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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 1 min ago

YuiTo Cheng

2,4064937

asked 1 hour ago

mavericmaveric

91912

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

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 1 min ago

YuiTo Cheng

2,4064937

asked 1 hour ago

mavericmaveric

91912

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

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 1 min ago

YuiTo Cheng

2,4064937

asked 1 hour ago

mavericmaveric

91912

$endgroup$

share|cite|improve this question

edited 1 min ago

YuiTo Cheng

2,4064937

asked 1 hour ago

mavericmaveric

91912

share|cite|improve this question

edited 1 min ago

YuiTo Cheng

2,4064937

asked 1 hour ago

mavericmaveric

91912

edited 1 min ago

YuiTo Cheng

2,4064937

edited 1 min ago

YuiTo Cheng

2,4064937

asked 1 hour ago

mavericmaveric

91912

asked 1 hour ago

mavericmaveric

91912

2 Answers
2

active

oldest

votes

5

$begingroup$

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 24 mins ago

answered 33 mins ago

Maria MazurMaria Mazur

49.9k1361125

$endgroup$

add a comment | 

2

$begingroup$

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 31 mins ago

Mark FischlerMark Fischler

34.1k12552

$endgroup$

add a comment | 

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5

$begingroup$

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 24 mins ago

answered 33 mins ago

Maria MazurMaria Mazur

49.9k1361125

$endgroup$

add a comment | 

5

$begingroup$

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 24 mins ago

answered 33 mins ago

Maria MazurMaria Mazur

49.9k1361125

$endgroup$

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 24 mins ago

answered 33 mins ago

Maria MazurMaria Mazur

49.9k1361125

$endgroup$

share|cite|improve this answer

edited 24 mins ago

answered 33 mins ago

Maria MazurMaria Mazur

49.9k1361125

answered 33 mins ago

Maria MazurMaria Mazur

49.9k1361125

answered 33 mins ago

Maria MazurMaria Mazur

49.9k1361125

2

$begingroup$

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 31 mins ago

Mark FischlerMark Fischler

34.1k12552

$endgroup$

add a comment | 

2

$begingroup$

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 31 mins ago

Mark FischlerMark Fischler

34.1k12552

$endgroup$

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 31 mins ago

Mark FischlerMark Fischler

34.1k12552

$endgroup$

share|cite|improve this answer

answered 31 mins ago

Mark FischlerMark Fischler

34.1k12552

answered 31 mins ago

Mark FischlerMark Fischler

34.1k12552

answered 31 mins ago

Mark FischlerMark Fischler

34.1k12552