Proving this mapping is equal to the identity
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Question:
Let $f, g$ be mappings of a set $S$ into itself. Assume that $f^2 = g^2 = I$ and that $f circ g = g circ f$.
Prove that $ (f circ g)^2 = I $. Prove that $ (f circ g)^3 = I $.
My Answer:
$ (f circ g)^2 = (f circ g) circ (f circ g) = f circ gcirc fcirc g = f^2 circ g^2 = I circ I= I $
$ (f circ g)^3 = (f circ g) circ (f circ g) circ (f circ g) = f circ gcirc fcirc g circ fcirc g = f^2 circ g^2 circ f circ g= I circ I circ fcirc g= f circ g ; ...$
I've worked out the first part of the question and reduced the second part of a the question to the case of showing that $ (f circ g) = I $, but I can't get any further than that. I also noted down that $ f = f^{-1} $ and $ g = g^{-1} $. I'm not sure whether this plays into it. Could someone point me in the right direction or point out if I'm making a mistake. I almost want to say this is a mistake in the book as I know there a couple in this book but I don't want to give up without knowing for sure. If it's of any help, this is from Serge Lang's Basic Mathematics.
Thank you
algebra-precalculus functions
asked Jan 6 at 23:43
peachespeaches
153
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add a comment |
$begingroup$
Question:
Let $f, g$ be mappings of a set $S$ into itself. Assume that $f^2 = g^2 = I$ and that $f circ g = g circ f$.
Prove that $ (f circ g)^2 = I $. Prove that $ (f circ g)^3 = I $.
My Answer:
$ (f circ g)^2 = (f circ g) circ (f circ g) = f circ gcirc fcirc g = f^2 circ g^2 = I circ I= I $
$ (f circ g)^3 = (f circ g) circ (f circ g) circ (f circ g) = f circ gcirc fcirc g circ fcirc g = f^2 circ g^2 circ f circ g= I circ I circ fcirc g= f circ g ; ...$
I've worked out the first part of the question and reduced the second part of a the question to the case of showing that $ (f circ g) = I $, but I can't get any further than that. I also noted down that $ f = f^{-1} $ and $ g = g^{-1} $. I'm not sure whether this plays into it. Could someone point me in the right direction or point out if I'm making a mistake. I almost want to say this is a mistake in the book as I know there a couple in this book but I don't want to give up without knowing for sure. If it's of any help, this is from Serge Lang's Basic Mathematics.
Thank you
algebra-precalculus functions
asked Jan 6 at 23:43
peachespeaches
153
$endgroup$
add a comment |
$begingroup$
Question:
Let $f, g$ be mappings of a set $S$ into itself. Assume that $f^2 = g^2 = I$ and that $f circ g = g circ f$.
Prove that $ (f circ g)^2 = I $. Prove that $ (f circ g)^3 = I $.
My Answer:
$ (f circ g)^2 = (f circ g) circ (f circ g) = f circ gcirc fcirc g = f^2 circ g^2 = I circ I= I $
$ (f circ g)^3 = (f circ g) circ (f circ g) circ (f circ g) = f circ gcirc fcirc g circ fcirc g = f^2 circ g^2 circ f circ g= I circ I circ fcirc g= f circ g ; ...$
I've worked out the first part of the question and reduced the second part of a the question to the case of showing that $ (f circ g) = I $, but I can't get any further than that. I also noted down that $ f = f^{-1} $ and $ g = g^{-1} $. I'm not sure whether this plays into it. Could someone point me in the right direction or point out if I'm making a mistake. I almost want to say this is a mistake in the book as I know there a couple in this book but I don't want to give up without knowing for sure. If it's of any help, this is from Serge Lang's Basic Mathematics.
Thank you
algebra-precalculus functions
asked Jan 6 at 23:43
peachespeaches
153
$endgroup$
Question:
Let $f, g$ be mappings of a set $S$ into itself. Assume that $f^2 = g^2 = I$ and that $f circ g = g circ f$.
