Is $lfloor2sqrtn-lfloor sqrtn rfloorrfloor <lfloor2sqrtn rfloor$ true for $ngeq 0$? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)$lfloor sqrt n+sqrt n+1+sqrtn+2+sqrtn+3+sqrtn+4rfloor=lfloorsqrt 25n+49rfloor$ is true?Is this proof that $lfloor x rfloor geq n leftlfloor fracxn rightrfloor$ correct?How prove this$lfloor sqrt2x-lfloorsqrt2xrfloorrfloor=lfloorfracsqrt8x+1-12rfloor$$lfloor xrfloor + lfloor yrfloor leq lfloor x+yrfloor$ for every pair of numbers of $x$ and $y$Solve $lfloor sqrt x rfloor = lfloor x/2 rfloor$ for real $x$Is there a simple way of proving that $lfloorsqrtnrfloor+lfloorsqrt4n+1rfloor = lfloorfrac32 lfloor sqrt4n+1 rfloorrfloor$?Prove that $lfloor 2x rfloor + lfloor 2y rfloor geq lfloor x rfloor + lfloor y rfloor + lfloor x+y rfloor$ for all real $x$ and $y$.Prove $lfloor 2x rfloor + lfloor 2y rfloor geq lfloor x rfloor + lfloor yrfloor+lfloor x+yrfloor$Prove or disprove: $phi: mathbbN to mathbbNtext, phi(n) = lfloor n cdot | sin( sqrt2 cdot n ) | rfloor$ is surjectiveCounterexample for floor function: $lfloor x+y rfloor geq lfloor x rfloor + lfloor y rfloor $
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Is $lfloor2sqrtn-lfloor sqrtn rfloorrfloor
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)$lfloor sqrt n+sqrt n+1+sqrtn+2+sqrtn+3+sqrtn+4rfloor=lfloorsqrt 25n+49rfloor$ is true?Is this proof that $lfloor x rfloor geq n leftlfloor fracxn rightrfloor$ correct?How prove this$lfloor sqrt2x-lfloorsqrt2xrfloorrfloor=lfloorfracsqrt8x+1-12rfloor$$lfloor xrfloor + lfloor yrfloor leq lfloor x+yrfloor$ for every pair of numbers of $x$ and $y$Solve $lfloor sqrt x rfloor = lfloor x/2 rfloor$ for real $x$Is there a simple way of proving that $lfloorsqrtnrfloor+lfloorsqrt4n+1rfloor = lfloorfrac32 lfloor sqrt4n+1 rfloorrfloor$?Prove that $lfloor 2x rfloor + lfloor 2y rfloor geq lfloor x rfloor + lfloor y rfloor + lfloor x+y rfloor$ for all real $x$ and $y$.Prove $lfloor 2x rfloor + lfloor 2y rfloor geq lfloor x rfloor + lfloor yrfloor+lfloor x+yrfloor$Prove or disprove: $phi: mathbbN to mathbbNtext, phi(n) = lfloor n cdot | sin( sqrt2 cdot n ) | rfloor$ is surjectiveCounterexample for floor function: $lfloor x+y rfloor geq lfloor x rfloor + lfloor y rfloor $
$begingroup$
I have reasons to believe that $lfloor2sqrtn-lfloor sqrtn rfloorrfloor <lfloor2sqrtn rfloor$ for $ngeq 0$. How could I go about proving (or disproving) this?
discrete-mathematics inequality real-numbers floor-function
asked Apr 2 at 8:18
DavidDavid
619417
$endgroup$
add a comment |
$begingroup$
I have reasons to believe that $lfloor2sqrtn-lfloor sqrtn rfloorrfloor <lfloor2sqrtn rfloor$ for $ngeq 0$. How could I go about proving (or disproving) this?
discrete-mathematics inequality real-numbers floor-function
asked Apr 2 at 8:18
DavidDavid
619417
$endgroup$
$begingroup$
I am afraid, you have accepted the answer too earlier. All you have to do is to change $<$ with $leq$.
