Mixing
Find the supA of $A= { x in R : |x||x+1| < 2 }$
$begingroup$
Let us define a set $A = { x in R : |x||x+1| < 2 } $. Which is the $supA$?
My solution:
We know that: $|x||x+1| = |x(x+1)| = |x^2 +x|$
Then
$ -2 < x^2 + x < 2 $
Is $supA=2$ ?
$supA = 2$ if $forall e > 0$ there is a $x in A$ such that
$x^2 + x > 2 -e$
Let us define $e=|x|$ (First Question)
Then
$x^2 + x> 2 - |x| Leftrightarrow ... Leftrightarrow |x| > -x^2 - x + 2 Leftrightarrow $
begin{cases}
x > - x^2 -x + 2 Leftrightarrow x^2 + 2x > 2\
x < x^2 + x - 2 Leftrightarrow x^2 > 2
end{cases}
So, we have two cases:
if $x^2>2$ then $ x > sqrt 2$
We replace $x=sqrt 2$ in the first inequality, so:
$|(sqrt 2)^2 + sqrt 2| = |2 + sqrt 2| > 2$
On the other hand if $x^2 + 2x -2 >0$
We find the roots of $x^2 + 2x - 2 = 0$
and if we replace $x = -1 + sqrt 3$ in the first inequality we have:
$|(sqrt 3)^2 + sqrt 3| > 2 $
So, there isn't any $xin A$ when $e=|x|$. Consequently, there isn't $supA$
First Question: Is the selection of e correct ? Could we find anoother e ?
Second Question: Is my solution correct ? Is there any easier way to solve the
exercise ?
calculus
asked Jan 26 at 21:24
Dimitris DimitriadisDimitris Dimitriadis
548
$endgroup$
add a comment |
$begingroup$
Let us define a set $A = { x in R : |x||x+1| < 2 } $. Which is the $supA$?
My solution:
We know that: $|x||x+1| = |x(x+1)| = |x^2 +x|$
Then
$ -2 < x^2 + x < 2 $
Is $supA=2$ ?
$supA = 2$ if $forall e > 0$ there is a $x in A$ such that
$x^2 + x > 2 -e$
Let us define $e=|x|$ (First Question)
Then
$x^2 + x> 2 - |x| Leftrightarrow ... Leftrightarrow |x| > -x^2 - x + 2 Leftrightarrow $
begin{cases}
x > - x^2 -x + 2 Leftrightarrow x^2 + 2x > 2\
x < x^2 + x - 2 Leftrightarrow x^2 > 2
end{cases}
So, we have two cases:
if $x^2>2$ then $ x > sqrt 2$
We replace $x=sqrt 2$ in the first inequality, so:
$|(sqrt 2)^2 + sqrt 2| = |2 + sqrt 2| > 2$
On the other hand if $x^2 + 2x -2 >0$
We find the roots of $x^2 + 2x - 2 = 0$
and if we replace $x = -1 + sqrt 3$ in the first inequality we have:
$|(sqrt 3)^2 + sqrt 3| > 2 $
So, there isn't any $xin A$ when $e=|x|$. Consequently, there isn't $supA$
First Question: Is the selection of e correct ? Could we find anoother e ?
Second Question: Is my solution correct ? Is there any easier way to solve the
exercise ?
calculus
asked Jan 26 at 21:24
Dimitris DimitriadisDimitris Dimitriadis
548
$endgroup$
add a comment |
$begingroup$
Let us define a set $A = { x in R : |x||x+1| < 2 } $. Which is the $supA$?
My solution:
We know that: $|x||x+1| = |x(x+1)| = |x^2 +x|$
Then
$ -2 < x^2 + x < 2 $
Is $supA=2$ ?
$supA = 2$ if $forall e > 0$ there is a $x in A$ such that
$x^2 + x > 2 -e$
Let us define $e=|x|$ (First Question)
Then
$x^2 + x> 2 - |x| Leftrightarrow ... Leftrightarrow |x| > -x^2 - x + 2 Leftrightarrow $
begin{cases}
x > - x^2 -x + 2 Leftrightarrow x^2 + 2x > 2\
x < x^2 + x - 2 Leftrightarrow x^2 > 2
end{cases}
So, we have two cases:
if $x^2>2$ then $ x > sqrt 2$
We replace $x=sqrt 2$ in the first inequality, so:
$|(sqrt 2)^2 + sqrt 2| = |2 + sqrt 2| > 2$
On the other hand if $x^2 + 2x -2 >0$
We find the roots of $x^2 + 2x - 2 = 0$
and if we replace $x = -1 + sqrt 3$ in the first inequality we have:
$|(sqrt 3)^2 + sqrt 3| > 2 $
So, there isn't any $xin A$ when $e=|x|$. Consequently, there isn't $supA$
First Question: Is the selection of e correct ? Could we find anoother e ?
