Mixing
Approve that $f = frac{x}{1 - x^2}$ is an injective function
$begingroup$
Assume that $f: R setminus {-1,1} to R$ and $f = frac{x}{1-x^2}$. Approve that $f$ is an injective function.
My solution:
Based on the theory: for each $x,y in R setminus {-1,1 } $ if $ f(x) = f(y) $ then $x=y$
$ x - xy^2 = y -yx^2 Leftrightarrow x - y = xy^2 - x^2y Leftrightarrow x-y = xy(y^2 - x^2) Leftrightarrow x-y = xy(y-x)(y+x) Leftrightarrow (y-x)[xy(y+x) +1 ) = 0$
Two cases:
1) $x = y$
or
2) $(xy(y+x) +1 ) = 0$
Edit:
$ x - xy^2 = y -yx^2 Leftrightarrow x - y = xy^2 - x^2y Leftrightarrow x-y = xy(y - x) $
Two cases:
1) $x = y$
or
2) $xy = -1$
We reject the second. So, f is injective function
My question:
1) Can we reject the second case? Please explain!
Answer of the Community
No, we can not reject the second case !
calculus
asked Jan 12 at 12:48
Dimitris DimitriadisDimitris Dimitriadis
478
$endgroup$
add a comment |
$begingroup$
Assume that $f: R setminus {-1,1} to R$ and $f = frac{x}{1-x^2}$. Approve that $f$ is an injective function.
My solution:
Based on the theory: for each $x,y in R setminus {-1,1 } $ if $ f(x) = f(y) $ then $x=y$
$ x - xy^2 = y -yx^2 Leftrightarrow x - y = xy^2 - x^2y Leftrightarrow x-y = xy(y^2 - x^2) Leftrightarrow x-y = xy(y-x)(y+x) Leftrightarrow (y-x)[xy(y+x) +1 ) = 0$
Two cases:
1) $x = y$
or
2) $(xy(y+x) +1 ) = 0$
Edit:
$ x - xy^2 = y -yx^2 Leftrightarrow x - y = xy^2 - x^2y Leftrightarrow x-y = xy(y - x) $
Two cases:
1) $x = y$
or
2) $xy = -1$
We reject the second. So, f is injective function
My question:
1) Can we reject the second case? Please explain!
Answer of the Community
No, we can not reject the second case !
calculus
asked Jan 12 at 12:48
Dimitris DimitriadisDimitris Dimitriadis
478
$endgroup$
$begingroup$
$xy^2-x^2y = xy(y-x)$ no squares on the right hand side...
$endgroup$
– Yanko
Jan 12 at 12:50
$begingroup$
Thank you for your comment
$endgroup$
– Dimitris Dimitriadis
Jan 12 at 13:10
add a comment |
$begingroup$
Assume that $f: R setminus {-1,1} to R$ and $f = frac{x}{1-x^2}$. Approve that $f$ is an injective function.
My solution:
Based on the theory: for each $x,y in R setminus {-1,1 } $ if $ f(x) = f(y) $ then $x=y$
$ x - xy^2 = y -yx^2 Leftrightarrow x - y = xy^2 - x^2y Leftrightarrow x-y = xy(y^2 - x^2) Leftrightarrow x-y = xy(y-x)(y+x) Leftrightarrow (y-x)[xy(y+x) +1 ) = 0$
Two cases:
1) $x = y$
or
2) $(xy(y+x) +1 ) = 0$
Edit:
$ x - xy^2 = y -yx^2 Leftrightarrow x - y = xy^2 - x^2y Leftrightarrow x-y = xy(y - x) $
Two cases:
1) $x = y$
or
2) $xy = -1$
We reject the second. So, f is injective function
My question:
1) Can we reject the second case? Please explain!
Answer of the Community
No, we can not reject the second case !
calculus
asked Jan 12 at 12:48
Dimitris DimitriadisDimitris Dimitriadis
478
$endgroup$
Assume that $f: R setminus {-1,1} to R$ and $f = frac{x}{1-x^2}$. Approve that $f$ is an injective function.