$endgroup$
– rtybase
Apr 2 at 8:51
$begingroup$
Did you even try $n=0$ !?
$endgroup$
– Yves Daoust
Apr 2 at 9:51
add a comment |
$begingroup$
I have reasons to believe that $lfloor2sqrtn-lfloor sqrtn rfloorrfloor <lfloor2sqrtn rfloor$ for $ngeq 0$. How could I go about proving (or disproving) this?
discrete-mathematics inequality real-numbers floor-function
asked Apr 2 at 8:18
DavidDavid
619417
$endgroup$
I have reasons to believe that $lfloor2sqrtn-lfloor sqrtn rfloorrfloor <lfloor2sqrtn rfloor$ for $ngeq 0$. How could I go about proving (or disproving) this?
discrete-mathematics inequality real-numbers floor-function
discrete-mathematics inequality real-numbers floor-function
asked Apr 2 at 8:18
DavidDavid
619417
asked Apr 2 at 8:18
DavidDavid
619417
asked Apr 2 at 8:18
DavidDavid
619417
asked Apr 2 at 8:18
DavidDavid
619417
asked Apr 2 at 8:18
DavidDavid
619417
619417
$begingroup$
I am afraid, you have accepted the answer too earlier. All you have to do is to change $<$ with $leq$.
$endgroup$
– rtybase
Apr 2 at 8:51
$begingroup$
Did you even try $n=0$ !?
$endgroup$
– Yves Daoust
Apr 2 at 9:51
add a comment |
$begingroup$
I am afraid, you have accepted the answer too earlier. All you have to do is to change $<$ with $leq$.
$endgroup$
– rtybase
Apr 2 at 8:51
$begingroup$
Did you even try $n=0$ !?
$endgroup$
– Yves Daoust
Apr 2 at 9:51
$begingroup$
I am afraid, you have accepted the answer too earlier. All you have to do is to change $<$ with $leq$.
$endgroup$
– rtybase
Apr 2 at 8:51
$begingroup$
I am afraid, you have accepted the answer too earlier. All you have to do is to change $<$ with $leq$.
$endgroup$
– rtybase
Apr 2 at 8:51
$begingroup$
Did you even try $n=0$ !?
$endgroup$
– Yves Daoust
Apr 2 at 9:51
$begingroup$
Did you even try $n=0$ !?
$endgroup$
– Yves Daoust
Apr 2 at 9:51
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You could start by checking the inequality for small values of $n$. You would find that it is false already for $n=2$.
answered Apr 2 at 8:32
LithoLitho
3,5181716
$endgroup$
$begingroup$
For $n=2$ LHS=RHS, it's a corner case, other than that, it seems true
$endgroup$
– rtybase
Apr 2 at 8:46
$begingroup$
@rtybase: there are infinitely many equalities. See my answer.
$endgroup$
– Yves Daoust
Apr 2 at 9:24
$begingroup$
@YvesDaoust, yes, true ... I am not sure if that's what OP was after (strict $<$ or $leq$). $n=0$ is easier to check than $n=2$.
$endgroup$
– rtybase
Apr 2 at 9:33
$begingroup$
@rtybase: non strict is immediate.
$endgroup$
– Yves Daoust
Apr 2 at 9:36
1
$begingroup$
@rtybase: I don't think so. You are hypothesizing that the question could be with $le$. But as its stands, there is no ambiguity: it is wrong. The OP should indeed have observed that it doesn't work with $n=0$ (which he willingly included in the domain).
$endgroup$
– Yves Daoust
Apr 2 at 9:42
|
show 1 more comment
$begingroup$
$$lfloor2sqrt xrfloor$$ is a non-decreasing function so it is obvious that
$$leftlfloor2sqrtn-lfloorsqrt nrfloorrightrfloorle lfloor2sqrt nrfloor$$ holds.
