Mixing
Let f : R → R be a function, such that $|f(x)−f(y)|≥5|x−y| :forall :x, yin mathbb{R}$. Show that $f$...
Intro to Math Proofs course
Know basic concepts of Injection functions (one-to-one)
formal-proofs
edited Nov 21 '18 at 17:55
Yadati Kiran
1,694619
asked Nov 21 '18 at 17:46
smithsmith
93
closed as off-topic by Henrik, user10354138, Saad, A. Pongrácz, Kavi Rama Murthy Nov 22 '18 at 7:50
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Henrik, user10354138, Saad, A. Pongrácz, Kavi Rama Murthy
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
Intro to Math Proofs course
Know basic concepts of Injection functions (one-to-one)
formal-proofs
edited Nov 21 '18 at 17:55
Yadati Kiran
1,694619
asked Nov 21 '18 at 17:46
smithsmith
93
closed as off-topic by Henrik, user10354138, Saad, A. Pongrácz, Kavi Rama Murthy Nov 22 '18 at 7:50
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Henrik, user10354138, Saad, A. Pongrácz, Kavi Rama Murthy
If this question can be reworded to fit the rules in the help center, please edit the question.
Welcome to MSE! What have you tried?
– MisterRiemann
Nov 21 '18 at 17:46
Tried doing a contrapostive example (i.e for every x1, x2 in A x1 does not equal x2 implies that f(x1) does not equal f(x2) but i am not sure how to use this in this case with 2 variables
– smith
Nov 21 '18 at 17:50
2
How about using the definition directly: Assume that $f(x)=f(y)$ and show that this implies $x=y$.
– MisterRiemann
Nov 21 '18 at 17:52
That should work but I am unsure how to format or start the answer with 2 variables or how to use the inequality to show x=y.
– smith
Nov 21 '18 at 17:59
1
Literally just write it, lol. If $f(x)=f(y)$ then literally write that in your inequality and see what you get.
– Zachary Selk
Nov 21 '18 at 18:01
add a comment |
Intro to Math Proofs course
Know basic concepts of Injection functions (one-to-one)
formal-proofs
edited Nov 21 '18 at 17:55
Yadati Kiran
1,694619
asked Nov 21 '18 at 17:46
smithsmith
93
Intro to Math Proofs course
Know basic concepts of Injection functions (one-to-one)
formal-proofs
formal-proofs
edited Nov 21 '18 at 17:55
Yadati Kiran
1,694619
asked Nov 21 '18 at 17:46
smithsmith
93
edited Nov 21 '18 at 17:55
Yadati Kiran
1,694619
asked Nov 21 '18 at 17:46
smithsmith
93
edited Nov 21 '18 at 17:55
Yadati Kiran
1,694619
edited Nov 21 '18 at 17:55
Yadati Kiran
1,694619
edited Nov 21 '18 at 17:55
Yadati Kiran
1,694619
1,694619
asked Nov 21 '18 at 17:46
smithsmith
93
asked Nov 21 '18 at 17:46
smithsmith
93
asked Nov 21 '18 at 17:46
smithsmith
93
93
closed as off-topic by Henrik, user10354138, Saad, A. Pongrácz, Kavi Rama Murthy Nov 22 '18 at 7:50
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Henrik, user10354138, Saad, A. Pongrácz, Kavi Rama Murthy
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Henrik, user10354138, Saad, A. Pongrácz, Kavi Rama Murthy Nov 22 '18 at 7:50
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Henrik, user10354138, Saad, A. Pongrácz, Kavi Rama Murthy
If this question can be reworded to fit the rules in the help center, please edit the question.
Welcome to MSE! What have you tried?
– MisterRiemann
Nov 21 '18 at 17:46
Tried doing a contrapostive example (i.e for every x1, x2 in A x1 does not equal x2 implies that f(x1) does not equal f(x2) but i am not sure how to use this in this case with 2 variables
– smith
Nov 21 '18 at 17:50
2
How about using the definition directly: Assume that $f(x)=f(y)$ and show that this implies $x=y$.
– MisterRiemann
Nov 21 '18 at 17:52
That should work but I am unsure how to format or start the answer with 2 variables or how to use the inequality to show x=y.
– smith
Nov 21 '18 at 17:59
1
Literally just write it, lol. If $f(x)=f(y)$ then literally write that in your inequality and see what you get.
– Zachary Selk
Nov 21 '18 at 18:01
add a comment |
Welcome to MSE! What have you tried?
– MisterRiemann
Nov 21 '18 at 17:46
Tried doing a contrapostive example (i.e for every x1, x2 in A x1 does not equal x2 implies that f(x1) does not equal f(x2) but i am not sure how to use this in this case with 2 variables
– smith
Nov 21 '18 at 17:50
2
How about using the definition directly: Assume that $f(x)=f(y)$ and show that this implies $x=y$.