Prove that $ (f circ g)^2 = I $. Prove that $ (f circ g)^3 = I $.
My Answer:
$ (f circ g)^2 = (f circ g) circ (f circ g) = f circ gcirc fcirc g = f^2 circ g^2 = I circ I= I $
$ (f circ g)^3 = (f circ g) circ (f circ g) circ (f circ g) = f circ gcirc fcirc g circ fcirc g = f^2 circ g^2 circ f circ g= I circ I circ fcirc g= f circ g ; ...$
I've worked out the first part of the question and reduced the second part of a the question to the case of showing that $ (f circ g) = I $, but I can't get any further than that. I also noted down that $ f = f^{-1} $ and $ g = g^{-1} $. I'm not sure whether this plays into it. Could someone point me in the right direction or point out if I'm making a mistake. I almost want to say this is a mistake in the book as I know there a couple in this book but I don't want to give up without knowing for sure. If it's of any help, this is from Serge Lang's Basic Mathematics.
Thank you
algebra-precalculus functions
algebra-precalculus functions
asked Jan 6 at 23:43
peachespeaches
153
asked Jan 6 at 23:43
peachespeaches
153
asked Jan 6 at 23:43
peachespeaches
153
asked Jan 6 at 23:43
peachespeaches
153
asked Jan 6 at 23:43
peachespeaches
153
153
add a comment |
add a comment |
1 Answer
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The second part seems false: take $S=mathbb{R}^2$, $f(x,y)=(x,-y)$, $g(x,y)=(-x,y)$.
answered Jan 7 at 0:00
MindlackMindlack
4,910211
$endgroup$
$begingroup$
I think you're right. Seems to check the three requirements and definietly does not hold. Thank you for your help!
$endgroup$
– peaches
Jan 7 at 0:15
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The second part seems false: take $S=mathbb{R}^2$, $f(x,y)=(x,-y)$, $g(x,y)=(-x,y)$.
answered Jan 7 at 0:00
MindlackMindlack
4,910211
$endgroup$
$begingroup$
I think you're right. Seems to check the three requirements and definietly does not hold. Thank you for your help!
$endgroup$
– peaches
Jan 7 at 0:15
add a comment |
$begingroup$
The second part seems false: take $S=mathbb{R}^2$, $f(x,y)=(x,-y)$, $g(x,y)=(-x,y)$.
answered Jan 7 at 0:00
MindlackMindlack
4,910211
$endgroup$
$begingroup$
I think you're right. Seems to check the three requirements and definietly does not hold. Thank you for your help!
$endgroup$
– peaches
Jan 7 at 0:15
add a comment |
$begingroup$
The second part seems false: take $S=mathbb{R}^2$, $f(x,y)=(x,-y)$, $g(x,y)=(-x,y)$.
answered Jan 7 at 0:00
MindlackMindlack
4,910211
$endgroup$
The second part seems false: take $S=mathbb{R}^2$, $f(x,y)=(x,-y)$, $g(x,y)=(-x,y)$.
answered Jan 7 at 0:00
MindlackMindlack
4,910211
answered Jan 7 at 0:00
MindlackMindlack
4,910211
answered Jan 7 at 0:00
MindlackMindlack
4,910211
answered Jan 7 at 0:00
MindlackMindlack
4,910211
4,910211
$begingroup$
I think you're right. Seems to check the three requirements and definietly does not hold. Thank you for your help!
$endgroup$
– peaches
Jan 7 at 0:15
add a comment |
$begingroup$
I think you're right. Seems to check the three requirements and definietly does not hold. Thank you for your help!
$endgroup$
– peaches
Jan 7 at 0:15
$begingroup$
I think you're right. Seems to check the three requirements and definietly does not hold. Thank you for your help!
$endgroup$
– peaches
Jan 7 at 0:15
$begingroup$
I think you're right. Seems to check the three requirements and definietly does not hold. Thank you for your help!
$endgroup$
– peaches
Jan 7 at 0:15
add a comment |
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