Mixing
Image of $f: Ato B$, $;f(x)=|x/2|$, for $xin mathbb Q$ such that $-12leq xleq 4$.
$begingroup$
Let $A subseteq mathbb{R}, B subset mathbb{R}^{+} cup {0}$, and $A={{xinmathbb{Q}mid -12leq x leq4}}.;$ Find $f(A)$, where $f(x)= |frac x2|$.
How to start? $mathbb{Q}$ is the set of rational numbers, so in $A$ will be $1/2,1/3,1/4$... to $4$ and in the negative to $-12$? There will be a lot of numbers to write! Please help.
Is it $$A={−12,−11,−10,−9,−8,−7,−6,−5,−4,−3,−2,−1,0,1,2,3,4}$$ or no?
functions discrete-mathematics
edited Jan 17 at 23:45
jordan_glen
1
asked Jan 17 at 17:23
ViktorViktor
477
$endgroup$
add a comment |
$begingroup$
Let $A subseteq mathbb{R}, B subset mathbb{R}^{+} cup {0}$, and $A={{xinmathbb{Q}mid -12leq x leq4}}.;$ Find $f(A)$, where $f(x)= |frac x2|$.
How to start? $mathbb{Q}$ is the set of rational numbers, so in $A$ will be $1/2,1/3,1/4$... to $4$ and in the negative to $-12$? There will be a lot of numbers to write! Please help.
Is it $$A={−12,−11,−10,−9,−8,−7,−6,−5,−4,−3,−2,−1,0,1,2,3,4}$$ or no?
functions discrete-mathematics
edited Jan 17 at 23:45
jordan_glen
1
asked Jan 17 at 17:23
ViktorViktor
477
$endgroup$
$begingroup$
Is the set $B$ used at all?
$endgroup$
– pwerth
Jan 17 at 17:37
$begingroup$
Yes there is one more request $f^{-1}(B)$
$endgroup$
– Viktor
Jan 17 at 17:38
add a comment |
$begingroup$
Let $A subseteq mathbb{R}, B subset mathbb{R}^{+} cup {0}$, and $A={{xinmathbb{Q}mid -12leq x leq4}}.;$ Find $f(A)$, where $f(x)= |frac x2|$.
How to start? $mathbb{Q}$ is the set of rational numbers, so in $A$ will be $1/2,1/3,1/4$... to $4$ and in the negative to $-12$? There will be a lot of numbers to write! Please help.
Is it $$A={−12,−11,−10,−9,−8,−7,−6,−5,−4,−3,−2,−1,0,1,2,3,4}$$ or no?
functions discrete-mathematics
edited Jan 17 at 23:45
jordan_glen
1
asked Jan 17 at 17:23
ViktorViktor
477
$endgroup$
Let $A subseteq mathbb{R}, B subset mathbb{R}^{+} cup {0}$, and $A={{xinmathbb{Q}mid -12leq x leq4}}.;$ Find $f(A)$, where $f(x)= |frac x2|$.
How to start? $mathbb{Q}$ is the set of rational numbers, so in $A$ will be $1/2,1/3,1/4$... to $4$ and in the negative to $-12$? There will be a lot of numbers to write! Please help.
Is it $$A={−12,−11,−10,−9,−8,−7,−6,−5,−4,−3,−2,−1,0,1,2,3,4}$$ or no?
functions discrete-mathematics
functions discrete-mathematics
edited Jan 17 at 23:45
jordan_glen
1
asked Jan 17 at 17:23
ViktorViktor
477
edited Jan 17 at 23:45
jordan_glen
1
asked Jan 17 at 17:23
ViktorViktor
477
edited Jan 17 at 23:45
jordan_glen
1
edited Jan 17 at 23:45
jordan_glen
1
edited Jan 17 at 23:45
jordan_glen
1
1
asked Jan 17 at 17:23
ViktorViktor
477
asked Jan 17 at 17:23
ViktorViktor
477
asked Jan 17 at 17:23
ViktorViktor
477
477
$begingroup$
Is the set $B$ used at all?
$endgroup$
– pwerth
Jan 17 at 17:37
$begingroup$
Yes there is one more request $f^{-1}(B)$
$endgroup$
– Viktor
Jan 17 at 17:38
add a comment |
$begingroup$
Is the set $B$ used at all?
$endgroup$
– pwerth
Jan 17 at 17:37
$begingroup$
Yes there is one more request $f^{-1}(B)$
$endgroup$
– Viktor
Jan 17 at 17:38
$begingroup$
Is the set $B$ used at all?
$endgroup$
– pwerth
Jan 17 at 17:37
$begingroup$
Is the set $B$ used at all?