The interesting cases are equality, i.e.
$$leftlfloor2sqrtn-lfloorsqrt nrfloorrightrfloor=lfloor2sqrt nrfloor.$$
The function $lfloor2sqrt xrfloor$ remains the constant $m$ for
$$m^2le4nle(m+1)^2-1.$$
So we need
$$m^2le 4(n-lfloorsqrt nrfloor)le m^2+2m.$$
As
$$leftlfloorfrac m2rightrfloorlelfloorsqrt nrfloorleleftlfloorfrac sqrtm(m+2)2rightrfloor,$$ by addition and taking the tightest bounds,
$$m^2+4leftlfloorfrac m2rightrfloorle 4nle m^2+2m,$$ which is not void for even $m$.
Every $m=2k$, i.e. $n=k(k+1)$ makes the equality, as
$$k(k+1)-leftlfloorsqrtk(k+1)rightrfloor=k^2$$ and
$$leftlfloor2sqrtk(k+1)rightrfloor=leftlfloor2sqrtk^2rightrfloor=2k.$$
There are no other solutions.
answered Apr 2 at 9:16
Yves DaoustYves Daoust
133k676232
$endgroup$
1
$begingroup$
Nice answer! (+1) ... to a badly formulated question
$endgroup$
– rtybase
Apr 2 at 9:41
$begingroup$
@rtybase: I was fearing that the inequalities couldn't be sorted out. But in the end it worked.
$endgroup$
– Yves Daoust
Apr 2 at 9:45
add a comment |
$begingroup$
Hint. It is true if you replace $<$ with $leq$ and use the fact that
Proposition. If $xleq y Rightarrow lfloor x rfloor leq lfloor y rfloor$
From that proposition, $forall ngeq 0$
$$0leq2sqrtn-lfloor sqrtn rfloor leq 2sqrtn iff lfloor sqrtn rfloor geq 0$$
and as a result
$$lfloor2sqrtn-lfloor sqrtn rfloorrfloor leqlfloor2sqrtn rfloor $$
answered Apr 2 at 9:01
rtybasertybase
11.7k31534
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You could start by checking the inequality for small values of $n$. You would find that it is false already for $n=2$.
answered Apr 2 at 8:32
LithoLitho
3,5181716
$endgroup$
$begingroup$
For $n=2$ LHS=RHS, it's a corner case, other than that, it seems true
$endgroup$
– rtybase
Apr 2 at 8:46
$begingroup$
@rtybase: there are infinitely many equalities. See my answer.
$endgroup$
– Yves Daoust
Apr 2 at 9:24
$begingroup$
@YvesDaoust, yes, true ... I am not sure if that's what OP was after (strict $<$ or $leq$). $n=0$ is easier to check than $n=2$.
$endgroup$
– rtybase
Apr 2 at 9:33
$begingroup$
@rtybase: non strict is immediate.
$endgroup$
– Yves Daoust
Apr 2 at 9:36
1
$begingroup$
@rtybase: I don't think so. You are hypothesizing that the question could be with $le$. But as its stands, there is no ambiguity: it is wrong. The OP should indeed have observed that it doesn't work with $n=0$ (which he willingly included in the domain).
$endgroup$
– Yves Daoust
Apr 2 at 9:42
|
show 1 more comment
$begingroup$
You could start by checking the inequality for small values of $n$. You would find that it is false already for $n=2$.
answered Apr 2 at 8:32
LithoLitho
3,5181716
$endgroup$
$begingroup$
For $n=2$ LHS=RHS, it's a corner case, other than that, it seems true
$endgroup$
– rtybase
Apr 2 at 8:46
$begingroup$
@rtybase: there are infinitely many equalities. See my answer.
$endgroup$
– Yves Daoust
Apr 2 at 9:24
$begingroup$
@YvesDaoust, yes, true ... I am not sure if that's what OP was after (strict $<$ or $leq$). $n=0$ is easier to check than $n=2$.
$endgroup$
– rtybase
Apr 2 at 9:33
$begingroup$
@rtybase: non strict is immediate.
