Discrete math - The ceiling of a real number x, denoted by$ ⌈𝑥⌉$, is the unique integer that satisfies the inequality Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Permutation and CombinationHow to get $sqrt k + frac1sqrtk+1$ in the form $fracsqrtk^2 + 1sqrtk+1$?Hierarchy of Mathematics BreakdownDiscrete math - Prove that a tree with n nodes must have exactly n - 1 edges?How to find a Direct Proof given 3 integersSummation simplification explanationGeneral solution to discrete dynamical system.Solve for $x$ in $fracsqrt x-1x-1>frac4^3/22^4$Mathematical Induction step $2^n < n!$Proving $(p land lnot q) rightarrow p$ is a tautology using logical equivalences
First paper to introduce the "principal-agent problem"
Why not use the yoke to control yaw, as well as pitch and roll?
Discrete math - The ceiling of a real number x, denoted by ⌈𝑥⌉, is the unique integer that satisfies the inequality
How many time has Arya actually used Needle?
malloc in main() or malloc in another function: allocating memory for a struct and its members
Why did Bronn offer to be Tyrion Lannister's champion in trial by combat?
How to get a flat-head nail out of a piece of wood?
What should one know about term logic before studying propositional and predicate logic?
What helicopter has the most rotor blades?
Is it OK if I do not take the receipt in Germany?
Noise in Eigenvalues plot
How do I find my Spellcasting Ability for my D&D character?
Is a copyright notice with a non-existent name be invalid?
How does the body cool itself in a stillsuit?
Getting representations of the Lie group out of representations of its Lie algebra
Inverse square law not accurate for non-point masses?
An isoperimetric-type inequality inside a cube
Any stored/leased 737s that could substitute for grounded MAXs?
Vertical ranges of Column Plots in 12
Did pre-Columbian Americans know the spherical shape of the Earth?
Shimano 105 brifters (5800) and Avid BB5 compatibility
Weaponising the Grasp-at-a-Distance spell
Can two people see the same photon?
Short story about astronauts fertilizing soil with their own bodies
Discrete math - The ceiling of a real number x, denoted by$ ⌈𝑥⌉$, is the unique integer that satisfies the inequality
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Permutation and CombinationHow to get $sqrt k + frac1sqrtk+1$ in the form $fracsqrtk^2 + 1sqrtk+1$?Hierarchy of Mathematics BreakdownDiscrete math - Prove that a tree with n nodes must have exactly n - 1 edges?How to find a Direct Proof given 3 integersSummation simplification explanationGeneral solution to discrete dynamical system.Solve for $x$ in $fracsqrt x-1x-1>frac4^3/22^4$Mathematical Induction step $2^n < n!$Proving $(p land lnot q) rightarrow p$ is a tautology using logical equivalences
$begingroup$
I have a discrete math question below with a solution written by my teacher. I'm really lost as to what answers I'm trying to find exactly. I don't understand how the teacher got 1 and -1 for the first two values, and I also don't understand how and why the last line turns $⌈(𝟔.𝟑 − ⌈𝟏.𝟕⌉)⌉$ into $⌈𝟔.𝟑 − 𝟐⌉ = ⌈𝟒.𝟑⌉ = 𝟓$
I would appreciate it if someone could walk me through the solution.
discrete-mathematics inequality
$endgroup$
add a comment |
$begingroup$
I have a discrete math question below with a solution written by my teacher. I'm really lost as to what answers I'm trying to find exactly. I don't understand how the teacher got 1 and -1 for the first two values, and I also don't understand how and why the last line turns $⌈(𝟔.𝟑 − ⌈𝟏.𝟕⌉)⌉$ into $⌈𝟔.𝟑 − 𝟐⌉ = ⌈𝟒.𝟑⌉ = 𝟓$
I would appreciate it if someone could walk me through the solution.
discrete-mathematics inequality
$endgroup$
$begingroup$
Here's what I see.
