Mixing
Obtaining $sqrt{r}$ from $r$ with quadratics
$begingroup$
A real number $r>0$ is written. If there is a number $a$ written, we are allowed to write down $a+1$. If there are numbers $a,b$ written (possibly $a=b$ but they must be written twice in that case), we are allowed to write down the (0, 1, or 2) real roots of $x^2+ax+b$. Is it true that we can always eventually write down the number $sqrt{r}$?
A particular case where this is certainly possible is if $sqrt{r}$ is an integer. Starting with $r$, we can write down numbers $2s$ and $s^2$ for some integer $s>0$. The root of $x^2+2sx+s^2$ is $x=-s$, so we can now get all integers at least $-s$. This includes all positive integers.
algebra-precalculus quadratics
asked Mar 27 '18 at 9:29
user11550user11550
29319
$endgroup$
add a comment |
$begingroup$
A real number $r>0$ is written. If there is a number $a$ written, we are allowed to write down $a+1$. If there are numbers $a,b$ written (possibly $a=b$ but they must be written twice in that case), we are allowed to write down the (0, 1, or 2) real roots of $x^2+ax+b$. Is it true that we can always eventually write down the number $sqrt{r}$?
A particular case where this is certainly possible is if $sqrt{r}$ is an integer. Starting with $r$, we can write down numbers $2s$ and $s^2$ for some integer $s>0$. The root of $x^2+2sx+s^2$ is $x=-s$, so we can now get all integers at least $-s$. This includes all positive integers.
algebra-precalculus quadratics
asked Mar 27 '18 at 9:29
user11550user11550
29319
$endgroup$
$begingroup$
It also works for $r=2$: Take $a = 4$ and $b = 2$ and note that $(sqrt 2 - 2)^2 + a (sqrt 2 -2 ) + b = 0$. Hence you can 'write down' $sqrt 2 -2 $ and consequently $sqrt r = sqrt 2$.
$endgroup$
– Jan Bohr
Mar 27 '18 at 9:53
$begingroup$
(I think the question would be easier to understand if phrased in terms of sets: Let $r > 0$ a real number and denote $Ssubset mathbb{R}$ the set inductively defined by: (1) $rin S$, (2) $xin S Rightarrow x +1 in S$, (3) $a,bin S Rightarrow$ reals roots of $x^2+a x + b$ lie in $S$. Question: Is $sqrt{r}in S$?)
$endgroup$
– Jan Bohr
Mar 27 '18 at 9:59
add a comment |
$begingroup$
A real number $r>0$ is written. If there is a number $a$ written, we are allowed to write down $a+1$. If there are numbers $a,b$ written (possibly $a=b$ but they must be written twice in that case), we are allowed to write down the (0, 1, or 2) real roots of $x^2+ax+b$. Is it true that we can always eventually write down the number $sqrt{r}$?
A particular case where this is certainly possible is if $sqrt{r}$ is an integer. Starting with $r$, we can write down numbers $2s$ and $s^2$ for some integer $s>0$. The root of $x^2+2sx+s^2$ is $x=-s$, so we can now get all integers at least $-s$. This includes all positive integers.
algebra-precalculus quadratics
asked Mar 27 '18 at 9:29
user11550user11550
29319
$endgroup$
A real number $r>0$ is written. If there is a number $a$ written, we are allowed to write down $a+1$. If there are numbers $a,b$ written (possibly $a=b$ but they must be written twice in that case), we are allowed to write down the (0, 1, or 2) real roots of $x^2+ax+b$. Is it true that we can always eventually write down the number $sqrt{r}$?
A particular case where this is certainly possible is if $sqrt{r}$ is an integer. Starting with $r$, we can write down numbers $2s$ and $s^2$ for some integer $s>0$. The root of $x^2+2sx+s^2$ is $x=-s$, so we can now get all integers at least $-s$. This includes all positive integers.
algebra-precalculus quadratics
algebra-precalculus quadratics
asked Mar 27 '18 at 9:29
user11550user11550
29319
asked Mar 27 '18 at 9:29
user11550user11550
29319
asked Mar 27 '18 at 9:29
user11550user11550
29319
asked Mar 27 '18 at 9:29
user11550user11550
29319
asked Mar 27 '18 at 9:29
user11550user11550
29319
29319
$begingroup$
It also works for $r=2$: Take $a = 4$ and $b = 2$ and note that $(sqrt 2 - 2)^2 + a (sqrt 2 -2 ) + b = 0$. Hence you can 'write down' $sqrt 2 -2 $ and consequently $sqrt r = sqrt 2$.