$endgroup$
– Yves Daoust
Apr 2 at 9:36
1
$begingroup$
@rtybase: I don't think so. You are hypothesizing that the question could be with $le$. But as its stands, there is no ambiguity: it is wrong. The OP should indeed have observed that it doesn't work with $n=0$ (which he willingly included in the domain).
$endgroup$
– Yves Daoust
Apr 2 at 9:42
|
show 1 more comment
$begingroup$
You could start by checking the inequality for small values of $n$. You would find that it is false already for $n=2$.
answered Apr 2 at 8:32
LithoLitho
3,5181716
$endgroup$
You could start by checking the inequality for small values of $n$. You would find that it is false already for $n=2$.
answered Apr 2 at 8:32
LithoLitho
3,5181716
answered Apr 2 at 8:32
LithoLitho
3,5181716
answered Apr 2 at 8:32
LithoLitho
3,5181716
answered Apr 2 at 8:32
LithoLitho
3,5181716
3,5181716
$begingroup$
For $n=2$ LHS=RHS, it's a corner case, other than that, it seems true
$endgroup$
– rtybase
Apr 2 at 8:46
$begingroup$
@rtybase: there are infinitely many equalities. See my answer.
$endgroup$
– Yves Daoust
Apr 2 at 9:24
$begingroup$
@YvesDaoust, yes, true ... I am not sure if that's what OP was after (strict $<$ or $leq$). $n=0$ is easier to check than $n=2$.
$endgroup$
– rtybase
Apr 2 at 9:33
$begingroup$
@rtybase: non strict is immediate.
$endgroup$
– Yves Daoust
Apr 2 at 9:36
1
$begingroup$
@rtybase: I don't think so. You are hypothesizing that the question could be with $le$. But as its stands, there is no ambiguity: it is wrong. The OP should indeed have observed that it doesn't work with $n=0$ (which he willingly included in the domain).
$endgroup$
– Yves Daoust
Apr 2 at 9:42
|
show 1 more comment
$begingroup$
For $n=2$ LHS=RHS, it's a corner case, other than that, it seems true
$endgroup$
– rtybase
Apr 2 at 8:46
$begingroup$
@rtybase: there are infinitely many equalities. See my answer.
$endgroup$
– Yves Daoust
Apr 2 at 9:24
$begingroup$
@YvesDaoust, yes, true ... I am not sure if that's what OP was after (strict $<$ or $leq$). $n=0$ is easier to check than $n=2$.
$endgroup$
– rtybase
Apr 2 at 9:33
$begingroup$
@rtybase: non strict is immediate.
$endgroup$
– Yves Daoust
Apr 2 at 9:36
1
$begingroup$
@rtybase: I don't think so. You are hypothesizing that the question could be with $le$. But as its stands, there is no ambiguity: it is wrong. The OP should indeed have observed that it doesn't work with $n=0$ (which he willingly included in the domain).
$endgroup$
– Yves Daoust
Apr 2 at 9:42
$begingroup$
For $n=2$ LHS=RHS, it's a corner case, other than that, it seems true
$endgroup$
– rtybase
Apr 2 at 8:46
$begingroup$
For $n=2$ LHS=RHS, it's a corner case, other than that, it seems true
$endgroup$
– rtybase
Apr 2 at 8:46
$begingroup$
@rtybase: there are infinitely many equalities. See my answer.
$endgroup$
– Yves Daoust
Apr 2 at 9:24
$begingroup$
@rtybase: there are infinitely many equalities. See my answer.
$endgroup$
– Yves Daoust
Apr 2 at 9:24
$begingroup$
@YvesDaoust, yes, true ... I am not sure if that's what OP was after (strict $<$ or $leq$). $n=0$ is easier to check than $n=2$.
$endgroup$
– rtybase
Apr 2 at 9:33
$begingroup$
@YvesDaoust, yes, true ... I am not sure if that's what OP was after (strict $<$ or $leq$). $n=0$ is easier to check than $n=2$.
