Proof by contradiction for one to one relation
$begingroup$
Prove the following statement by contradiction:
Let $f:A to B$. Then $f^{−1}:Range(f) to A$ implies $f$ is one-to-one.
I can deduce why the statement must be true but i am having trouble using a proof by contradiction.
My proof so far assumes that $f$ is not one to one and $f^{−1}:Range(f) to A$,
Which implies that the inverse function in not a function, which is a contradiction.
FSOC assume that $f:A to B$ is not 1 to 1 and $f^{-1}:Ran(f) to A$
$therefore exists y in B | y=f(a_1 )=f(a_2 )$,where $a_1 neq a_2$
$therefore exists a_1$ and $exists a_2 in Dom(f)$ where $a_1 neq a_2 | f(a_1 )=f(a_2 )=y_1$
$(therefore f^{-1} (Ran(f))$ is not a function because it does not pass the vertical line test.#)
Is this proof acceptable?
discrete-mathematics proof-writing
edited Nov 28 '18 at 6:35
platty
3,370320
asked Nov 28 '18 at 6:30
Kedar YadavKedar Yadav
42
$endgroup$
add a comment |
$begingroup$
Prove the following statement by contradiction:
Let $f:A to B$. Then $f^{−1}:Range(f) to A$ implies $f$ is one-to-one.
I can deduce why the statement must be true but i am having trouble using a proof by contradiction.
My proof so far assumes that $f$ is not one to one and $f^{−1}:Range(f) to A$,
Which implies that the inverse function in not a function, which is a contradiction.
FSOC assume that $f:A to B$ is not 1 to 1 and $f^{-1}:Ran(f) to A$
$therefore exists y in B | y=f(a_1 )=f(a_2 )$,where $a_1 neq a_2$
$therefore exists a_1$ and $exists a_2 in Dom(f)$ where $a_1 neq a_2 | f(a_1 )=f(a_2 )=y_1$
$(therefore f^{-1} (Ran(f))$ is not a function because it does not pass the vertical line test.#)
Is this proof acceptable?
discrete-mathematics proof-writing
edited Nov 28 '18 at 6:35
platty
3,370320
asked Nov 28 '18 at 6:30
Kedar YadavKedar Yadav
42
$endgroup$
add a comment |
$begingroup$
Prove the following statement by contradiction:
Let $f:A to B$. Then $f^{−1}:Range(f) to A$ implies $f$ is one-to-one.
I can deduce why the statement must be true but i am having trouble using a proof by contradiction.
My proof so far assumes that $f$ is not one to one and $f^{−1}:Range(f) to A$,
Which implies that the inverse function in not a function, which is a contradiction.
FSOC assume that $f:A to B$ is not 1 to 1 and $f^{-1}:Ran(f) to A$
$therefore exists y in B | y=f(a_1 )=f(a_2 )$,where $a_1 neq a_2$
$therefore exists a_1$ and $exists a_2 in Dom(f)$ where $a_1 neq a_2 | f(a_1 )=f(a_2 )=y_1$
$(therefore f^{-1} (Ran(f))$ is not a function because it does not pass the vertical line test.#)
Is this proof acceptable?
discrete-mathematics proof-writing
edited Nov 28 '18 at 6:35
platty
3,370320
asked Nov 28 '18 at 6:30
Kedar YadavKedar Yadav
42
$endgroup$
Prove the following statement by contradiction:
Let $f:A to B$. Then $f^{−1}:Range(f) to A$ implies $f$ is one-to-one.
I can deduce why the statement must be true but i am having trouble using a proof by contradiction.
My proof so far assumes that $f$ is not one to one and $f^{−1}:Range(f) to A$,
Which implies that the inverse function in not a function, which is a contradiction.
FSOC assume that $f:A to B$ is not 1 to 1 and $f^{-1}:Ran(f) to A$
$therefore exists y in B | y=f(a_1 )=f(a_2 )$,where $a_1 neq a_2$
$therefore exists a_1$ and $exists a_2 in Dom(f)$ where $a_1 neq a_2 | f(a_1 )=f(a_2 )=y_1$
$(therefore f^{-1} (Ran(f))$ is not a function because it does not pass the vertical line test.#)
Is this proof acceptable?
discrete-mathematics proof-writing
discrete-mathematics proof-writing
edited Nov 28 '18 at 6:35
platty
3,370320
asked Nov 28 '18 at 6:30
Kedar YadavKedar Yadav
42
edited Nov 28 '18 at 6:35
platty
3,370320
asked Nov 28 '18 at 6:30
Kedar YadavKedar Yadav
42
edited Nov 28 '18 at 6:35
platty
3,370320
edited Nov 28 '18 at 6:35
platty
3,370320
edited Nov 28 '18 at 6:35
platty
3,370320
3,370320
asked Nov 28 '18 at 6:30
Kedar YadavKedar Yadav
42
asked Nov 28 '18 at 6:30
Kedar YadavKedar Yadav
42
asked Nov 28 '18 at 6:30
Kedar YadavKedar Yadav
42
42
add a comment |
add a comment |
1 Answer
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Yes, it is acceptable. Note that from your second point, it follows that $exists y in Range(f)$ such that $ f^{-1}(y) = a_{1} = a_{2}$, which contradicts the fact that $f^{-1}$ is a function.
answered Nov 28 '18 at 6:46
user1430user1430
507
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add a comment |
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1 Answer
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$begingroup$
Yes, it is acceptable. Note that from your second point, it follows that $exists y in Range(f)$ such that $ f^{-1}(y) = a_{1} = a_{2}$, which contradicts the fact that $f^{-1}$ is a function.
answered Nov 28 '18 at 6:46
user1430user1430
507
$endgroup$
add a comment |
$begingroup$
Yes, it is acceptable. Note that from your second point, it follows that $exists y in Range(f)$ such that $ f^{-1}(y) = a_{1} = a_{2}$, which contradicts the fact that $f^{-1}$ is a function.
answered Nov 28 '18 at 6:46
user1430user1430
507
$endgroup$
add a comment |
$begingroup$
Yes, it is acceptable. Note that from your second point, it follows that $exists y in Range(f)$ such that $ f^{-1}(y) = a_{1} = a_{2}$, which contradicts the fact that $f^{-1}$ is a function.
answered Nov 28 '18 at 6:46
user1430user1430
507
$endgroup$
Yes, it is acceptable. Note that from your second point, it follows that $exists y in Range(f)$ such that $ f^{-1}(y) = a_{1} = a_{2}$, which contradicts the fact that $f^{-1}$ is a function.
answered Nov 28 '18 at 6:46
user1430user1430
507
answered Nov 28 '18 at 6:46
user1430user1430
507
answered Nov 28 '18 at 6:46
user1430user1430
507
answered Nov 28 '18 at 6:46
user1430user1430
507
507
add a comment |
add a comment |
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