$endgroup$
– Shaun
1 hour ago
$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
1 hour ago
1
$begingroup$
Let there be an integer valued function $f$ such that $$f(x) -1 < x leq f(x), forall x in Bbb R.$$ Then we have $$x leq f(x) < x+1.$$ This shows that $$f(x)=lfloor f(x) rfloor = lfloor x rfloor.$$
$endgroup$
– Dbchatto67
1 hour ago
$begingroup$
The above argument proves the uniqueness of $lceil x rceil.$
$endgroup$
– Dbchatto67
1 hour ago
$begingroup$
Sorry in the above argument $lfloor f(x) rfloor$ should be replaced by $lceil f(x) rceil$
$endgroup$
– Dbchatto67
1 hour ago
add a comment |
$begingroup$
I have a discrete math question below with a solution written by my teacher. I'm really lost as to what answers I'm trying to find exactly. I don't understand how the teacher got 1 and -1 for the first two values, and I also don't understand how and why the last line turns $⌈(𝟔.𝟑 − ⌈𝟏.𝟕⌉)⌉$ into $⌈𝟔.𝟑 − 𝟐⌉ = ⌈𝟒.𝟑⌉ = 𝟓$
I would appreciate it if someone could walk me through the solution.
discrete-mathematics inequality
$endgroup$
I have a discrete math question below with a solution written by my teacher. I'm really lost as to what answers I'm trying to find exactly. I don't understand how the teacher got 1 and -1 for the first two values, and I also don't understand how and why the last line turns $⌈(𝟔.𝟑 − ⌈𝟏.𝟕⌉)⌉$ into $⌈𝟔.𝟑 − 𝟐⌉ = ⌈𝟒.𝟑⌉ = 𝟓$
I would appreciate it if someone could walk me through the solution.
discrete-mathematics inequality
discrete-mathematics inequality
edited 7 mins ago
YuiTo Cheng
2,65641037
2,65641037
asked 1 hour ago
GilmoreGirlingGilmoreGirling
465
465
$begingroup$
Here's what I see.
$endgroup$
– Shaun
1 hour ago
$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
1 hour ago
1
$begingroup$
Let there be an integer valued function $f$ such that $$f(x) -1 < x leq f(x), forall x in Bbb R.$$ Then we have $$x leq f(x) < x+1.$$ This shows that $$f(x)=lfloor f(x) rfloor = lfloor x rfloor.$$
$endgroup$
– Dbchatto67
1 hour ago
$begingroup$
The above argument proves the uniqueness of $lceil x rceil.$
$endgroup$
– Dbchatto67
1 hour ago
$begingroup$
Sorry in the above argument $lfloor f(x) rfloor$ should be replaced by $lceil f(x) rceil$
$endgroup$
– Dbchatto67
1 hour ago
add a comment |
$begingroup$
Here's what I see.
$endgroup$
– Shaun
1 hour ago
$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
1 hour ago
1
$begingroup$
Let there be an integer valued function $f$ such that $$f(x) -1 < x leq f(x), forall x in Bbb R.$$ Then we have $$x leq f(x) < x+1.$$ This shows that $$f(x)=lfloor f(x) rfloor = lfloor x rfloor.$$
$endgroup$
– Dbchatto67
1 hour ago
$begingroup$
The above argument proves the uniqueness of $lceil x rceil.$
$endgroup$
– Dbchatto67
1 hour ago
$begingroup$
Sorry in the above argument $lfloor f(x) rfloor$ should be replaced by $lceil f(x) rceil$
$endgroup$
– Dbchatto67
1 hour ago
$begingroup$
Here's what I see.
$endgroup$
– Shaun
1 hour ago
$begingroup$
Here's what I see.