$endgroup$
– rtybase
Apr 2 at 9:33
$begingroup$
@rtybase: non strict is immediate.
$endgroup$
– Yves Daoust
Apr 2 at 9:36
$begingroup$
@rtybase: non strict is immediate.
$endgroup$
– Yves Daoust
Apr 2 at 9:36
1
1
$begingroup$
@rtybase: I don't think so. You are hypothesizing that the question could be with $le$. But as its stands, there is no ambiguity: it is wrong. The OP should indeed have observed that it doesn't work with $n=0$ (which he willingly included in the domain).
$endgroup$
– Yves Daoust
Apr 2 at 9:42
$begingroup$
@rtybase: I don't think so. You are hypothesizing that the question could be with $le$. But as its stands, there is no ambiguity: it is wrong. The OP should indeed have observed that it doesn't work with $n=0$ (which he willingly included in the domain).
$endgroup$
– Yves Daoust
Apr 2 at 9:42
|
show 1 more comment
$begingroup$
$$lfloor2sqrt xrfloor$$ is a non-decreasing function so it is obvious that
$$leftlfloor2sqrtn-lfloorsqrt nrfloorrightrfloorle lfloor2sqrt nrfloor$$ holds.
The interesting cases are equality, i.e.
$$leftlfloor2sqrtn-lfloorsqrt nrfloorrightrfloor=lfloor2sqrt nrfloor.$$
The function $lfloor2sqrt xrfloor$ remains the constant $m$ for
$$m^2le4nle(m+1)^2-1.$$
So we need
$$m^2le 4(n-lfloorsqrt nrfloor)le m^2+2m.$$
As
$$leftlfloorfrac m2rightrfloorlelfloorsqrt nrfloorleleftlfloorfrac sqrtm(m+2)2rightrfloor,$$ by addition and taking the tightest bounds,
$$m^2+4leftlfloorfrac m2rightrfloorle 4nle m^2+2m,$$ which is not void for even $m$.
Every $m=2k$, i.e. $n=k(k+1)$ makes the equality, as
$$k(k+1)-leftlfloorsqrtk(k+1)rightrfloor=k^2$$ and
$$leftlfloor2sqrtk(k+1)rightrfloor=leftlfloor2sqrtk^2rightrfloor=2k.$$
There are no other solutions.
answered Apr 2 at 9:16
Yves DaoustYves Daoust
133k676232
$endgroup$
1
$begingroup$
Nice answer! (+1) ... to a badly formulated question
$endgroup$
– rtybase
Apr 2 at 9:41
$begingroup$
@rtybase: I was fearing that the inequalities couldn't be sorted out. But in the end it worked.
$endgroup$
– Yves Daoust
Apr 2 at 9:45
add a comment |
$begingroup$
$$lfloor2sqrt xrfloor$$ is a non-decreasing function so it is obvious that
$$leftlfloor2sqrtn-lfloorsqrt nrfloorrightrfloorle lfloor2sqrt nrfloor$$ holds.
The interesting cases are equality, i.e.
$$leftlfloor2sqrtn-lfloorsqrt nrfloorrightrfloor=lfloor2sqrt nrfloor.$$
The function $lfloor2sqrt xrfloor$ remains the constant $m$ for
$$m^2le4nle(m+1)^2-1.$$
So we need
$$m^2le 4(n-lfloorsqrt nrfloor)le m^2+2m.$$
As
$$leftlfloorfrac m2rightrfloorlelfloorsqrt nrfloorleleftlfloorfrac sqrtm(m+2)2rightrfloor,$$ by addition and taking the tightest bounds,
$$m^2+4leftlfloorfrac m2rightrfloorle 4nle m^2+2m,$$ which is not void for even $m$.
