse.

    To prove that the inverse for each element is unique I let A, B, C \in \mathcal{P}(M_n), where C and B are inverses of A. Then

    AB=e and AC=e. Therefore AB=AC \iff B=C.

    Is it correct?
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  7. #7
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by demode View Post
    I think A \cup B = A \cap B iff A=B, which means every element is its own inverse.

    To prove that the inverse for each element is unique I let A, B, C \in \mathcal{P}(M_n), where C and B are inverses of A. Then

    AB=e and AC=e. Therefore AB=AC \iff B=C.

    Is it correct?
    Yes, that is correct.
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