– MisterRiemann
Nov 21 '18 at 17:52
That should work but I am unsure how to format or start the answer with 2 variables or how to use the inequality to show x=y.
– smith
Nov 21 '18 at 17:59
1
Literally just write it, lol. If $f(x)=f(y)$ then literally write that in your inequality and see what you get.
– Zachary Selk
Nov 21 '18 at 18:01
Welcome to MSE! What have you tried?
– MisterRiemann
Nov 21 '18 at 17:46
Welcome to MSE! What have you tried?
– MisterRiemann
Nov 21 '18 at 17:46
Tried doing a contrapostive example (i.e for every x1, x2 in A x1 does not equal x2 implies that f(x1) does not equal f(x2) but i am not sure how to use this in this case with 2 variables
– smith
Nov 21 '18 at 17:50
Tried doing a contrapostive example (i.e for every x1, x2 in A x1 does not equal x2 implies that f(x1) does not equal f(x2) but i am not sure how to use this in this case with 2 variables
– smith
Nov 21 '18 at 17:50
2
2
How about using the definition directly: Assume that $f(x)=f(y)$ and show that this implies $x=y$.
– MisterRiemann
Nov 21 '18 at 17:52
How about using the definition directly: Assume that $f(x)=f(y)$ and show that this implies $x=y$.
– MisterRiemann
Nov 21 '18 at 17:52
That should work but I am unsure how to format or start the answer with 2 variables or how to use the inequality to show x=y.
– smith
Nov 21 '18 at 17:59
That should work but I am unsure how to format or start the answer with 2 variables or how to use the inequality to show x=y.
– smith
Nov 21 '18 at 17:59
1
1
Literally just write it, lol. If $f(x)=f(y)$ then literally write that in your inequality and see what you get.
– Zachary Selk
Nov 21 '18 at 18:01
Literally just write it, lol. If $f(x)=f(y)$ then literally write that in your inequality and see what you get.
– Zachary Selk
Nov 21 '18 at 18:01
add a comment |
2 Answers
2
active
oldest
votes
To prove that a funtion is injective you've got to show that $f(x)=f(y) Rightarrow x=y$.
Now try to prove this by contradiction, meaning if $f(x)-f(y)=0$ and $xneq y$ that is $|f(x)-f(y)|=0 geq 5|x-y|$, now, since $xneq y Rightarrow |x-y|>0$, which implies that $0$ is bigger than five times some positive integer, which is absurd. Now, it follows that is $f(x)=f(y)$ then $x=y$
answered Nov 21 '18 at 18:17
LuislandlagaLuislandlaga
113
Very clear and helpful. Thank you
– smith
Nov 21 '18 at 18:19
add a comment |
Shortcut approach:
Given:$$|f(x)−f(y)|≥5|x−y| :forall :x, yin mathbb{R}$$
By definition of injective functions: $$f(x)=f(y) hspace{0.5cm}text{implies} hspace{0.5cm}x=y $$
So, if you put $f(x)=f(y)$, your LHS becomes $0$.
So needs to be RHS, which is only possible if $x=y$.
Proved!
answered Nov 21 '18 at 18:26
ideaidea
2,15441025
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
To prove that a funtion is injective you've got to show that $f(x)=f(y) Rightarrow x=y$.
Now try to prove this by contradiction, meaning if $f(x)-f(y)=0$ and $xneq y$ that is $|f(x)-f(y)|=0 geq 5|x-y|$, now, since $xneq y Rightarrow |x-y|>0$, which implies that $0$ is bigger than five times some positive integer, which is absurd. Now, it follows that is $f(x)=f(y)$ then $x=y$
answered Nov 21 '18 at 18:17
LuislandlagaLuislandlaga
113
Very clear and helpful. Thank you
– smith
Nov 21 '18 at 18:19
add a comment |
To prove that a funtion is injective you've got to show that $f(x)=f(y) Rightarrow x=y$.
Now try to prove this by contradiction, meaning if $f(x)-f(y)=0$ and $xneq y$ that is $|f(x)-f(y)|=0 geq 5|x-y|$, now, since $xneq y Rightarrow |x-y|>0$, which implies that $0$ is bigger than five times some positive integer, which is absurd. Now, it follows that is $f(x)=f(y)$ then $x=y$
answered Nov 21 '18 at 18:17
LuislandlagaLuislandlaga
113
Very clear and helpful. Thank you
– smith
Nov 21 '18 at 18:19
add a comment |
To prove that a funtion is injective you've got to show that $f(x)=f(y) Rightarrow x=y$.