Every $m=2k$, i.e. $n=k(k+1)$ makes the equality, as
$$k(k+1)-leftlfloorsqrtk(k+1)rightrfloor=k^2$$ and
$$leftlfloor2sqrtk(k+1)rightrfloor=leftlfloor2sqrtk^2rightrfloor=2k.$$
There are no other solutions.
answered Apr 2 at 9:16
Yves DaoustYves Daoust
133k676232
$endgroup$
1
$begingroup$
Nice answer! (+1) ... to a badly formulated question
$endgroup$
– rtybase
Apr 2 at 9:41
$begingroup$
@rtybase: I was fearing that the inequalities couldn't be sorted out. But in the end it worked.
$endgroup$
– Yves Daoust
Apr 2 at 9:45
add a comment |
$begingroup$
$$lfloor2sqrt xrfloor$$ is a non-decreasing function so it is obvious that
$$leftlfloor2sqrtn-lfloorsqrt nrfloorrightrfloorle lfloor2sqrt nrfloor$$ holds.
The interesting cases are equality, i.e.
$$leftlfloor2sqrtn-lfloorsqrt nrfloorrightrfloor=lfloor2sqrt nrfloor.$$
The function $lfloor2sqrt xrfloor$ remains the constant $m$ for
$$m^2le4nle(m+1)^2-1.$$
So we need
$$m^2le 4(n-lfloorsqrt nrfloor)le m^2+2m.$$
As
$$leftlfloorfrac m2rightrfloorlelfloorsqrt nrfloorleleftlfloorfrac sqrtm(m+2)2rightrfloor,$$ by addition and taking the tightest bounds,
$$m^2+4leftlfloorfrac m2rightrfloorle 4nle m^2+2m,$$ which is not void for even $m$.
Every $m=2k$, i.e. $n=k(k+1)$ makes the equality, as
$$k(k+1)-leftlfloorsqrtk(k+1)rightrfloor=k^2$$ and
$$leftlfloor2sqrtk(k+1)rightrfloor=leftlfloor2sqrtk^2rightrfloor=2k.$$
There are no other solutions.
answered Apr 2 at 9:16
Yves DaoustYves Daoust
133k676232
$endgroup$
$$lfloor2sqrt xrfloor$$ is a non-decreasing function so it is obvious that
$$leftlfloor2sqrtn-lfloorsqrt nrfloorrightrfloorle lfloor2sqrt nrfloor$$ holds.
The interesting cases are equality, i.e.
$$leftlfloor2sqrtn-lfloorsqrt nrfloorrightrfloor=lfloor2sqrt nrfloor.$$
The function $lfloor2sqrt xrfloor$ remains the constant $m$ for
$$m^2le4nle(m+1)^2-1.$$
So we need
$$m^2le 4(n-lfloorsqrt nrfloor)le m^2+2m.$$
As
$$leftlfloorfrac m2rightrfloorlelfloorsqrt nrfloorleleftlfloorfrac sqrtm(m+2)2rightrfloor,$$ by addition and taking the tightest bounds,
$$m^2+4leftlfloorfrac m2rightrfloorle 4nle m^2+2m,$$ which is not void for even $m$.
Every $m=2k$, i.e. $n=k(k+1)$ makes the equality, as
$$k(k+1)-leftlfloorsqrtk(k+1)rightrfloor=k^2$$ and
$$leftlfloor2sqrtk(k+1)rightrfloor=leftlfloor2sqrtk^2rightrfloor=2k.$$
There are no other solutions.
answered Apr 2 at 9:16
Yves DaoustYves Daoust
133k676232
edited Apr 2 at 9:44
answered Apr 2 at 9:16
Yves DaoustYves Daoust
133k676232
answered Apr 2 at 9:16
Yves DaoustYves Daoust
133k676232
answered Apr 2 at 9:16
Yves DaoustYves Daoust
133k676232
133k676232
1
$begingroup$
Nice answer! (+1) ... to a badly formulated question
$endgroup$
– rtybase
Apr 2 at 9:41
$begingroup$
@rtybase: I was fearing that the inequalities couldn't be sorted out. But in the end it worked.