$endgroup$
– Shaun
1 hour ago
$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
1 hour ago
$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
1 hour ago
1
1
$begingroup$
Let there be an integer valued function $f$ such that $$f(x) -1 < x leq f(x), forall x in Bbb R.$$ Then we have $$x leq f(x) < x+1.$$ This shows that $$f(x)=lfloor f(x) rfloor = lfloor x rfloor.$$
$endgroup$
– Dbchatto67
1 hour ago
$begingroup$
Let there be an integer valued function $f$ such that $$f(x) -1 < x leq f(x), forall x in Bbb R.$$ Then we have $$x leq f(x) < x+1.$$ This shows that $$f(x)=lfloor f(x) rfloor = lfloor x rfloor.$$
$endgroup$
– Dbchatto67
1 hour ago
$begingroup$
The above argument proves the uniqueness of $lceil x rceil.$
$endgroup$
– Dbchatto67
1 hour ago
$begingroup$
The above argument proves the uniqueness of $lceil x rceil.$
$endgroup$
– Dbchatto67
1 hour ago
$begingroup$
Sorry in the above argument $lfloor f(x) rfloor$ should be replaced by $lceil f(x) rceil$
$endgroup$
– Dbchatto67
1 hour ago
$begingroup$
Sorry in the above argument $lfloor f(x) rfloor$ should be replaced by $lceil f(x) rceil$
$endgroup$
– Dbchatto67
1 hour ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
A drawing of the function $f(x)=lceil xrceil$ can help:
Taking for granted what the drawing says it's very easy to check any of the substitutions (e.g. clearly $lceil 2.4rceil=lceil 2.15rceil=lceil 2.836rceil=3$ simply reading it)
You can too check the truth of the substitutions right into the inequalities. E. g. $lceil1.7rceil=2$ satisfies $2-1<1.7leq2$
Finally, the very name helps: for a number some integer is its ceiling.
$endgroup$
add a comment |
$begingroup$
You just need to know that $lceilxrceil$ is the smallest integer that is greater than or equal to $x$.
Let $x = 0.1$.
Is $0 ≥ x$? No.
Is $1 ≥ x$? Yes.
Is $2 ≥ x$? Yes. However, it is not the smallest integer as $1$ also satisfies this condition.
Try this procedure with $x = -1.7$ and note that $-1.7 colorred≤ -1$.
$endgroup$
add a comment |
$begingroup$
Let there be an integer valued function $f$ such that $$f(x)-1 < x leq f(x), forall x in Bbb R.$$ Then we have $$x leq f(x) < x+1, forall x in Bbb R.$$ Now two cases may arise which are
$(1)$ $x in Bbb Z.$
$(2)$ $x notin Bbb Z.$
If $x in Bbb Z$ then since $f$ is an integer valued function with $f(x) in [x,x+1)$ it follows that $f(x) = x.$ But then we have $f(x) = lceil f(x) rceil = lceil x rceil.$
Now if $x notin Bbb Z$ then $exists$ a unique integer $n$ with $x<n<x+1.$ So $n$ is the least integer just exceeding $x.$ Therefore $$lceil x rceil = n. (1)$$ On the other hand since $f$ is integer valued with $f(x) in [x,x+1)$ and the only integer in the interval $[x,x+1)$ is $n$ so it also follows that $$f(x)=n. (2)$$ Combining $(1)$ and $(2)$ it follows that $$f(x) = lceil x rceil$$ as required.
This proves the uniqueness of the ceiling function $lceil x rceil.$
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);
;
$window.on('scroll', onScroll);
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3196739%2fdiscrete-math-the-ceiling-of-a-real-number-x-denoted-by-is-the-unique%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A drawing of the function $f(x)=lceil xrceil$ can help:
Taking for granted what the drawing says it's very easy to check any of the substitutions (e.g. clearly $lceil 2.4rceil=lceil 2.15rceil=lceil 2.836rceil=3$ simply reading it)
You can too check the truth of the substitutions right into the inequalities. E. g. $lceil1.7rceil=2$ satisfies $2-1<1.7leq2$
Finally, the very name helps: for a number some integer is its ceiling.
$endgroup$
add a comment |
$begingroup$
A drawing of the function $f(x)=lceil xrceil$ can help:
Taking for granted what the drawing says it's very easy to check any of the substitutions (e.g. clearly $lceil 2.4rceil=lceil 2.15rceil=lceil 2.836rceil=3$ simply reading it)
You can too check the truth of the substitutions right into the inequalities. E. g. $lceil1.7rceil=2$ satisfies $2-1<1.7leq2$
Finally, the very name helps: for a number some integer is its ceiling.
$endgroup$
add a comment |
$begingroup$
A drawing of the function $f(x)=lceil xrceil$ can help:
Taking for granted what the drawing says it's very easy to check any of the substitutions (e.g. clearly $lceil 2.4rceil=lceil 2.15rceil=lceil 2.836rceil=3$ simply reading it)
You can too check the truth of the substitutions right into the inequalities. E. g. $lceil1.7rceil=2$ satisfies $2-1<1.7leq2$
Finally, the very name helps: for a number some integer is its ceiling.