Now try to prove this by contradiction, meaning if $f(x)-f(y)=0$ and $xneq y$ that is $|f(x)-f(y)|=0 geq 5|x-y|$, now, since $xneq y Rightarrow |x-y|>0$, which implies that $0$ is bigger than five times some positive integer, which is absurd. Now, it follows that is $f(x)=f(y)$ then $x=y$
answered Nov 21 '18 at 18:17
LuislandlagaLuislandlaga
113
To prove that a funtion is injective you've got to show that $f(x)=f(y) Rightarrow x=y$.
Now try to prove this by contradiction, meaning if $f(x)-f(y)=0$ and $xneq y$ that is $|f(x)-f(y)|=0 geq 5|x-y|$, now, since $xneq y Rightarrow |x-y|>0$, which implies that $0$ is bigger than five times some positive integer, which is absurd. Now, it follows that is $f(x)=f(y)$ then $x=y$
answered Nov 21 '18 at 18:17
LuislandlagaLuislandlaga
113
answered Nov 21 '18 at 18:17
LuislandlagaLuislandlaga
113
answered Nov 21 '18 at 18:17
LuislandlagaLuislandlaga
113
answered Nov 21 '18 at 18:17
LuislandlagaLuislandlaga
113
113
Very clear and helpful. Thank you
– smith
Nov 21 '18 at 18:19
add a comment |
Very clear and helpful. Thank you
– smith
Nov 21 '18 at 18:19
Very clear and helpful. Thank you
– smith
Nov 21 '18 at 18:19
Very clear and helpful. Thank you
– smith
Nov 21 '18 at 18:19
add a comment |
Shortcut approach:
Given:$$|f(x)−f(y)|≥5|x−y| :forall :x, yin mathbb{R}$$
By definition of injective functions: $$f(x)=f(y) hspace{0.5cm}text{implies} hspace{0.5cm}x=y $$
So, if you put $f(x)=f(y)$, your LHS becomes $0$.
So needs to be RHS, which is only possible if $x=y$.
Proved!
answered Nov 21 '18 at 18:26
ideaidea
2,15441025
add a comment |
Shortcut approach:
Given:$$|f(x)−f(y)|≥5|x−y| :forall :x, yin mathbb{R}$$
By definition of injective functions: $$f(x)=f(y) hspace{0.5cm}text{implies} hspace{0.5cm}x=y $$
So, if you put $f(x)=f(y)$, your LHS becomes $0$.
So needs to be RHS, which is only possible if $x=y$.
Proved!
answered Nov 21 '18 at 18:26
ideaidea
2,15441025
add a comment |
Shortcut approach:
Given:$$|f(x)−f(y)|≥5|x−y| :forall :x, yin mathbb{R}$$
By definition of injective functions: $$f(x)=f(y) hspace{0.5cm}text{implies} hspace{0.5cm}x=y $$
So, if you put $f(x)=f(y)$, your LHS becomes $0$.
So needs to be RHS, which is only possible if $x=y$.
Proved!
answered Nov 21 '18 at 18:26
ideaidea
2,15441025
Shortcut approach:
Given:$$|f(x)−f(y)|≥5|x−y| :forall :x, yin mathbb{R}$$
By definition of injective functions: $$f(x)=f(y) hspace{0.5cm}text{implies} hspace{0.5cm}x=y $$
So, if you put $f(x)=f(y)$, your LHS becomes $0$.
So needs to be RHS, which is only possible if $x=y$.
Proved!
answered Nov 21 '18 at 18:26
ideaidea
2,15441025
answered Nov 21 '18 at 18:26
ideaidea
2,15441025
answered Nov 21 '18 at 18:26
ideaidea
2,15441025
answered Nov 21 '18 at 18:26
ideaidea
2,15441025
2,15441025
add a comment |
add a comment |
Welcome to MSE! What have you tried?
– MisterRiemann
Nov 21 '18 at 17:46
Tried doing a contrapostive example (i.e for every x1, x2 in A x1 does not equal x2 implies that f(x1) does not equal f(x2) but i am not sure how to use this in this case with 2 variables
– smith
Nov 21 '18 at 17:50
2
How about using the definition directly: Assume that $f(x)=f(y)$ and show that this implies $x=y$.
– MisterRiemann
Nov 21 '18 at 17:52
That should work but I am unsure how to format or start the answer with 2 variables or how to use the inequality to show x=y.
– smith
Nov 21 '18 at 17:59
1
Literally just write it, lol. If $f(x)=f(y)$ then literally write that in your inequality and see what you get.
– Zachary Selk
Nov 21 '18 at 18:01