$endgroup$
– Yves Daoust
Apr 2 at 9:45
add a comment |
1
$begingroup$
Nice answer! (+1) ... to a badly formulated question
$endgroup$
– rtybase
Apr 2 at 9:41
$begingroup$
@rtybase: I was fearing that the inequalities couldn't be sorted out. But in the end it worked.
$endgroup$
– Yves Daoust
Apr 2 at 9:45
1
1
$begingroup$
Nice answer! (+1) ... to a badly formulated question
$endgroup$
– rtybase
Apr 2 at 9:41
$begingroup$
Nice answer! (+1) ... to a badly formulated question
$endgroup$
– rtybase
Apr 2 at 9:41
$begingroup$
@rtybase: I was fearing that the inequalities couldn't be sorted out. But in the end it worked.
$endgroup$
– Yves Daoust
Apr 2 at 9:45
$begingroup$
@rtybase: I was fearing that the inequalities couldn't be sorted out. But in the end it worked.
$endgroup$
– Yves Daoust
Apr 2 at 9:45
add a comment |
$begingroup$
Hint. It is true if you replace $<$ with $leq$ and use the fact that
Proposition. If $xleq y Rightarrow lfloor x rfloor leq lfloor y rfloor$
From that proposition, $forall ngeq 0$
$$0leq2sqrtn-lfloor sqrtn rfloor leq 2sqrtn iff lfloor sqrtn rfloor geq 0$$
and as a result
$$lfloor2sqrtn-lfloor sqrtn rfloorrfloor leqlfloor2sqrtn rfloor $$
answered Apr 2 at 9:01
rtybasertybase
11.7k31534
$endgroup$
add a comment |
$begingroup$
Hint. It is true if you replace $<$ with $leq$ and use the fact that
Proposition. If $xleq y Rightarrow lfloor x rfloor leq lfloor y rfloor$
From that proposition, $forall ngeq 0$
$$0leq2sqrtn-lfloor sqrtn rfloor leq 2sqrtn iff lfloor sqrtn rfloor geq 0$$
and as a result
$$lfloor2sqrtn-lfloor sqrtn rfloorrfloor leqlfloor2sqrtn rfloor $$
answered Apr 2 at 9:01
rtybasertybase
11.7k31534
$endgroup$
add a comment |
$begingroup$
Hint. It is true if you replace $<$ with $leq$ and use the fact that
Proposition. If $xleq y Rightarrow lfloor x rfloor leq lfloor y rfloor$
From that proposition, $forall ngeq 0$
$$0leq2sqrtn-lfloor sqrtn rfloor leq 2sqrtn iff lfloor sqrtn rfloor geq 0$$
and as a result
$$lfloor2sqrtn-lfloor sqrtn rfloorrfloor leqlfloor2sqrtn rfloor $$
answered Apr 2 at 9:01
rtybasertybase
11.7k31534
$endgroup$
Hint. It is true if you replace $<$ with $leq$ and use the fact that
Proposition. If $xleq y Rightarrow lfloor x rfloor leq lfloor y rfloor$
From that proposition, $forall ngeq 0$
$$0leq2sqrtn-lfloor sqrtn rfloor leq 2sqrtn iff lfloor sqrtn rfloor geq 0$$
and as a result
$$lfloor2sqrtn-lfloor sqrtn rfloorrfloor leqlfloor2sqrtn rfloor $$
answered Apr 2 at 9:01
rtybasertybase
11.7k31534
answered Apr 2 at 9:01
rtybasertybase
11.7k31534
answered Apr 2 at 9:01
rtybasertybase
11.7k31534
answered Apr 2 at 9:01
rtybasertybase
11.7k31534
11.7k31534
add a comment |
add a comment |
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$begingroup$
I am afraid, you have accepted the answer too earlier. All you have to do is to change $<$ with $leq$.
$endgroup$
– rtybase
Apr 2 at 8:51
$begingroup$
Did you even try $n=0$ !?
$endgroup$
– Yves Daoust
Apr 2 at 9:51