$endgroup$
A drawing of the function $f(x)=lceil xrceil$ can help:
Taking for granted what the drawing says it's very easy to check any of the substitutions (e.g. clearly $lceil 2.4rceil=lceil 2.15rceil=lceil 2.836rceil=3$ simply reading it)
You can too check the truth of the substitutions right into the inequalities. E. g. $lceil1.7rceil=2$ satisfies $2-1<1.7leq2$
Finally, the very name helps: for a number some integer is its ceiling.
answered 1 hour ago
Rafa BudríaRafa Budría
6,0521825
6,0521825
add a comment |
add a comment |
$begingroup$
You just need to know that $lceilxrceil$ is the smallest integer that is greater than or equal to $x$.
Let $x = 0.1$.
Is $0 ≥ x$? No.
Is $1 ≥ x$? Yes.
Is $2 ≥ x$? Yes. However, it is not the smallest integer as $1$ also satisfies this condition.
Try this procedure with $x = -1.7$ and note that $-1.7 colorred≤ -1$.
$endgroup$
add a comment |
$begingroup$
You just need to know that $lceilxrceil$ is the smallest integer that is greater than or equal to $x$.
Let $x = 0.1$.
Is $0 ≥ x$? No.
Is $1 ≥ x$? Yes.
Is $2 ≥ x$? Yes. However, it is not the smallest integer as $1$ also satisfies this condition.
Try this procedure with $x = -1.7$ and note that $-1.7 colorred≤ -1$.
$endgroup$
add a comment |
$begingroup$
You just need to know that $lceilxrceil$ is the smallest integer that is greater than or equal to $x$.
Let $x = 0.1$.
Is $0 ≥ x$? No.
Is $1 ≥ x$? Yes.
Is $2 ≥ x$? Yes. However, it is not the smallest integer as $1$ also satisfies this condition.
Try this procedure with $x = -1.7$ and note that $-1.7 colorred≤ -1$.
$endgroup$
You just need to know that $lceilxrceil$ is the smallest integer that is greater than or equal to $x$.
Let $x = 0.1$.
Is $0 ≥ x$? No.
Is $1 ≥ x$? Yes.
Is $2 ≥ x$? Yes. However, it is not the smallest integer as $1$ also satisfies this condition.
Try this procedure with $x = -1.7$ and note that $-1.7 colorred≤ -1$.
answered 1 hour ago
Toby MakToby Mak
3,70011128
3,70011128
add a comment |
add a comment |
$begingroup$
Let there be an integer valued function $f$ such that $$f(x)-1 < x leq f(x), forall x in Bbb R.$$ Then we have $$x leq f(x) < x+1, forall x in Bbb R.$$ Now two cases may arise which are
$(1)$ $x in Bbb Z.$
$(2)$ $x notin Bbb Z.$
If $x in Bbb Z$ then since $f$ is an integer valued function with $f(x) in [x,x+1)$ it follows that $f(x) = x.$ But then we have $f(x) = lceil f(x) rceil = lceil x rceil.$
Now if $x notin Bbb Z$ then $exists$ a unique integer $n$ with $x<n<x+1.$ So $n$ is the least integer just exceeding $x.$ Therefore $$lceil x rceil = n. (1)$$ On the other hand since $f$ is integer valued with $f(x) in [x,x+1)$ and the only integer in the interval $[x,x+1)$ is $n$ so it also follows that $$f(x)=n. (2)$$ Combining $(1)$ and $(2)$ it follows that $$f(x) = lceil x rceil$$ as required.