My solution:
Based on the theory: for each $x,y in R setminus {-1,1 } $ if $ f(x) = f(y) $ then $x=y$
$ x - xy^2 = y -yx^2 Leftrightarrow x - y = xy^2 - x^2y Leftrightarrow x-y = xy(y^2 - x^2) Leftrightarrow x-y = xy(y-x)(y+x) Leftrightarrow (y-x)[xy(y+x) +1 ) = 0$
Two cases:
1) $x = y$
or
2) $(xy(y+x) +1 ) = 0$
Edit:
$ x - xy^2 = y -yx^2 Leftrightarrow x - y = xy^2 - x^2y Leftrightarrow x-y = xy(y - x) $
Two cases:
1) $x = y$
or
2) $xy = -1$
We reject the second. So, f is injective function
My question:
1) Can we reject the second case? Please explain!
Answer of the Community
No, we can not reject the second case !
calculus
calculus
asked Jan 12 at 12:48
Dimitris DimitriadisDimitris Dimitriadis
478
asked Jan 12 at 12:48
Dimitris DimitriadisDimitris Dimitriadis
478
edited Jan 12 at 13:18
Dimitris Dimitriadis
asked Jan 12 at 12:48
Dimitris DimitriadisDimitris Dimitriadis
478
asked Jan 12 at 12:48
Dimitris DimitriadisDimitris Dimitriadis
478
asked Jan 12 at 12:48
Dimitris DimitriadisDimitris Dimitriadis
478
478
$begingroup$
$xy^2-x^2y = xy(y-x)$ no squares on the right hand side...
$endgroup$
– Yanko
Jan 12 at 12:50
$begingroup$
Thank you for your comment
$endgroup$
– Dimitris Dimitriadis
Jan 12 at 13:10
add a comment |
$begingroup$
$xy^2-x^2y = xy(y-x)$ no squares on the right hand side...
$endgroup$
– Yanko
Jan 12 at 12:50
$begingroup$
Thank you for your comment
$endgroup$
– Dimitris Dimitriadis
Jan 12 at 13:10
$begingroup$
$xy^2-x^2y = xy(y-x)$ no squares on the right hand side...
$endgroup$
– Yanko
Jan 12 at 12:50
$begingroup$
$xy^2-x^2y = xy(y-x)$ no squares on the right hand side...
$endgroup$
– Yanko
Jan 12 at 12:50
$begingroup$
Thank you for your comment
$endgroup$
– Dimitris Dimitriadis
Jan 12 at 13:10
$begingroup$
Thank you for your comment
$endgroup$
– Dimitris Dimitriadis
Jan 12 at 13:10
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Going by the very definition, as you did:
$$f(x)=f(y)ifffrac x{1-x^2}=frac y{1-y^2}iff x-xy^2=y-x^2yiff $$
$$iff(x-y)=-xy(x-y)iffbegin{cases}x=y\or\xy=-1end{cases}$$
Thus, any pair of numbers $;x,,yinBbb Rsetminus{-1,1};$ s.t. $;xy=-1;$ give you a counterexample to injectivity. For example
$$x=-2,,y=frac12;,;text{and certainly:};;f(-2)=frac{-2}{1-4}=frac23=frac{frac12}{1-frac14} =fleft(frac12right)$$
and etc.
answered Jan 12 at 12:59
DonAntonioDonAntonio
178k1494230
$endgroup$
add a comment |
$begingroup$
It is not an injective function as we can see from horizontal line test.
answered Jan 12 at 12:52
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
$begingroup$
It equals for $x,y$ such that $xcdot y = -1$.
$endgroup$
– Yanko
Jan 12 at 12:52
add a comment |
$begingroup$
There's an error in your computation:
$$ x - y = xy^2 - x^2y iff x-y = xycolor{red}{(y - x)}iffbegin{cases}x=y \ xy=-1iff y=-frac1x, ; xne 0
end{cases}$$
answered Jan 12 at 13:07
BernardBernard
121k740116
$endgroup$
$begingroup$
I edited. Thank you
$endgroup$
– Dimitris Dimitriadis
Jan 12 at 13:10
$begingroup$
You can not say that $y= frac{1}{x}$ because $x$ can be $0$
$endgroup$
– Dimitris Dimitriadis
Jan 12 at 13:12
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Going by the very definition, as you did:
$$f(x)=f(y)ifffrac x{1-x^2}=frac y{1-y^2}iff x-xy^2=y-x^2yiff $$
$$iff(x-y)=-xy(x-y)iffbegin{cases}x=y\or\xy=-1end{cases}$$
Thus, any pair of numbers $;x,,yinBbb Rsetminus{-1,1};$ s.t. $;xy=-1;$ give you a counterexample to injectivity. For example
$$x=-2,,y=frac12;,;text{and certainly:};;f(-2)=frac{-2}{1-4}=frac23=frac{frac12}{1-frac14} =fleft(frac12right)$$
and etc.