Second Question: Is my solution correct ? Is there any easier way to solve the
exercise ?
calculus
asked Jan 26 at 21:24
Dimitris DimitriadisDimitris Dimitriadis
548
$endgroup$
Let us define a set $A = { x in R : |x||x+1| < 2 } $. Which is the $supA$?
My solution:
We know that: $|x||x+1| = |x(x+1)| = |x^2 +x|$
Then
$ -2 < x^2 + x < 2 $
Is $supA=2$ ?
$supA = 2$ if $forall e > 0$ there is a $x in A$ such that
$x^2 + x > 2 -e$
Let us define $e=|x|$ (First Question)
Then
$x^2 + x> 2 - |x| Leftrightarrow ... Leftrightarrow |x| > -x^2 - x + 2 Leftrightarrow $
begin{cases}
x > - x^2 -x + 2 Leftrightarrow x^2 + 2x > 2\
x < x^2 + x - 2 Leftrightarrow x^2 > 2
end{cases}
So, we have two cases:
if $x^2>2$ then $ x > sqrt 2$
We replace $x=sqrt 2$ in the first inequality, so:
$|(sqrt 2)^2 + sqrt 2| = |2 + sqrt 2| > 2$
On the other hand if $x^2 + 2x -2 >0$
We find the roots of $x^2 + 2x - 2 = 0$
and if we replace $x = -1 + sqrt 3$ in the first inequality we have:
$|(sqrt 3)^2 + sqrt 3| > 2 $
So, there isn't any $xin A$ when $e=|x|$. Consequently, there isn't $supA$
First Question: Is the selection of e correct ? Could we find anoother e ?
Second Question: Is my solution correct ? Is there any easier way to solve the
exercise ?
calculus
calculus
asked Jan 26 at 21:24
Dimitris DimitriadisDimitris Dimitriadis
548
asked Jan 26 at 21:24
Dimitris DimitriadisDimitris Dimitriadis
548
asked Jan 26 at 21:24
Dimitris DimitriadisDimitris Dimitriadis
548
asked Jan 26 at 21:24
Dimitris DimitriadisDimitris Dimitriadis
548
asked Jan 26 at 21:24
Dimitris DimitriadisDimitris Dimitriadis
548
548
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: Use
$$x^2+x<2 iff (x+1/2)^2 < 9/4 iff |x+1/2|<3/2$$
to prove that $sup (A) = 1$.
answered Jan 26 at 21:31
Math LoverMath Lover
14.1k31437
$endgroup$
$begingroup$
If A is bounded from top, ther must be a supA. It is bounded from top cause let say every X>10 is not in the group, so 10 is a good bound. So 10 doesnt have to the Sup, but there is one.
$endgroup$
– Shaq
Jan 26 at 22:23
add a comment |
$begingroup$
Let's try and simplify the solution. You're correct that the condition translates to
$$
-2<x^2+x<2
$$
The inequality $x^2+x-2<0$ is satisfied on the interval $(-2,1)$; the inequality $x^2+x+2>0$ is satisfied for every $x$. Thus your set $A$ is $A=(-2,1)$.
So no, your solution is incorrect, I'm afraid.
answered Jan 26 at 22:34
egregegreg
184k1486206
$endgroup$
$begingroup$
Thank you ! it was very useful. So, I correctly find that $supA neq 2 $, but I couldn't find the solution because I ignored that $A = (-2,1)$.
$endgroup$
– Dimitris Dimitriadis
Jan 26 at 22:42
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Use
$$x^2+x<2 iff (x+1/2)^2 < 9/4 iff |x+1/2|<3/2$$
to prove that $sup (A) = 1$.
answered Jan 26 at 21:31
Math LoverMath Lover
14.1k31437
$endgroup$
$begingroup$
If A is bounded from top, ther must be a supA. It is bounded from top cause let say every X>10 is not in the group, so 10 is a good bound. So 10 doesnt have to the Sup, but there is one.
$endgroup$
– Shaq
Jan 26 at 22:23
add a comment |
$begingroup$
Hint: Use
$$x^2+x<2 iff (x+1/2)^2 < 9/4 iff |x+1/2|<3/2$$
to prove that $sup (A) = 1$.
answered Jan 26 at 21:31
Math LoverMath Lover
14.1k31437
$endgroup$
$begingroup$
If A is bounded from top, ther must be a supA. It is bounded from top cause let say every X>10 is not in the group, so 10 is a good bound. So 10 doesnt have to the Sup, but there is one.
$endgroup$
– Shaq
Jan 26 at 22:23
add a comment |
$begingroup$
Hint: Use
$$x^2+x<2 iff (x+1/2)^2 < 9/4 iff |x+1/2|<3/2$$
to prove that $sup (A) = 1$.
answered Jan 26 at 21:31
Math LoverMath Lover
14.1k31437
$endgroup$
Hint: Use
$$x^2+x<2 iff (x+1/2)^2 < 9/4 iff |x+1/2|<3/2$$
to prove that $sup (A) = 1$.
answered Jan 26 at 21:31
Math LoverMath Lover
14.1k31437
answered Jan 26 at 21:31
Math LoverMath Lover
14.1k31437
answered Jan 26 at 21:31
Math LoverMath Lover
14.1k31437
answered Jan 26 at 21:31
Math LoverMath Lover
14.1k31437
14.1k31437
$begingroup$
If A is bounded from top, ther must be a supA. It is bounded from top cause let say every X>10 is not in the group, so 10 is a good bound. So 10 doesnt have to the Sup, but there is one.
