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Solving $ 2

4

1

$begingroup$

How to solve inequalities in which we have quadratic terms and greatest integer function.
$$ 2< x^2 -[x]<5$$ [.] is greatest integer function.

Do we need to break into the cases as [0,1), [1,2) and so on?

functions inequality ceiling-function

share|cite|improve this question

edited yesterday

Maria Mazur

50k1361125

asked yesterday

mavericmaveric

91912

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, that is exactly what you have to do.
  $endgroup$
  – Kavi Rama Murthy
  yesterday
  

  
  
  
  

add a comment | 

4

1

$begingroup$

How to solve inequalities in which we have quadratic terms and greatest integer function.
$$ 2< x^2 -[x]<5$$ [.] is greatest integer function.

Do we need to break into the cases as [0,1), [1,2) and so on?

functions inequality ceiling-function

share|cite|improve this question

edited yesterday

Maria Mazur

50k1361125

asked yesterday

mavericmaveric

91912

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, that is exactly what you have to do.
  $endgroup$
  – Kavi Rama Murthy
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

How to solve inequalities in which we have quadratic terms and greatest integer function.
$$ 2< x^2 -[x]<5$$ [.] is greatest integer function.

Do we need to break into the cases as [0,1), [1,2) and so on?

functions inequality ceiling-function

share|cite|improve this question

edited yesterday

Maria Mazur

50k1361125

asked yesterday

mavericmaveric

91912

$endgroup$

share|cite|improve this question

edited yesterday

Maria Mazur

50k1361125

asked yesterday

mavericmaveric

91912

share|cite|improve this question

edited yesterday

Maria Mazur

50k1361125

asked yesterday

mavericmaveric

91912

edited yesterday

Maria Mazur

50k1361125

edited yesterday

Maria Mazur

50k1361125

asked yesterday

mavericmaveric

91912

asked yesterday

mavericmaveric

91912

2 Answers
2

active

oldest

votes

7

$begingroup$

Hint: use $$x-1<[x]leq x$$ and then solve some quadratic inequalites like

$$ x^2-5<[x] implies x^2-5<ximplies x^2-x-6<0$$

so $xin (-2,3)$ and so on...

• If $xin (-2,-1)$ then $[x]=-2 $ so $0<x^2<3$ so $-sqrt{3}<x<sqrt{3}$ so $xin(-sqrt{3},-1)$
  


• If $xin [-1,0)$ ...

share|cite|improve this answer

edited yesterday

answered yesterday

Maria MazurMaria Mazur

50k1361125

$endgroup$

add a comment | 

4

$begingroup$

While breaking up into integer intervals always works, sometimes it is too much work.

Let's say you instead had $$
213 < x^2 - lfloor x rfloor < 505 $$
You really don't want to consider $24$ cases individually.

So you get clever and change this to a pair of simultaneous inequalities: Let $x = n + y$ with $n in Bbb Z$ and $0 leq y < 1$. Then the inequalities are
$$
left{ begin{array}{l} 213 < (n+y)^2 - n < 505 \ 0 leq y < 1 end{array} right.
$$
Then $$213 < n^2 + (2y-1)n +y^2 , 0 leq y < 1 implies 213 < n^2 + (1)n + 1
$$
and this shows that $n geq 15$ (you can get that with one application of the quadratic formula), cutting out a lot of the work. Similarly,
$$ n^2 + (2y-1)n +y^2 < 505 , 0 leq y < 1 implies n^2 + (-1)n + 0 < 505
$$
and this shows that $n < 23$.

Finally, you can also justify only examining the allowed values of $y$ in the boundary cases ($n = 15$ and $n = 22$); in between, any value of $y$ works. (This might not be the case for cubic expressions, for example.)

