Math Proof: Surjective/Injctive
Q: Let f : R → R be bijective. Define g : R → R as g(x) = |x|. Then g ◦ f is neither surjective nor injective
Pf: Let f: R → R be bijective. g: R1 → R2 as g(x)=|x|. Consider x1=2 and x2=2. g(x1)-g(2)=|2|=2=2=|-2|=g(-2)=g(x2). Since g(x1)=g(x2) then g is not injective. Now consider n=-2. -2 does not exist in R2 since R2>=0 by the definition of absolute value and there does not exist and x in R1 s.t. f(x)=-2. As a result, g is not surjective. So how do we prove that g ◦ f is neither surjective nor injective?
functions proof-verification elementary-set-theory proof-explanation
edited Nov 29 at 20:19
Martin Sleziak
44.6k7115270
asked Nov 29 at 19:27
Robert Klein
65
add a comment |
Q: Let f : R → R be bijective. Define g : R → R as g(x) = |x|. Then g ◦ f is neither surjective nor injective
Pf: Let f: R → R be bijective. g: R1 → R2 as g(x)=|x|. Consider x1=2 and x2=2. g(x1)-g(2)=|2|=2=2=|-2|=g(-2)=g(x2). Since g(x1)=g(x2) then g is not injective. Now consider n=-2. -2 does not exist in R2 since R2>=0 by the definition of absolute value and there does not exist and x in R1 s.t. f(x)=-2. As a result, g is not surjective. So how do we prove that g ◦ f is neither surjective nor injective?
functions proof-verification elementary-set-theory proof-explanation
edited Nov 29 at 20:19
Martin Sleziak
44.6k7115270
asked Nov 29 at 19:27
Robert Klein
65
1
In your proof, x$_1$ = x$_2$. Also what is R$_1$ and R$_2$?
– Joel Pereira
Nov 29 at 19:34
add a comment |
Q: Let f : R → R be bijective. Define g : R → R as g(x) = |x|. Then g ◦ f is neither surjective nor injective
Pf: Let f: R → R be bijective. g: R1 → R2 as g(x)=|x|. Consider x1=2 and x2=2. g(x1)-g(2)=|2|=2=2=|-2|=g(-2)=g(x2). Since g(x1)=g(x2) then g is not injective. Now consider n=-2. -2 does not exist in R2 since R2>=0 by the definition of absolute value and there does not exist and x in R1 s.t. f(x)=-2. As a result, g is not surjective. So how do we prove that g ◦ f is neither surjective nor injective?
functions proof-verification elementary-set-theory proof-explanation
edited Nov 29 at 20:19
Martin Sleziak
44.6k7115270
asked Nov 29 at 19:27
Robert Klein
65
Q: Let f : R → R be bijective. Define g : R → R as g(x) = |x|. Then g ◦ f is neither surjective nor injective
Pf: Let f: R → R be bijective. g: R1 → R2 as g(x)=|x|. Consider x1=2 and x2=2. g(x1)-g(2)=|2|=2=2=|-2|=g(-2)=g(x2). Since g(x1)=g(x2) then g is not injective. Now consider n=-2. -2 does not exist in R2 since R2>=0 by the definition of absolute value and there does not exist and x in R1 s.t. f(x)=-2. As a result, g is not surjective. So how do we prove that g ◦ f is neither surjective nor injective?
functions proof-verification elementary-set-theory proof-explanation
functions proof-verification elementary-set-theory proof-explanation
edited Nov 29 at 20:19
Martin Sleziak
44.6k7115270
asked Nov 29 at 19:27
Robert Klein
65
edited Nov 29 at 20:19
Martin Sleziak
44.6k7115270
asked Nov 29 at 19:27
Robert Klein
65
edited Nov 29 at 20:19
Martin Sleziak
44.6k7115270
edited Nov 29 at 20:19
Martin Sleziak
44.6k7115270
edited Nov 29 at 20:19
Martin Sleziak
44.6k7115270
44.6k7115270
asked Nov 29 at 19:27
Robert Klein
65
asked Nov 29 at 19:27
Robert Klein
65
asked Nov 29 at 19:27
Robert Klein
65
65
1
In your proof, x$_1$ = x$_2$. Also what is R$_1$ and R$_2$?
– Joel Pereira
Nov 29 at 19:34
add a comment |
1
In your proof, x$_1$ = x$_2$. Also what is R$_1$ and R$_2$?
– Joel Pereira
Nov 29 at 19:34
1
1
In your proof, x$_1$ = x$_2$. Also what is R$_1$ and R$_2$?
– Joel Pereira
Nov 29 at 19:34
In your proof, x$_1$ = x$_2$. Also what is R$_1$ and R$_2$?
– Joel Pereira
Nov 29 at 19:34
add a comment |
1 Answer
1
active
oldest
votes
hint
for a real $x$, let
$$h(x)=gcirc f(x)=|f(x)|$$
thus
$$h(x_1)=h(x_2)implies |f(x_1)|=|f(x_2)|$$
$$implies f(x_1)=pm f(x_2)$$
If $f$ is positive, then $gcirc f$ is bijective.
answered Nov 29 at 19:42
hamam_Abdallah
37.8k21634
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
hint
for a real $x$, let
$$h(x)=gcirc f(x)=|f(x)|$$
thus
$$h(x_1)=h(x_2)implies |f(x_1)|=|f(x_2)|$$
$$implies f(x_1)=pm f(x_2)$$
If $f$ is positive, then $gcirc f$ is bijective.
answered Nov 29 at 19:42
hamam_Abdallah
37.8k21634
add a comment |
hint
for a real $x$, let
$$h(x)=gcirc f(x)=|f(x)|$$
thus
$$h(x_1)=h(x_2)implies |f(x_1)|=|f(x_2)|$$
$$implies f(x_1)=pm f(x_2)$$
If $f$ is positive, then $gcirc f$ is bijective.
answered Nov 29 at 19:42
hamam_Abdallah
37.8k21634
add a comment |
hint
for a real $x$, let
$$h(x)=gcirc f(x)=|f(x)|$$
thus
$$h(x_1)=h(x_2)implies |f(x_1)|=|f(x_2)|$$
$$implies f(x_1)=pm f(x_2)$$
If $f$ is positive, then $gcirc f$ is bijective.
answered Nov 29 at 19:42
hamam_Abdallah
37.8k21634
hint
for a real $x$, let
$$h(x)=gcirc f(x)=|f(x)|$$
thus
$$h(x_1)=h(x_2)implies |f(x_1)|=|f(x_2)|$$
$$implies f(x_1)=pm f(x_2)$$
If $f$ is positive, then $gcirc f$ is bijective.
answered Nov 29 at 19:42
hamam_Abdallah
37.8k21634
answered Nov 29 at 19:42
hamam_Abdallah
37.8k21634
answered Nov 29 at 19:42
hamam_Abdallah
37.8k21634
answered Nov 29 at 19:42
hamam_Abdallah
37.8k21634
37.8k21634
add a comment |
add a comment |
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In your proof, x$_1$ = x$_2$. Also what is R$_1$ and R$_2$?
– Joel Pereira
Nov 29 at 19:34