Define a function 𝒇:𝑹 − {𝟎} → 𝑹 − {𝟐} by the formula 𝑓(𝑥) = (2𝑥+1)/x for all...
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Define a function 𝒇:𝑹 − {𝟎} → 𝑹 − {𝟐} by the formula 𝑓(𝑥) = (2𝑥+1)/x for all non-zero real numbers x.
The 2019 Stack Overflow Developer Survey Results Are InProving whether functions are one-to-one and onto.Consider the function h where $h(x,y) = (x+y,x-y)$, $h : mathbb Ntimes mathbb Nto mathbb Ntimesmathbb N$Function with single formula that have non zero value only in one small rangeHow to display one to one correspondence for all bit strings not containing the bit O?Investigating the bijectivity of $ 2 x + |cos(x)| $.Using a formula, define a function $f:Ato B$ which is surjective but not injective.Bijection between the sets $ A={x^{2}: 0<x<1 } text{ and }B={x^{3}: 1<x<2 } $Function with domain all real numbers and range $(0,1)$Prove that $f$ is ontoExample of Composition of 2 functions onto or one one but that both function need not onto or one-one.
$begingroup$
1) Prove that f is one-to-one.
2) Prove that f is onto.
To prove that a function is one to one. I can use the approach g(a) = g(b) right?
So it will become (2a+1)/a = (2b+1)/b
b(2a + 1) = a(2a + 1)
2ab + b = 2ab + a
b = a, or a = b
Is this right?
Any help would be appreciated.
functions
asked Mar 21 at 4:09
apex_123apex_123
12
$endgroup$
|
show 6 more comments
$begingroup$
1) Prove that f is one-to-one.
2) Prove that f is onto.
To prove that a function is one to one. I can use the approach g(a) = g(b) right?
So it will become (2a+1)/a = (2b+1)/b
b(2a + 1) = a(2a + 1)
2ab + b = 2ab + a
b = a, or a = b
Is this right?
Any help would be appreciated.
functions
asked Mar 21 at 4:09
apex_123apex_123
12
$endgroup$
$begingroup$
You know the definition of a function being one-to-one, right? Please state it so that I can confirm it is correct. What happened when you tried to apply that definition?
$endgroup$
– астон вілла олоф мэллбэрг
Mar 21 at 4:15
$begingroup$
hey @астонвіллаолофмэллбэрг, I have edited my post. Can you check if I have the right approach?
$endgroup$
– apex_123
Mar 21 at 4:18
$begingroup$
Use MathJax for readability, it will also attract more attention to your question. Also, your explanation for one-one is correct. How about the onto explanation?
$endgroup$
– астон вілла олоф мэллбэрг
Mar 21 at 4:23
$begingroup$
The definition of onto is that every element of y is mapped to x and no number is left out. But I don't know how to prove it mathematically for this question
$endgroup$
– apex_123
Mar 21 at 4:30
$begingroup$
Yes, so the idea is : given a $y in mathbb R setminus {2}$ , we want to find an $x$ such that $x$ maps to $y$. This means $frac{2x+1}{x} = y$. Now, make $x$ the subject of this equation i.e. above we have written $y$ in terms of $x$, right? Now use techniques you know to make $x$ one side of the equation and the other side only depending on $y$. Then, such an $x$ will map to $y$, right? Try this.
$endgroup$
– астон вілла олоф мэллбэрг
Mar 21 at 4:33
|
show 6 more comments
$begingroup$
1) Prove that f is one-to-one.
2) Prove that f is onto.
To prove that a function is one to one. I can use the approach g(a) = g(b) right?
So it will become (2a+1)/a = (2b+1)/b
b(2a + 1) = a(2a + 1)
2ab + b = 2ab + a
b = a, or a = b
Is this right?
Any help would be appreciated.
functions
asked Mar 21 at 4:09
apex_123apex_123
12
$endgroup$
1) Prove that f is one-to-one.
2) Prove that f is onto.
To prove that a function is one to one. I can use the approach g(a) = g(b) right?
So it will become (2a+1)/a = (2b+1)/b
b(2a + 1) = a(2a + 1)
2ab + b = 2ab + a
b = a, or a = b
Is this right?
Any help would be appreciated.
functions
functions
asked Mar 21 at 4:09
apex_123apex_123
12
asked Mar 21 at 4:09
apex_123apex_123
12
edited Mar 21 at 4:18
apex_123
asked Mar 21 at 4:09
apex_123apex_123
12
asked Mar 21 at 4:09
apex_123apex_123
12
asked Mar 21 at 4:09
apex_123apex_123
12
12
$begingroup$
You know the definition of a function being one-to-one, right? Please state it so that I can confirm it is correct. What happened when you tried to apply that definition?
$endgroup$
– астон вілла олоф мэллбэрг
Mar 21 at 4:15
$begingroup$
hey @астонвіллаолофмэллбэрг, I have edited my post. Can you check if I have the right approach?
$endgroup$
– apex_123
Mar 21 at 4:18
$begingroup$
Use MathJax for readability, it will also attract more attention to your question. Also, your explanation for one-one is correct. How about the onto explanation?
$endgroup$
– астон вілла олоф мэллбэрг
Mar 21 at 4:23
$begingroup$
The definition of onto is that every element of y is mapped to x and no number is left out. But I don't know how to prove it mathematically for this question
$endgroup$
– apex_123
Mar 21 at 4:30
$begingroup$
Yes, so the idea is : given a $y in mathbb R setminus {2}$ , we want to find an $x$ such that $x$ maps to $y$. This means $frac{2x+1}{x} = y$. Now, make $x$ the subject of this equation i.e. above we have written $y$ in terms of $x$, right? Now use techniques you know to make $x$ one side of the equation and the other side only depending on $y$. Then, such an $x$ will map to $y$, right? Try this.