$endgroup$
– Shaq
Jan 26 at 22:23
add a comment |
$begingroup$
If A is bounded from top, ther must be a supA. It is bounded from top cause let say every X>10 is not in the group, so 10 is a good bound. So 10 doesnt have to the Sup, but there is one.
$endgroup$
– Shaq
Jan 26 at 22:23
$begingroup$
If A is bounded from top, ther must be a supA. It is bounded from top cause let say every X>10 is not in the group, so 10 is a good bound. So 10 doesnt have to the Sup, but there is one.
$endgroup$
– Shaq
Jan 26 at 22:23
$begingroup$
If A is bounded from top, ther must be a supA. It is bounded from top cause let say every X>10 is not in the group, so 10 is a good bound. So 10 doesnt have to the Sup, but there is one.
$endgroup$
– Shaq
Jan 26 at 22:23
add a comment |
$begingroup$
Let's try and simplify the solution. You're correct that the condition translates to
$$
-2<x^2+x<2
$$
The inequality $x^2+x-2<0$ is satisfied on the interval $(-2,1)$; the inequality $x^2+x+2>0$ is satisfied for every $x$. Thus your set $A$ is $A=(-2,1)$.
So no, your solution is incorrect, I'm afraid.
answered Jan 26 at 22:34
egregegreg
184k1486206
$endgroup$
$begingroup$
Thank you ! it was very useful. So, I correctly find that $supA neq 2 $, but I couldn't find the solution because I ignored that $A = (-2,1)$.
$endgroup$
– Dimitris Dimitriadis
Jan 26 at 22:42
add a comment |
$begingroup$
Let's try and simplify the solution. You're correct that the condition translates to
$$
-2<x^2+x<2
$$
The inequality $x^2+x-2<0$ is satisfied on the interval $(-2,1)$; the inequality $x^2+x+2>0$ is satisfied for every $x$. Thus your set $A$ is $A=(-2,1)$.
So no, your solution is incorrect, I'm afraid.
answered Jan 26 at 22:34
egregegreg
184k1486206
$endgroup$
$begingroup$
Thank you ! it was very useful. So, I correctly find that $supA neq 2 $, but I couldn't find the solution because I ignored that $A = (-2,1)$.
$endgroup$
– Dimitris Dimitriadis
Jan 26 at 22:42
add a comment |
$begingroup$
Let's try and simplify the solution. You're correct that the condition translates to
$$
-2<x^2+x<2
$$
The inequality $x^2+x-2<0$ is satisfied on the interval $(-2,1)$; the inequality $x^2+x+2>0$ is satisfied for every $x$. Thus your set $A$ is $A=(-2,1)$.
So no, your solution is incorrect, I'm afraid.
answered Jan 26 at 22:34
egregegreg
184k1486206
$endgroup$
Let's try and simplify the solution. You're correct that the condition translates to
$$
-2<x^2+x<2
$$
The inequality $x^2+x-2<0$ is satisfied on the interval $(-2,1)$; the inequality $x^2+x+2>0$ is satisfied for every $x$. Thus your set $A$ is $A=(-2,1)$.
So no, your solution is incorrect, I'm afraid.
answered Jan 26 at 22:34
egregegreg
184k1486206
answered Jan 26 at 22:34
egregegreg
184k1486206
answered Jan 26 at 22:34
egregegreg
184k1486206
answered Jan 26 at 22:34
egregegreg
184k1486206
184k1486206
$begingroup$
Thank you ! it was very useful. So, I correctly find that $supA neq 2 $, but I couldn't find the solution because I ignored that $A = (-2,1)$.
$endgroup$
– Dimitris Dimitriadis
Jan 26 at 22:42
add a comment |
$begingroup$
Thank you ! it was very useful. So, I correctly find that $supA neq 2 $, but I couldn't find the solution because I ignored that $A = (-2,1)$.
$endgroup$
– Dimitris Dimitriadis
Jan 26 at 22:42
$begingroup$
Thank you ! it was very useful. So, I correctly find that $supA neq 2 $, but I couldn't find the solution because I ignored that $A = (-2,1)$.
$endgroup$
– Dimitris Dimitriadis
Jan 26 at 22:42
$begingroup$
Thank you ! it was very useful. So, I correctly find that $supA neq 2 $, but I couldn't find the solution because I ignored that $A = (-2,1)$.
$endgroup$
– Dimitris Dimitriadis
Jan 26 at 22:42
add a comment |
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