This proves the uniqueness of the ceiling function $lceil x rceil.$
$endgroup$
add a comment |
$begingroup$
Let there be an integer valued function $f$ such that $$f(x)-1 < x leq f(x), forall x in Bbb R.$$ Then we have $$x leq f(x) < x+1, forall x in Bbb R.$$ Now two cases may arise which are
$(1)$ $x in Bbb Z.$
$(2)$ $x notin Bbb Z.$
If $x in Bbb Z$ then since $f$ is an integer valued function with $f(x) in [x,x+1)$ it follows that $f(x) = x.$ But then we have $f(x) = lceil f(x) rceil = lceil x rceil.$
Now if $x notin Bbb Z$ then $exists$ a unique integer $n$ with $x<n<x+1.$ So $n$ is the least integer just exceeding $x.$ Therefore $$lceil x rceil = n. (1)$$ On the other hand since $f$ is integer valued with $f(x) in [x,x+1)$ and the only integer in the interval $[x,x+1)$ is $n$ so it also follows that $$f(x)=n. (2)$$ Combining $(1)$ and $(2)$ it follows that $$f(x) = lceil x rceil$$ as required.
This proves the uniqueness of the ceiling function $lceil x rceil.$
$endgroup$
add a comment |
$begingroup$
Let there be an integer valued function $f$ such that $$f(x)-1 < x leq f(x), forall x in Bbb R.$$ Then we have $$x leq f(x) < x+1, forall x in Bbb R.$$ Now two cases may arise which are
$(1)$ $x in Bbb Z.$
$(2)$ $x notin Bbb Z.$
If $x in Bbb Z$ then since $f$ is an integer valued function with $f(x) in [x,x+1)$ it follows that $f(x) = x.$ But then we have $f(x) = lceil f(x) rceil = lceil x rceil.$
Now if $x notin Bbb Z$ then $exists$ a unique integer $n$ with $x<n<x+1.$ So $n$ is the least integer just exceeding $x.$ Therefore $$lceil x rceil = n. (1)$$ On the other hand since $f$ is integer valued with $f(x) in [x,x+1)$ and the only integer in the interval $[x,x+1)$ is $n$ so it also follows that $$f(x)=n. (2)$$ Combining $(1)$ and $(2)$ it follows that $$f(x) = lceil x rceil$$ as required.
This proves the uniqueness of the ceiling function $lceil x rceil.$
$endgroup$
Let there be an integer valued function $f$ such that $$f(x)-1 < x leq f(x), forall x in Bbb R.$$ Then we have $$x leq f(x) < x+1, forall x in Bbb R.$$ Now two cases may arise which are
$(1)$ $x in Bbb Z.$
$(2)$ $x notin Bbb Z.$
If $x in Bbb Z$ then since $f$ is an integer valued function with $f(x) in [x,x+1)$ it follows that $f(x) = x.$ But then we have $f(x) = lceil f(x) rceil = lceil x rceil.$
Now if $x notin Bbb Z$ then $exists$ a unique integer $n$ with $x<n<x+1.$ So $n$ is the least integer just exceeding $x.$ Therefore $$lceil x rceil = n. (1)$$ On the other hand since $f$ is integer valued with $f(x) in [x,x+1)$ and the only integer in the interval $[x,x+1)$ is $n$ so it also follows that $$f(x)=n. (2)$$ Combining $(1)$ and $(2)$ it follows that $$f(x) = lceil x rceil$$ as required.
This proves the uniqueness of the ceiling function $lceil x rceil.$
answered 40 mins ago
Dbchatto67Dbchatto67
3,571626
3,571626
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);
;
$window.on('scroll', onScroll);
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3196739%2fdiscrete-math-the-ceiling-of-a-real-number-x-denoted-by-is-the-unique%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);
;
$window.on('scroll', onScroll);
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);
;
$window.on('scroll', onScroll);
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);
;
$window.on('scroll', onScroll);
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Here's what I see.
$endgroup$
– Shaun
1 hour ago
$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
1 hour ago
1
$begingroup$
Let there be an integer valued function $f$ such that $$f(x) -1 < x leq f(x), forall x in Bbb R.$$ Then we have $$x leq f(x) < x+1.$$ This shows that $$f(x)=lfloor f(x) rfloor = lfloor x rfloor.$$
$endgroup$
– Dbchatto67
1 hour ago
$begingroup$
The above argument proves the uniqueness of $lceil x rceil.$
$endgroup$
– Dbchatto67
1 hour ago
$begingroup$
Sorry in the above argument $lfloor f(x) rfloor$ should be replaced by $lceil f(x) rceil$
$endgroup$
– Dbchatto67
1 hour ago