So for example, on the low end, you would need to solve for $y$ in
$$
213 < (15+y)^2 - 15 = 210 - 30 y + y^2 \
y^2 -30 y -3 > 0\
y > frac12( sqrt{912} -30 ) \
x > (sqrt{228} - 15) + 15 implies x > sqrt{228}
$$
and similarly you need to consider the case of $n=22$ to get the upper limit for $x$.

share|cite|improve this answer

answered yesterday

Mark FischlerMark Fischler

34.2k12552

$endgroup$

add a comment | 

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7

$begingroup$

Hint: use $$x-1<[x]leq x$$ and then solve some quadratic inequalites like

$$ x^2-5<[x] implies x^2-5<ximplies x^2-x-6<0$$

so $xin (-2,3)$ and so on...

• If $xin (-2,-1)$ then $[x]=-2 $ so $0<x^2<3$ so $-sqrt{3}<x<sqrt{3}$ so $xin(-sqrt{3},-1)$
  


• If $xin [-1,0)$ ...

share|cite|improve this answer

edited yesterday

answered yesterday

Maria MazurMaria Mazur

50k1361125

$endgroup$

add a comment | 

7

$begingroup$

Hint: use $$x-1<[x]leq x$$ and then solve some quadratic inequalites like

$$ x^2-5<[x] implies x^2-5<ximplies x^2-x-6<0$$

so $xin (-2,3)$ and so on...

• If $xin (-2,-1)$ then $[x]=-2 $ so $0<x^2<3$ so $-sqrt{3}<x<sqrt{3}$ so $xin(-sqrt{3},-1)$
  


• If $xin [-1,0)$ ...

share|cite|improve this answer

edited yesterday

answered yesterday

Maria MazurMaria Mazur

50k1361125

$endgroup$

add a comment | 

$begingroup$

Hint: use $$x-1<[x]leq x$$ and then solve some quadratic inequalites like

$$ x^2-5<[x] implies x^2-5<ximplies x^2-x-6<0$$

so $xin (-2,3)$ and so on...

• If $xin (-2,-1)$ then $[x]=-2 $ so $0<x^2<3$ so $-sqrt{3}<x<sqrt{3}$ so $xin(-sqrt{3},-1)$
  


• If $xin [-1,0)$ ...

share|cite|improve this answer

edited yesterday

answered yesterday

Maria MazurMaria Mazur

50k1361125

$endgroup$

share|cite|improve this answer

edited yesterday

answered yesterday

Maria MazurMaria Mazur

50k1361125

answered yesterday

Maria MazurMaria Mazur

50k1361125

answered yesterday

Maria MazurMaria Mazur

50k1361125

4

$begingroup$

While breaking up into integer intervals always works, sometimes it is too much work.

Let's say you instead had $$
213 < x^2 - lfloor x rfloor < 505 $$
You really don't want to consider $24$ cases individually.

So you get clever and change this to a pair of simultaneous inequalities: Let $x = n + y$ with $n in Bbb Z$ and $0 leq y < 1$. Then the inequalities are
$$
left{ begin{array}{l} 213 < (n+y)^2 - n < 505 \ 0 leq y < 1 end{array} right.
$$
Then $$213 < n^2 + (2y-1)n +y^2 , 0 leq y < 1 implies 213 < n^2 + (1)n + 1
$$
and this shows that $n geq 15$ (you can get that with one application of the quadratic formula), cutting out a lot of the work. Similarly,
$$ n^2 + (2y-1)n +y^2 < 505 , 0 leq y < 1 implies n^2 + (-1)n + 0 < 505
$$
and this shows that $n < 23$.

Finally, you can also justify only examining the allowed values of $y$ in the boundary cases ($n = 15$ and $n = 22$); in between, any value of $y$ works. (This might not be the case for cubic expressions, for example.)