$endgroup$
– астон вілла олоф мэллбэрг
Mar 21 at 4:33
|
show 6 more comments
$begingroup$
You know the definition of a function being one-to-one, right? Please state it so that I can confirm it is correct. What happened when you tried to apply that definition?
$endgroup$
– астон вілла олоф мэллбэрг
Mar 21 at 4:15
$begingroup$
hey @астонвіллаолофмэллбэрг, I have edited my post. Can you check if I have the right approach?
$endgroup$
– apex_123
Mar 21 at 4:18
$begingroup$
Use MathJax for readability, it will also attract more attention to your question. Also, your explanation for one-one is correct. How about the onto explanation?
$endgroup$
– астон вілла олоф мэллбэрг
Mar 21 at 4:23
$begingroup$
The definition of onto is that every element of y is mapped to x and no number is left out. But I don't know how to prove it mathematically for this question
$endgroup$
– apex_123
Mar 21 at 4:30
$begingroup$
Yes, so the idea is : given a $y in mathbb R setminus {2}$ , we want to find an $x$ such that $x$ maps to $y$. This means $frac{2x+1}{x} = y$. Now, make $x$ the subject of this equation i.e. above we have written $y$ in terms of $x$, right? Now use techniques you know to make $x$ one side of the equation and the other side only depending on $y$. Then, such an $x$ will map to $y$, right? Try this.
$endgroup$
– астон вілла олоф мэллбэрг
Mar 21 at 4:33
$begingroup$
You know the definition of a function being one-to-one, right? Please state it so that I can confirm it is correct. What happened when you tried to apply that definition?
$endgroup$
– астон вілла олоф мэллбэрг
Mar 21 at 4:15
$begingroup$
You know the definition of a function being one-to-one, right? Please state it so that I can confirm it is correct. What happened when you tried to apply that definition?
$endgroup$
– астон вілла олоф мэллбэрг
Mar 21 at 4:15
$begingroup$
hey @астонвіллаолофмэллбэрг, I have edited my post. Can you check if I have the right approach?
$endgroup$
– apex_123
Mar 21 at 4:18
$begingroup$
hey @астонвіллаолофмэллбэрг, I have edited my post. Can you check if I have the right approach?
$endgroup$
– apex_123
Mar 21 at 4:18
$begingroup$
Use MathJax for readability, it will also attract more attention to your question. Also, your explanation for one-one is correct. How about the onto explanation?
$endgroup$
– астон вілла олоф мэллбэрг
Mar 21 at 4:23
$begingroup$
Use MathJax for readability, it will also attract more attention to your question. Also, your explanation for one-one is correct. How about the onto explanation?
$endgroup$
– астон вілла олоф мэллбэрг
Mar 21 at 4:23
$begingroup$
The definition of onto is that every element of y is mapped to x and no number is left out. But I don't know how to prove it mathematically for this question
$endgroup$
– apex_123
Mar 21 at 4:30
$begingroup$
The definition of onto is that every element of y is mapped to x and no number is left out. But I don't know how to prove it mathematically for this question
$endgroup$
– apex_123
Mar 21 at 4:30
$begingroup$
Yes, so the idea is : given a $y in mathbb R setminus {2}$ , we want to find an $x$ such that $x$ maps to $y$. This means $frac{2x+1}{x} = y$. Now, make $x$ the subject of this equation i.e. above we have written $y$ in terms of $x$, right? Now use techniques you know to make $x$ one side of the equation and the other side only depending on $y$. Then, such an $x$ will map to $y$, right? Try this.
$endgroup$
– астон вілла олоф мэллбэрг
Mar 21 at 4:33
$begingroup$
Yes, so the idea is : given a $y in mathbb R setminus {2}$ , we want to find an $x$ such that $x$ maps to $y$. This means $frac{2x+1}{x} = y$. Now, make $x$ the subject of this equation i.e. above we have written $y$ in terms of $x$, right? Now use techniques you know to make $x$ one side of the equation and the other side only depending on $y$. Then, such an $x$ will map to $y$, right? Try this.
$endgroup$
– астон вілла олоф мэллбэрг
Mar 21 at 4:33
|
show 6 more comments
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$begingroup$
You know the definition of a function being one-to-one, right? Please state it so that I can confirm it is correct. What happened when you tried to apply that definition?
$endgroup$
– астон вілла олоф мэллбэрг
Mar 21 at 4:15
$begingroup$
hey @астонвіллаолофмэллбэрг, I have edited my post. Can you check if I have the right approach?
$endgroup$
– apex_123
Mar 21 at 4:18
$begingroup$
Use MathJax for readability, it will also attract more attention to your question. Also, your explanation for one-one is correct. How about the onto explanation?
$endgroup$
– астон вілла олоф мэллбэрг
Mar 21 at 4:23
$begingroup$
The definition of onto is that every element of y is mapped to x and no number is left out. But I don't know how to prove it mathematically for this question
$endgroup$
– apex_123
Mar 21 at 4:30
$begingroup$
Yes, so the idea is : given a $y in mathbb R setminus {2}$ , we want to find an $x$ such that $x$ maps to $y$. This means $frac{2x+1}{x} = y$. Now, make $x$ the subject of this equation i.e. above we have written $y$ in terms of $x$, right? Now use techniques you know to make $x$ one side of the equation and the other side only depending on $y$. Then, such an $x$ will map to $y$, right? Try this.
$endgroup$
– астон вілла олоф мэллбэрг
Mar 21 at 4:33