Let f : R → R be a function, such that $|f(x)−f(y)|≥5|x−y| :forall :x, yin mathbb{R}$. Show that $f$...
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Intro to Math Proofs course
Know basic concepts of Injection functions (one-to-one)
formal-proofs
edited Nov 21 at 17:55
Yadati Kiran
964316
asked Nov 21 at 17:46
smith
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put on hold as off-topic by Henrik, user10354138, user302797, A. Pongrácz, Kavi Rama Murthy 2 days ago
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, user302797, A. Pongrácz, Kavi Rama Murthy
If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
-1
down vote
favorite
Intro to Math Proofs course
Know basic concepts of Injection functions (one-to-one)
formal-proofs
edited Nov 21 at 17:55
Yadati Kiran
964316
asked Nov 21 at 17:46
smith
91
New contributor
smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Henrik, user10354138, user302797, A. Pongrácz, Kavi Rama Murthy 2 days ago
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, user302797, 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 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 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 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 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 at 18:01
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Intro to Math Proofs course
Know basic concepts of Injection functions (one-to-one)
formal-proofs
edited Nov 21 at 17:55
Yadati Kiran
964316
asked Nov 21 at 17:46
smith
91
New contributor
smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Intro to Math Proofs course
Know basic concepts of Injection functions (one-to-one)
formal-proofs
formal-proofs
edited Nov 21 at 17:55
Yadati Kiran
964316
asked Nov 21 at 17:46
smith
91
New contributor
smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 21 at 17:55
Yadati Kiran
964316
asked Nov 21 at 17:46
smith
91
New contributor
smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 21 at 17:55
Yadati Kiran
964316
edited Nov 21 at 17:55
Yadati Kiran
964316
edited Nov 21 at 17:55
Yadati Kiran
964316
964316
asked Nov 21 at 17:46
smith
91
New contributor
smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 21 at 17:46
smith
91
asked Nov 21 at 17:46
smith
91
91
New contributor
smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Henrik, user10354138, user302797, A. Pongrácz, Kavi Rama Murthy 2 days ago
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, user302797, A. Pongrácz, Kavi Rama Murthy
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by Henrik, user10354138, user302797, A. Pongrácz, Kavi Rama Murthy 2 days ago
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, user302797, 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 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 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 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 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 at 18:01
add a comment |
Welcome to MSE! What have you tried?
– MisterRiemann
Nov 21 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 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 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 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 at 18:01
Welcome to MSE! What have you tried?
– MisterRiemann
Nov 21 at 17:46
Welcome to MSE! What have you tried?
– MisterRiemann
Nov 21 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 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 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 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 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 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 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 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 at 18:01
add a comment |
2 Answers
2
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up vote
1
down vote
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 at 18:17
Luislandlaga
113
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Luislandlaga is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Very clear and helpful. Thank you
– smith
Nov 21 at 18:19
add a comment |
up vote
1
down vote
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 at 18:26
idea
2,20121024
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
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 at 18:17
Luislandlaga
113
New contributor
Luislandlaga is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Very clear and helpful. Thank you
– smith
Nov 21 at 18:19
add a comment |
up vote
1
down vote
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 at 18:17
Luislandlaga
113
New contributor
Luislandlaga is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Very clear and helpful. Thank you
– smith
Nov 21 at 18:19
add a comment |
up vote
1
down vote
up vote
1
down vote
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 at 18:17
Luislandlaga
113
New contributor
Luislandlaga is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 at 18:17
Luislandlaga
113
New contributor
Luislandlaga is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Nov 21 at 18:17
Luislandlaga
113
New contributor
Luislandlaga is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Nov 21 at 18:17
Luislandlaga
113
answered Nov 21 at 18:17
Luislandlaga
113
113
New contributor
Luislandlaga is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Luislandlaga is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Luislandlaga is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Very clear and helpful. Thank you
– smith
Nov 21 at 18:19
add a comment |
Very clear and helpful. Thank you
– smith
Nov 21 at 18:19
Very clear and helpful. Thank you
– smith
Nov 21 at 18:19
Very clear and helpful. Thank you
– smith
Nov 21 at 18:19
add a comment |
up vote
1
down vote
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 at 18:26
idea
2,20121024
add a comment |
up vote
1
down vote
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 at 18:26
idea
2,20121024
add a comment |
up vote
1
down vote
up vote
1
down vote
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 at 18:26
idea
2,20121024
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 at 18:26
idea
2,20121024
answered Nov 21 at 18:26
idea
2,20121024
answered Nov 21 at 18:26
idea
2,20121024
answered Nov 21 at 18:26
idea
2,20121024
2,20121024
add a comment |
add a comment |
Welcome to MSE! What have you tried?
– MisterRiemann
Nov 21 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 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 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 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 at 18:01