$endgroup$
– pwerth
Jan 17 at 17:37
$begingroup$
Yes there is one more request $f^{-1}(B)$
$endgroup$
– Viktor
Jan 17 at 17:38
$begingroup$
Yes there is one more request $f^{-1}(B)$
$endgroup$
– Viktor
Jan 17 at 17:38
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
By definition, the image of $f$ is
$${|x/2|:xinmathbb{Q}cap[-12,4]}.$$
I think it will be easier to see what this reduces to if you notice that
begin{align}
{|x/2|:xinmathbb{Q}cap[-12,4]}&= {x/2:xinmathbb{Q}cap[0,12]}
end{align}
This is because $f(x)=left|x/2right| geq 0, forall x in mathbb Qcap [-12, 4]$, so we can just assume we're working on a set of non-negative numbers, and then the absolute value bars aren't needed.
I claim that
$${x/2:xinmathbb{Q}cap[0,12]}= mathbb{Q}cap[0,6].$$
To show these sets are equal, observe that the left-hand-side is clearly contained in the RHS because if we take a rational number in $[0,12]$ and divide it by $2$, it is now in $[0,6]$ and still rational. Conversely, given any rational number $qin[0,6]$, $2q$ is a rational number in $[0,12]$ and $q=2q/2$. Therefore the image of $f$ is the set of rational numbers between $0$ and $6$. Hence,
$$f(A)=mathbb{Q}cap[0,6]$$
edited Jan 17 at 23:45
jordan_glen
1
answered Jan 17 at 17:35
pwerthpwerth
3,243417
$endgroup$
$begingroup$
Tell me is it $A={{-12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1,2,3,4}}$ or no?
$endgroup$
– Viktor
Jan 17 at 19:28
$begingroup$
No, $A$ is every rational number between $-12$ and $4$. You only listed the integers in that range. You can't possibly write down all the rationals since there are infinitely many of them, which is why we use set notation
$endgroup$
– pwerth
Jan 17 at 20:06
1
$begingroup$
Indeed, Viktor, please read through this answer carefully: In your own statement of the problem, Viktor, we have $xA = {xin mathbb Qmid -12leq xleq 4}$. That means $A$ contains all rational numbers in the interval $[-12, 4] subset mathbb Q$. While $A$ does contain the intergers you list, $A$ contains countably infinite many of rational numbers, of which there are only finitely many integers. Note that $f(A) = { |x/2|mid xin mathbb Q cap [-12, 4]}$, and @pwerth helps you to narrow that down.
$endgroup$
– jordan_glen
Jan 17 at 23:04
add a comment |
Your Answer
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1 Answer
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1 Answer
1
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oldest
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oldest
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active
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votes
$begingroup$
By definition, the image of $f$ is
$${|x/2|:xinmathbb{Q}cap[-12,4]}.$$
I think it will be easier to see what this reduces to if you notice that
begin{align}
{|x/2|:xinmathbb{Q}cap[-12,4]}&= {x/2:xinmathbb{Q}cap[0,12]}
end{align}
This is because $f(x)=left|x/2right| geq 0, forall x in mathbb Qcap [-12, 4]$, so we can just assume we're working on a set of non-negative numbers, and then the absolute value bars aren't needed.
I claim that
$${x/2:xinmathbb{Q}cap[0,12]}= mathbb{Q}cap[0,6].$$
To show these sets are equal, observe that the left-hand-side is clearly contained in the RHS because if we take a rational number in $[0,12]$ and divide it by $2$, it is now in $[0,6]$ and still rational. Conversely, given any rational number $qin[0,6]$, $2q$ is a rational number in $[0,12]$ and $q=2q/2$. Therefore the image of $f$ is the set of rational numbers between $0$ and $6$. Hence,
$$f(A)=mathbb{Q}cap[0,6]$$
edited Jan 17 at 23:45
jordan_glen
1
answered Jan 17 at 17:35
pwerthpwerth
3,243417
$endgroup$
$begingroup$
Tell me is it $A={{-12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1,2,3,4}}$ or no?
$endgroup$
– Viktor
Jan 17 at 19:28
$begingroup$
No, $A$ is every rational number between $-12$ and $4$. You only listed the integers in that range. You can't possibly write down all the rationals since there are infinitely many of them, which is why we use set notation
$endgroup$
– pwerth
Jan 17 at 20:06
1
$begingroup$
Indeed, Viktor, please read through this answer carefully: In your own statement of the problem, Viktor, we have $xA = {xin mathbb Qmid -12leq xleq 4}$. That means $A$ contains all rational numbers in the interval $[-12, 4] subset mathbb Q$. While $A$ does contain the intergers you list, $A$ contains countably infinite many of rational numbers, of which there are only finitely many integers. Note that $f(A) = { |x/2|mid xin mathbb Q cap [-12, 4]}$, and @pwerth helps you to narrow that down.