So for example, on the low end, you would need to solve for $y$ in
$$
213 < (15+y)^2 - 15 = 210 - 30 y + y^2 \
y^2 -30 y -3 > 0\
y > frac12( sqrt{912} -30 ) \
x > (sqrt{228} - 15) + 15 implies x > sqrt{228}
$$
and similarly you need to consider the case of $n=22$ to get the upper limit for $x$.

share|cite|improve this answer

answered yesterday

Mark FischlerMark Fischler

34.2k12552

$endgroup$

add a comment | 

4

$begingroup$

While breaking up into integer intervals always works, sometimes it is too much work.

Let's say you instead had $$
213 < x^2 - lfloor x rfloor < 505 $$
You really don't want to consider $24$ cases individually.

So you get clever and change this to a pair of simultaneous inequalities: Let $x = n + y$ with $n in Bbb Z$ and $0 leq y < 1$. Then the inequalities are
$$
left{ begin{array}{l} 213 < (n+y)^2 - n < 505 \ 0 leq y < 1 end{array} right.
$$
Then $$213 < n^2 + (2y-1)n +y^2 , 0 leq y < 1 implies 213 < n^2 + (1)n + 1
$$
and this shows that $n geq 15$ (you can get that with one application of the quadratic formula), cutting out a lot of the work. Similarly,
$$ n^2 + (2y-1)n +y^2 < 505 , 0 leq y < 1 implies n^2 + (-1)n + 0 < 505
$$
and this shows that $n < 23$.

Finally, you can also justify only examining the allowed values of $y$ in the boundary cases ($n = 15$ and $n = 22$); in between, any value of $y$ works. (This might not be the case for cubic expressions, for example.)

So for example, on the low end, you would need to solve for $y$ in
$$
213 < (15+y)^2 - 15 = 210 - 30 y + y^2 \
y^2 -30 y -3 > 0\
y > frac12( sqrt{912} -30 ) \
x > (sqrt{228} - 15) + 15 implies x > sqrt{228}
$$
and similarly you need to consider the case of $n=22$ to get the upper limit for $x$.

share|cite|improve this answer

answered yesterday

Mark FischlerMark Fischler

34.2k12552

$endgroup$

add a comment | 

$begingroup$

While breaking up into integer intervals always works, sometimes it is too much work.

Let's say you instead had $$
213 < x^2 - lfloor x rfloor < 505 $$
You really don't want to consider $24$ cases individually.

So you get clever and change this to a pair of simultaneous inequalities: Let $x = n + y$ with $n in Bbb Z$ and $0 leq y < 1$. Then the inequalities are
$$
left{ begin{array}{l} 213 < (n+y)^2 - n < 505 \ 0 leq y < 1 end{array} right.
$$
Then $$213 < n^2 + (2y-1)n +y^2 , 0 leq y < 1 implies 213 < n^2 + (1)n + 1
$$
and this shows that $n geq 15$ (you can get that with one application of the quadratic formula), cutting out a lot of the work. Similarly,
$$ n^2 + (2y-1)n +y^2 < 505 , 0 leq y < 1 implies n^2 + (-1)n + 0 < 505
$$
and this shows that $n < 23$.

Finally, you can also justify only examining the allowed values of $y$ in the boundary cases ($n = 15$ and $n = 22$); in between, any value of $y$ works. (This might not be the case for cubic expressions, for example.)

So for example, on the low end, you would need to solve for $y$ in
$$
213 < (15+y)^2 - 15 = 210 - 30 y + y^2 \
y^2 -30 y -3 > 0\
y > frac12( sqrt{912} -30 ) \
x > (sqrt{228} - 15) + 15 implies x > sqrt{228}
$$
and similarly you need to consider the case of $n=22$ to get the upper limit for $x$.

share|cite|improve this answer

answered yesterday

Mark FischlerMark Fischler

34.2k12552

$endgroup$

share|cite|improve this answer

answered yesterday

Mark FischlerMark Fischler

34.2k12552

answered yesterday

Mark FischlerMark Fischler

34.2k12552

answered yesterday

Mark FischlerMark Fischler

34.2k12552