$endgroup$
– Jan Bohr
Mar 27 '18 at 9:53
$begingroup$
(I think the question would be easier to understand if phrased in terms of sets: Let $r > 0$ a real number and denote $Ssubset mathbb{R}$ the set inductively defined by: (1) $rin S$, (2) $xin S Rightarrow x +1 in S$, (3) $a,bin S Rightarrow$ reals roots of $x^2+a x + b$ lie in $S$. Question: Is $sqrt{r}in S$?)
$endgroup$
– Jan Bohr
Mar 27 '18 at 9:59
add a comment |
$begingroup$
It also works for $r=2$: Take $a = 4$ and $b = 2$ and note that $(sqrt 2 - 2)^2 + a (sqrt 2 -2 ) + b = 0$. Hence you can 'write down' $sqrt 2 -2 $ and consequently $sqrt r = sqrt 2$.
$endgroup$
– Jan Bohr
Mar 27 '18 at 9:53
$begingroup$
(I think the question would be easier to understand if phrased in terms of sets: Let $r > 0$ a real number and denote $Ssubset mathbb{R}$ the set inductively defined by: (1) $rin S$, (2) $xin S Rightarrow x +1 in S$, (3) $a,bin S Rightarrow$ reals roots of $x^2+a x + b$ lie in $S$. Question: Is $sqrt{r}in S$?)
$endgroup$
– Jan Bohr
Mar 27 '18 at 9:59
$begingroup$
It also works for $r=2$: Take $a = 4$ and $b = 2$ and note that $(sqrt 2 - 2)^2 + a (sqrt 2 -2 ) + b = 0$. Hence you can 'write down' $sqrt 2 -2 $ and consequently $sqrt r = sqrt 2$.
$endgroup$
– Jan Bohr
Mar 27 '18 at 9:53
$begingroup$
It also works for $r=2$: Take $a = 4$ and $b = 2$ and note that $(sqrt 2 - 2)^2 + a (sqrt 2 -2 ) + b = 0$. Hence you can 'write down' $sqrt 2 -2 $ and consequently $sqrt r = sqrt 2$.
$endgroup$
– Jan Bohr
Mar 27 '18 at 9:53
$begingroup$
(I think the question would be easier to understand if phrased in terms of sets: Let $r > 0$ a real number and denote $Ssubset mathbb{R}$ the set inductively defined by: (1) $rin S$, (2) $xin S Rightarrow x +1 in S$, (3) $a,bin S Rightarrow$ reals roots of $x^2+a x + b$ lie in $S$. Question: Is $sqrt{r}in S$?)
$endgroup$
– Jan Bohr
Mar 27 '18 at 9:59
$begingroup$
(I think the question would be easier to understand if phrased in terms of sets: Let $r > 0$ a real number and denote $Ssubset mathbb{R}$ the set inductively defined by: (1) $rin S$, (2) $xin S Rightarrow x +1 in S$, (3) $a,bin S Rightarrow$ reals roots of $x^2+a x + b$ lie in $S$. Question: Is $sqrt{r}in S$?)
$endgroup$
– Jan Bohr
Mar 27 '18 at 9:59
add a comment |
1 Answer
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$begingroup$
Given $r>0$, we have $r+1$.
So, we have the roots of $x^2+(r+1)x+r=(x+r)(x+1)$, which are $-r$ and $-1$.
Adding $1$ to $-1$, we have $0$, and thus we have the roots of $x^2+0x-r$, so we have $sqrt{r}$.
answered Mar 27 '18 at 10:02
ChrisChris
305128
$endgroup$
1
$begingroup$
Nice. You need a few $ symbols on the last line.
$endgroup$
– quasi
Mar 27 '18 at 10:50
$begingroup$
The key was getting 0. Nice.
$endgroup$
– marty cohen
Feb 3 at 6:20
add a comment |
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$begingroup$
Given $r>0$, we have $r+1$.