answered Jan 12 at 12:59
DonAntonioDonAntonio
178k1494230
$endgroup$
add a comment |
$begingroup$
Going by the very definition, as you did:
$$f(x)=f(y)ifffrac x{1-x^2}=frac y{1-y^2}iff x-xy^2=y-x^2yiff $$
$$iff(x-y)=-xy(x-y)iffbegin{cases}x=y\or\xy=-1end{cases}$$
Thus, any pair of numbers $;x,,yinBbb Rsetminus{-1,1};$ s.t. $;xy=-1;$ give you a counterexample to injectivity. For example
$$x=-2,,y=frac12;,;text{and certainly:};;f(-2)=frac{-2}{1-4}=frac23=frac{frac12}{1-frac14} =fleft(frac12right)$$
and etc.
answered Jan 12 at 12:59
DonAntonioDonAntonio
178k1494230
$endgroup$
add a comment |
$begingroup$
Going by the very definition, as you did:
$$f(x)=f(y)ifffrac x{1-x^2}=frac y{1-y^2}iff x-xy^2=y-x^2yiff $$
$$iff(x-y)=-xy(x-y)iffbegin{cases}x=y\or\xy=-1end{cases}$$
Thus, any pair of numbers $;x,,yinBbb Rsetminus{-1,1};$ s.t. $;xy=-1;$ give you a counterexample to injectivity. For example
$$x=-2,,y=frac12;,;text{and certainly:};;f(-2)=frac{-2}{1-4}=frac23=frac{frac12}{1-frac14} =fleft(frac12right)$$
and etc.
answered Jan 12 at 12:59
DonAntonioDonAntonio
178k1494230
$endgroup$
Going by the very definition, as you did:
$$f(x)=f(y)ifffrac x{1-x^2}=frac y{1-y^2}iff x-xy^2=y-x^2yiff $$
$$iff(x-y)=-xy(x-y)iffbegin{cases}x=y\or\xy=-1end{cases}$$
Thus, any pair of numbers $;x,,yinBbb Rsetminus{-1,1};$ s.t. $;xy=-1;$ give you a counterexample to injectivity. For example
$$x=-2,,y=frac12;,;text{and certainly:};;f(-2)=frac{-2}{1-4}=frac23=frac{frac12}{1-frac14} =fleft(frac12right)$$
and etc.
answered Jan 12 at 12:59
DonAntonioDonAntonio
178k1494230
answered Jan 12 at 12:59
DonAntonioDonAntonio
178k1494230
answered Jan 12 at 12:59
DonAntonioDonAntonio
178k1494230
answered Jan 12 at 12:59
DonAntonioDonAntonio
178k1494230
178k1494230
add a comment |
add a comment |
$begingroup$
It is not an injective function as we can see from horizontal line test.
answered Jan 12 at 12:52
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
$begingroup$
It equals for $x,y$ such that $xcdot y = -1$.
$endgroup$
– Yanko
Jan 12 at 12:52
add a comment |
$begingroup$
It is not an injective function as we can see from horizontal line test.
answered Jan 12 at 12:52
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
$begingroup$
It equals for $x,y$ such that $xcdot y = -1$.
$endgroup$
– Yanko
Jan 12 at 12:52
add a comment |
$begingroup$
It is not an injective function as we can see from horizontal line test.
answered Jan 12 at 12:52
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
It is not an injective function as we can see from horizontal line test.
answered Jan 12 at 12:52
Siong Thye GohSiong Thye Goh
101k1466118
answered Jan 12 at 12:52
Siong Thye GohSiong Thye Goh
101k1466118
answered Jan 12 at 12:52
Siong Thye GohSiong Thye Goh
101k1466118
answered Jan 12 at 12:52
Siong Thye GohSiong Thye Goh
101k1466118
101k1466118
$begingroup$
It equals for $x,y$ such that $xcdot y = -1$.