$endgroup$
– jordan_glen
Jan 17 at 23:04
add a comment |
$begingroup$
By definition, the image of $f$ is
$${|x/2|:xinmathbb{Q}cap[-12,4]}.$$
I think it will be easier to see what this reduces to if you notice that
begin{align}
{|x/2|:xinmathbb{Q}cap[-12,4]}&= {x/2:xinmathbb{Q}cap[0,12]}
end{align}
This is because $f(x)=left|x/2right| geq 0, forall x in mathbb Qcap [-12, 4]$, so we can just assume we're working on a set of non-negative numbers, and then the absolute value bars aren't needed.
I claim that
$${x/2:xinmathbb{Q}cap[0,12]}= mathbb{Q}cap[0,6].$$
To show these sets are equal, observe that the left-hand-side is clearly contained in the RHS because if we take a rational number in $[0,12]$ and divide it by $2$, it is now in $[0,6]$ and still rational. Conversely, given any rational number $qin[0,6]$, $2q$ is a rational number in $[0,12]$ and $q=2q/2$. Therefore the image of $f$ is the set of rational numbers between $0$ and $6$. Hence,
$$f(A)=mathbb{Q}cap[0,6]$$
edited Jan 17 at 23:45
jordan_glen
1
answered Jan 17 at 17:35
pwerthpwerth
3,243417
$endgroup$
$begingroup$
Tell me is it $A={{-12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1,2,3,4}}$ or no?
$endgroup$
– Viktor
Jan 17 at 19:28
$begingroup$
No, $A$ is every rational number between $-12$ and $4$. You only listed the integers in that range. You can't possibly write down all the rationals since there are infinitely many of them, which is why we use set notation
$endgroup$
– pwerth
Jan 17 at 20:06
1
$begingroup$
Indeed, Viktor, please read through this answer carefully: In your own statement of the problem, Viktor, we have $xA = {xin mathbb Qmid -12leq xleq 4}$. That means $A$ contains all rational numbers in the interval $[-12, 4] subset mathbb Q$. While $A$ does contain the intergers you list, $A$ contains countably infinite many of rational numbers, of which there are only finitely many integers. Note that $f(A) = { |x/2|mid xin mathbb Q cap [-12, 4]}$, and @pwerth helps you to narrow that down.
$endgroup$
– jordan_glen
Jan 17 at 23:04
add a comment |
$begingroup$
By definition, the image of $f$ is
$${|x/2|:xinmathbb{Q}cap[-12,4]}.$$
I think it will be easier to see what this reduces to if you notice that
begin{align}
{|x/2|:xinmathbb{Q}cap[-12,4]}&= {x/2:xinmathbb{Q}cap[0,12]}
end{align}
This is because $f(x)=left|x/2right| geq 0, forall x in mathbb Qcap [-12, 4]$, so we can just assume we're working on a set of non-negative numbers, and then the absolute value bars aren't needed.
I claim that
$${x/2:xinmathbb{Q}cap[0,12]}= mathbb{Q}cap[0,6].$$
To show these sets are equal, observe that the left-hand-side is clearly contained in the RHS because if we take a rational number in $[0,12]$ and divide it by $2$, it is now in $[0,6]$ and still rational. Conversely, given any rational number $qin[0,6]$, $2q$ is a rational number in $[0,12]$ and $q=2q/2$. Therefore the image of $f$ is the set of rational numbers between $0$ and $6$. Hence,
$$f(A)=mathbb{Q}cap[0,6]$$
edited Jan 17 at 23:45
jordan_glen
1
answered Jan 17 at 17:35
pwerthpwerth
3,243417
$endgroup$
By definition, the image of $f$ is
$${|x/2|:xinmathbb{Q}cap[-12,4]}.$$
I think it will be easier to see what this reduces to if you notice that
begin{align}
{|x/2|:xinmathbb{Q}cap[-12,4]}&= {x/2:xinmathbb{Q}cap[0,12]}
end{align}
This is because $f(x)=left|x/2right| geq 0, forall x in mathbb Qcap [-12, 4]$, so we can just assume we're working on a set of non-negative numbers, and then the absolute value bars aren't needed.
I claim that
$${x/2:xinmathbb{Q}cap[0,12]}= mathbb{Q}cap[0,6].$$
To show these sets are equal, observe that the left-hand-side is clearly contained in the RHS because if we take a rational number in $[0,12]$ and divide it by $2$, it is now in $[0,6]$ and still rational. Conversely, given any rational number $qin[0,6]$, $2q$ is a rational number in $[0,12]$ and $q=2q/2$. Therefore the image of $f$ is the set of rational numbers between $0$ and $6$. Hence,
$$f(A)=mathbb{Q}cap[0,6]$$
edited Jan 17 at 23:45
jordan_glen
1
answered Jan 17 at 17:35
pwerthpwerth
3,243417
edited Jan 17 at 23:45
jordan_glen
1
edited Jan 17 at 23:45
jordan_glen
1
edited Jan 17 at 23:45
jordan_glen
1
1
answered Jan 17 at 17:35
pwerthpwerth
3,243417
answered Jan 17 at 17:35
pwerthpwerth
3,243417
answered Jan 17 at 17:35
pwerthpwerth
3,243417
3,243417
$begingroup$
Tell me is it $A={{-12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1,2,3,4}}$ or no?