So, we have the roots of $x^2+(r+1)x+r=(x+r)(x+1)$, which are $-r$ and $-1$.
Adding $1$ to $-1$, we have $0$, and thus we have the roots of $x^2+0x-r$, so we have $sqrt{r}$.
answered Mar 27 '18 at 10:02
ChrisChris
305128
$endgroup$
1
$begingroup$
Nice. You need a few $ symbols on the last line.
$endgroup$
– quasi
Mar 27 '18 at 10:50
$begingroup$
The key was getting 0. Nice.
$endgroup$
– marty cohen
Feb 3 at 6:20
add a comment |
$begingroup$
Given $r>0$, we have $r+1$.
So, we have the roots of $x^2+(r+1)x+r=(x+r)(x+1)$, which are $-r$ and $-1$.
Adding $1$ to $-1$, we have $0$, and thus we have the roots of $x^2+0x-r$, so we have $sqrt{r}$.
answered Mar 27 '18 at 10:02
ChrisChris
305128
$endgroup$
1
$begingroup$
Nice. You need a few $ symbols on the last line.
$endgroup$
– quasi
Mar 27 '18 at 10:50
$begingroup$
The key was getting 0. Nice.
$endgroup$
– marty cohen
Feb 3 at 6:20
add a comment |
$begingroup$
Given $r>0$, we have $r+1$.
So, we have the roots of $x^2+(r+1)x+r=(x+r)(x+1)$, which are $-r$ and $-1$.
Adding $1$ to $-1$, we have $0$, and thus we have the roots of $x^2+0x-r$, so we have $sqrt{r}$.
answered Mar 27 '18 at 10:02
ChrisChris
305128
$endgroup$
Given $r>0$, we have $r+1$.
So, we have the roots of $x^2+(r+1)x+r=(x+r)(x+1)$, which are $-r$ and $-1$.
Adding $1$ to $-1$, we have $0$, and thus we have the roots of $x^2+0x-r$, so we have $sqrt{r}$.
answered Mar 27 '18 at 10:02
ChrisChris
305128
edited Feb 3 at 6:14
answered Mar 27 '18 at 10:02
ChrisChris
305128
answered Mar 27 '18 at 10:02
ChrisChris
305128
answered Mar 27 '18 at 10:02
ChrisChris
305128
305128
1
$begingroup$
Nice. You need a few $ symbols on the last line.
$endgroup$
– quasi
Mar 27 '18 at 10:50
$begingroup$
The key was getting 0. Nice.
$endgroup$
– marty cohen
Feb 3 at 6:20
add a comment |
1
$begingroup$
Nice. You need a few $ symbols on the last line.
$endgroup$
– quasi
Mar 27 '18 at 10:50
$begingroup$
The key was getting 0. Nice.
$endgroup$
– marty cohen
Feb 3 at 6:20
1
1
$begingroup$
Nice. You need a few $ symbols on the last line.
$endgroup$
– quasi
Mar 27 '18 at 10:50
$begingroup$
Nice. You need a few $ symbols on the last line.
$endgroup$
– quasi
Mar 27 '18 at 10:50
$begingroup$
The key was getting 0. Nice.
$endgroup$
– marty cohen
Feb 3 at 6:20
$begingroup$
The key was getting 0. Nice.
$endgroup$
– marty cohen
Feb 3 at 6:20
add a comment |
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$begingroup$
It also works for $r=2$: Take $a = 4$ and $b = 2$ and note that $(sqrt 2 - 2)^2 + a (sqrt 2 -2 ) + b = 0$. Hence you can 'write down' $sqrt 2 -2 $ and consequently $sqrt r = sqrt 2$.
$endgroup$
– Jan Bohr
Mar 27 '18 at 9:53
$begingroup$
(I think the question would be easier to understand if phrased in terms of sets: Let $r > 0$ a real number and denote $Ssubset mathbb{R}$ the set inductively defined by: (1) $rin S$, (2) $xin S Rightarrow x +1 in S$, (3) $a,bin S Rightarrow$ reals roots of $x^2+a x + b$ lie in $S$. Question: Is $sqrt{r}in S$?)
$endgroup$
– Jan Bohr
Mar 27 '18 at 9:59