$endgroup$
– Yanko
Jan 12 at 12:52
add a comment |
$begingroup$
It equals for $x,y$ such that $xcdot y = -1$.
$endgroup$
– Yanko
Jan 12 at 12:52
$begingroup$
It equals for $x,y$ such that $xcdot y = -1$.
$endgroup$
– Yanko
Jan 12 at 12:52
$begingroup$
It equals for $x,y$ such that $xcdot y = -1$.
$endgroup$
– Yanko
Jan 12 at 12:52
add a comment |
$begingroup$
There's an error in your computation:
$$ x - y = xy^2 - x^2y iff x-y = xycolor{red}{(y - x)}iffbegin{cases}x=y \ xy=-1iff y=-frac1x, ; xne 0
end{cases}$$
answered Jan 12 at 13:07
BernardBernard
121k740116
$endgroup$
$begingroup$
I edited. Thank you
$endgroup$
– Dimitris Dimitriadis
Jan 12 at 13:10
$begingroup$
You can not say that $y= frac{1}{x}$ because $x$ can be $0$
$endgroup$
– Dimitris Dimitriadis
Jan 12 at 13:12
add a comment |
$begingroup$
There's an error in your computation:
$$ x - y = xy^2 - x^2y iff x-y = xycolor{red}{(y - x)}iffbegin{cases}x=y \ xy=-1iff y=-frac1x, ; xne 0
end{cases}$$
answered Jan 12 at 13:07
BernardBernard
121k740116
$endgroup$
$begingroup$
I edited. Thank you
$endgroup$
– Dimitris Dimitriadis
Jan 12 at 13:10
$begingroup$
You can not say that $y= frac{1}{x}$ because $x$ can be $0$
$endgroup$
– Dimitris Dimitriadis
Jan 12 at 13:12
add a comment |
$begingroup$
There's an error in your computation:
$$ x - y = xy^2 - x^2y iff x-y = xycolor{red}{(y - x)}iffbegin{cases}x=y \ xy=-1iff y=-frac1x, ; xne 0
end{cases}$$
answered Jan 12 at 13:07
BernardBernard
121k740116
$endgroup$
There's an error in your computation:
$$ x - y = xy^2 - x^2y iff x-y = xycolor{red}{(y - x)}iffbegin{cases}x=y \ xy=-1iff y=-frac1x, ; xne 0
end{cases}$$
answered Jan 12 at 13:07
BernardBernard
121k740116
edited Jan 12 at 13:14
answered Jan 12 at 13:07
BernardBernard
121k740116
answered Jan 12 at 13:07
BernardBernard
121k740116
answered Jan 12 at 13:07
BernardBernard
121k740116
121k740116
$begingroup$
I edited. Thank you
$endgroup$
– Dimitris Dimitriadis
Jan 12 at 13:10
$begingroup$
You can not say that $y= frac{1}{x}$ because $x$ can be $0$
$endgroup$
– Dimitris Dimitriadis
Jan 12 at 13:12
add a comment |
$begingroup$
I edited. Thank you
$endgroup$
– Dimitris Dimitriadis
Jan 12 at 13:10
$begingroup$
You can not say that $y= frac{1}{x}$ because $x$ can be $0$
$endgroup$
– Dimitris Dimitriadis
Jan 12 at 13:12
$begingroup$
I edited. Thank you
$endgroup$
– Dimitris Dimitriadis
Jan 12 at 13:10
$begingroup$
I edited. Thank you
$endgroup$
– Dimitris Dimitriadis
Jan 12 at 13:10
$begingroup$
You can not say that $y= frac{1}{x}$ because $x$ can be $0$
$endgroup$
– Dimitris Dimitriadis
Jan 12 at 13:12
$begingroup$
You can not say that $y= frac{1}{x}$ because $x$ can be $0$
$endgroup$
– Dimitris Dimitriadis
Jan 12 at 13:12
add a comment |
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$begingroup$
$xy^2-x^2y = xy(y-x)$ no squares on the right hand side...
$endgroup$
– Yanko
Jan 12 at 12:50
$begingroup$
Thank you for your comment
$endgroup$
– Dimitris Dimitriadis
Jan 12 at 13:10