$endgroup$
– Viktor
Jan 17 at 19:28
$begingroup$
No, $A$ is every rational number between $-12$ and $4$. You only listed the integers in that range. You can't possibly write down all the rationals since there are infinitely many of them, which is why we use set notation
$endgroup$
– pwerth
Jan 17 at 20:06
1
$begingroup$
Indeed, Viktor, please read through this answer carefully: In your own statement of the problem, Viktor, we have $xA = {xin mathbb Qmid -12leq xleq 4}$. That means $A$ contains all rational numbers in the interval $[-12, 4] subset mathbb Q$. While $A$ does contain the intergers you list, $A$ contains countably infinite many of rational numbers, of which there are only finitely many integers. Note that $f(A) = { |x/2|mid xin mathbb Q cap [-12, 4]}$, and @pwerth helps you to narrow that down.
$endgroup$
– jordan_glen
Jan 17 at 23:04
add a comment |
$begingroup$
Tell me is it $A={{-12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1,2,3,4}}$ or no?
$endgroup$
– Viktor
Jan 17 at 19:28
$begingroup$
No, $A$ is every rational number between $-12$ and $4$. You only listed the integers in that range. You can't possibly write down all the rationals since there are infinitely many of them, which is why we use set notation
$endgroup$
– pwerth
Jan 17 at 20:06
1
$begingroup$
Indeed, Viktor, please read through this answer carefully: In your own statement of the problem, Viktor, we have $xA = {xin mathbb Qmid -12leq xleq 4}$. That means $A$ contains all rational numbers in the interval $[-12, 4] subset mathbb Q$. While $A$ does contain the intergers you list, $A$ contains countably infinite many of rational numbers, of which there are only finitely many integers. Note that $f(A) = { |x/2|mid xin mathbb Q cap [-12, 4]}$, and @pwerth helps you to narrow that down.
$endgroup$
– jordan_glen
Jan 17 at 23:04
$begingroup$
Tell me is it $A={{-12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1,2,3,4}}$ or no?
$endgroup$
– Viktor
Jan 17 at 19:28
$begingroup$
Tell me is it $A={{-12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1,2,3,4}}$ or no?
$endgroup$
– Viktor
Jan 17 at 19:28
$begingroup$
No, $A$ is every rational number between $-12$ and $4$. You only listed the integers in that range. You can't possibly write down all the rationals since there are infinitely many of them, which is why we use set notation
$endgroup$
– pwerth
Jan 17 at 20:06
$begingroup$
No, $A$ is every rational number between $-12$ and $4$. You only listed the integers in that range. You can't possibly write down all the rationals since there are infinitely many of them, which is why we use set notation
$endgroup$
– pwerth
Jan 17 at 20:06
1
1
$begingroup$
Indeed, Viktor, please read through this answer carefully: In your own statement of the problem, Viktor, we have $xA = {xin mathbb Qmid -12leq xleq 4}$. That means $A$ contains all rational numbers in the interval $[-12, 4] subset mathbb Q$. While $A$ does contain the intergers you list, $A$ contains countably infinite many of rational numbers, of which there are only finitely many integers. Note that $f(A) = { |x/2|mid xin mathbb Q cap [-12, 4]}$, and @pwerth helps you to narrow that down.
$endgroup$
– jordan_glen
Jan 17 at 23:04
$begingroup$
Indeed, Viktor, please read through this answer carefully: In your own statement of the problem, Viktor, we have $xA = {xin mathbb Qmid -12leq xleq 4}$. That means $A$ contains all rational numbers in the interval $[-12, 4] subset mathbb Q$. While $A$ does contain the intergers you list, $A$ contains countably infinite many of rational numbers, of which there are only finitely many integers. Note that $f(A) = { |x/2|mid xin mathbb Q cap [-12, 4]}$, and @pwerth helps you to narrow that down.
$endgroup$
– jordan_glen
Jan 17 at 23:04
add a comment |
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$begingroup$
Is the set $B$ used at all?
$endgroup$
– pwerth
Jan 17 at 17:37
$begingroup$
Yes there is one more request $f^{-1}(B)$
$endgroup$
– Viktor
Jan 17 at 17:38