Mixing
For a given type X, rigorously show that idX = (idX)^-1
$begingroup$
For a given type X, rigorously show that idX = (idX)^-1
I have done the following so far:
(x,y) ∈ idX
<=> identity relation
x=y ^ x ∈ X
.
.
.
.
y=x ^ y ∈ X
<=> identity relation
(y,x) ∈ idX
<=> relational inverse
(x,y) ∈ (idX)^-1
Basically I don't know how to connect the two together.
Thanks in advance
discrete-mathematics relations
asked Feb 2 at 18:28
xalalauxalalau
154
$endgroup$
add a comment |
$begingroup$
For a given type X, rigorously show that idX = (idX)^-1
I have done the following so far:
(x,y) ∈ idX
<=> identity relation
x=y ^ x ∈ X
.
.
.
.
y=x ^ y ∈ X
<=> identity relation
(y,x) ∈ idX
<=> relational inverse
(x,y) ∈ (idX)^-1
Basically I don't know how to connect the two together.
Thanks in advance
discrete-mathematics relations
asked Feb 2 at 18:28
xalalauxalalau
154
$endgroup$
add a comment |
$begingroup$
For a given type X, rigorously show that idX = (idX)^-1
I have done the following so far:
(x,y) ∈ idX
<=> identity relation
x=y ^ x ∈ X
.
.
.
.
y=x ^ y ∈ X
<=> identity relation
(y,x) ∈ idX
<=> relational inverse
(x,y) ∈ (idX)^-1
Basically I don't know how to connect the two together.
Thanks in advance
discrete-mathematics relations
asked Feb 2 at 18:28
xalalauxalalau
154
$endgroup$
For a given type X, rigorously show that idX = (idX)^-1
I have done the following so far:
(x,y) ∈ idX
<=> identity relation
x=y ^ x ∈ X
.
.
.
.
y=x ^ y ∈ X
<=> identity relation
(y,x) ∈ idX
<=> relational inverse
(x,y) ∈ (idX)^-1
Basically I don't know how to connect the two together.
Thanks in advance
discrete-mathematics relations
discrete-mathematics relations
asked Feb 2 at 18:28
xalalauxalalau
154
asked Feb 2 at 18:28
xalalauxalalau
154
asked Feb 2 at 18:28
xalalauxalalau
154
asked Feb 2 at 18:28
xalalauxalalau
154
asked Feb 2 at 18:28
xalalauxalalau
154
154
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
From $x=yland xin X$, we obtain $x=y$ and $xin X$. Using the equality $x=y$, we can substitute $x$ with $y$ in $xin X$ and thus obtain $yin X$. As equality is symmetric, $x=y$ implies $y=x$. So we have $y=x$ and $yin X$, hence also $y=xland yin X$.
answered Feb 2 at 18:33
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
$begingroup$
Uhh I see it now, didn't think it was that simple I was looking at which proofs to use
$endgroup$
– xalalau
Feb 2 at 18:34
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
From $x=yland xin X$, we obtain $x=y$ and $xin X$. Using the equality $x=y$, we can substitute $x$ with $y$ in $xin X$ and thus obtain $yin X$. As equality is symmetric, $x=y$ implies $y=x$. So we have $y=x$ and $yin X$, hence also $y=xland yin X$.
answered Feb 2 at 18:33
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
$begingroup$
Uhh I see it now, didn't think it was that simple I was looking at which proofs to use
$endgroup$
– xalalau
Feb 2 at 18:34
add a comment |
$begingroup$
From $x=yland xin X$, we obtain $x=y$ and $xin X$. Using the equality $x=y$, we can substitute $x$ with $y$ in $xin X$ and thus obtain $yin X$. As equality is symmetric, $x=y$ implies $y=x$. So we have $y=x$ and $yin X$, hence also $y=xland yin X$.
answered Feb 2 at 18:33
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
$begingroup$
Uhh I see it now, didn't think it was that simple I was looking at which proofs to use
$endgroup$
– xalalau
Feb 2 at 18:34
add a comment |
$begingroup$
From $x=yland xin X$, we obtain $x=y$ and $xin X$. Using the equality $x=y$, we can substitute $x$ with $y$ in $xin X$ and thus obtain $yin X$. As equality is symmetric, $x=y$ implies $y=x$. So we have $y=x$ and $yin X$, hence also $y=xland yin X$.
answered Feb 2 at 18:33
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
From $x=yland xin X$, we obtain $x=y$ and $xin X$. Using the equality $x=y$, we can substitute $x$ with $y$ in $xin X$ and thus obtain $yin X$. As equality is symmetric, $x=y$ implies $y=x$. So we have $y=x$ and $yin X$, hence also $y=xland yin X$.
answered Feb 2 at 18:33
Hagen von EitzenHagen von Eitzen
283k23273508
answered Feb 2 at 18:33
Hagen von EitzenHagen von Eitzen
283k23273508
answered Feb 2 at 18:33
Hagen von EitzenHagen von Eitzen
283k23273508
answered Feb 2 at 18:33
Hagen von EitzenHagen von Eitzen
283k23273508
283k23273508
$begingroup$
Uhh I see it now, didn't think it was that simple I was looking at which proofs to use
$endgroup$
– xalalau
Feb 2 at 18:34
add a comment |
$begingroup$
Uhh I see it now, didn't think it was that simple I was looking at which proofs to use
$endgroup$
– xalalau
Feb 2 at 18:34
$begingroup$
Uhh I see it now, didn't think it was that simple I was looking at which proofs to use
$endgroup$
– xalalau
Feb 2 at 18:34
$begingroup$
Uhh I see it now, didn't think it was that simple I was looking at which proofs to use
$endgroup$
– xalalau
Feb 2 at 18:34
add a comment |
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