Mixing
prove that the greatest integer function [x] is continuous at all points except at integer points?
$begingroup$
I was solving this function , now the question that arises is that I was solving this using an example i.e. A numerical value, but my teacher keeps saying that it's wrong or I have to solve it using constants such as k... Etc. Is this method wrong according to u?
i) f(x) = [x], for all x in R
==> By the definition of greatest integer function: If x lies between two successive integers, then f(x) = least integer of them.
ii) So, at x = 2, f(x) = [2] = 2 -------- (1)
Left side limit (x ---> 2-h): f(x) = [2 - h] = 1 ----- (2)
{Since (2 - h) lies between 1 & 2; and the least being 1}
Right side limit (x --> 2+h): f(x) = [2 + h] = 2 -------- (3)
{Since (2+h) lies between 2 & 3; and the least being 2}
iii) Thus from the above 3 equations, left side limit is not equal to right side limit.
So limit of the function does not exist.
Hence it is discontinuous at x = 2
So this is not derivable at x = 2
Hence Proved.
calculus functions continuity
asked Oct 3 '17 at 12:25
Jatt-Soorma Jatt-Soorma
1112
$endgroup$
add a comment |
$begingroup$
I was solving this function , now the question that arises is that I was solving this using an example i.e. A numerical value, but my teacher keeps saying that it's wrong or I have to solve it using constants such as k... Etc. Is this method wrong according to u?
i) f(x) = [x], for all x in R
==> By the definition of greatest integer function: If x lies between two successive integers, then f(x) = least integer of them.
ii) So, at x = 2, f(x) = [2] = 2 -------- (1)
Left side limit (x ---> 2-h): f(x) = [2 - h] = 1 ----- (2)
{Since (2 - h) lies between 1 & 2; and the least being 1}
Right side limit (x --> 2+h): f(x) = [2 + h] = 2 -------- (3)
{Since (2+h) lies between 2 & 3; and the least being 2}
iii) Thus from the above 3 equations, left side limit is not equal to right side limit.
So limit of the function does not exist.
Hence it is discontinuous at x = 2
So this is not derivable at x = 2
Hence Proved.
calculus functions continuity
asked Oct 3 '17 at 12:25
Jatt-Soorma Jatt-Soorma
1112
$endgroup$
$begingroup$
Your answer presentation is not correct. Because in these type of question answer should not be explained with help of an example but it should be explained mathematically...
$endgroup$
– Akshit Bansal
Jun 3 '18 at 4:42
add a comment |
$begingroup$
I was solving this function , now the question that arises is that I was solving this using an example i.e. A numerical value, but my teacher keeps saying that it's wrong or I have to solve it using constants such as k... Etc. Is this method wrong according to u?
i) f(x) = [x], for all x in R
==> By the definition of greatest integer function: If x lies between two successive integers, then f(x) = least integer of them.
ii) So, at x = 2, f(x) = [2] = 2 -------- (1)
Left side limit (x ---> 2-h): f(x) = [2 - h] = 1 ----- (2)
{Since (2 - h) lies between 1 & 2; and the least being 1}
Right side limit (x --> 2+h): f(x) = [2 + h] = 2 -------- (3)
{Since (2+h) lies between 2 & 3; and the least being 2}
iii) Thus from the above 3 equations, left side limit is not equal to right side limit.
So limit of the function does not exist.
Hence it is discontinuous at x = 2
So this is not derivable at x = 2
Hence Proved.
calculus functions continuity
asked Oct 3 '17 at 12:25
Jatt-Soorma Jatt-Soorma
1112
$endgroup$
I was solving this function , now the question that arises is that I was solving this using an example i.e. A numerical value, but my teacher keeps saying that it's wrong or I have to solve it using constants such as k... Etc. Is this method wrong according to u?
i) f(x) = [x], for all x in R
==> By the definition of greatest integer function: If x lies between two successive integers, then f(x) = least integer of them.
ii) So, at x = 2, f(x) = [2] = 2 -------- (1)
Left side limit (x ---> 2-h): f(x) = [2 - h] = 1 ----- (2)
{Since (2 - h) lies between 1 & 2; and the least being 1}
Right side limit (x --> 2+h): f(x) = [2 + h] = 2 -------- (3)
{Since (2+h) lies between 2 & 3; and the least being 2}
iii) Thus from the above 3 equations, left side limit is not equal to right side limit.
So limit of the function does not exist.
Hence it is discontinuous at x = 2
So this is not derivable at x = 2
Hence Proved.
calculus functions continuity
calculus functions continuity
asked Oct 3 '17 at 12:25
Jatt-Soorma Jatt-Soorma
1112
asked Oct 3 '17 at 12:25
Jatt-Soorma Jatt-Soorma
1112
asked Oct 3 '17 at 12:25
Jatt-Soorma Jatt-Soorma
1112
asked Oct 3 '17 at 12:25
Jatt-Soorma Jatt-Soorma
1112
asked Oct 3 '17 at 12:25
Jatt-Soorma Jatt-Soorma
1112
1112
$begingroup$
Your answer presentation is not correct. Because in these type of question answer should not be explained with help of an example but it should be explained mathematically...
$endgroup$
– Akshit Bansal
Jun 3 '18 at 4:42
add a comment |
$begingroup$
Your answer presentation is not correct. Because in these type of question answer should not be explained with help of an example but it should be explained mathematically...
$endgroup$
– Akshit Bansal
Jun 3 '18 at 4:42
$begingroup$
Your answer presentation is not correct. Because in these type of question answer should not be explained with help of an example but it should be explained mathematically...
$endgroup$
– Akshit Bansal
Jun 3 '18 at 4:42
$begingroup$
Your answer presentation is not correct. Because in these type of question answer should not be explained with help of an example but it should be explained mathematically...
$endgroup$
– Akshit Bansal
Jun 3 '18 at 4:42
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your argument is correct and valid. However, you still have a way to go, since...
You proved:
The function is discontinuous at $2$.
You have to prove:
- The function is discontinuous at all integer points.
- The function is continous at points that are not integers
answered Oct 3 '17 at 12:30
5xum5xum
91.4k394161
$endgroup$
$begingroup$
Is there any easy way to solve this function?
$endgroup$
– Jatt-Soorma
Oct 3 '17 at 12:34
$begingroup$
@Jatt-Soorma Depends on what you think is easy. To a graduate student of math, the statement you are trying to prove is probably obvious. To you, it is not, so you need to put some work into it. The proof that the function is discontinuous at all integer points is similar to the one you arleady made, for example.
$endgroup$
– 5xum
Oct 3 '17 at 12:36
$begingroup$
@Jatt-Soorma, Instead of only showing that it's discontinuous at $x=2$, adapt your argument for $x=k$ where $k$ is any integer. Furthermore, show that it is continuous in $[k+epsilon, k+1)$ where $kinBbb Z$ and $0lt epsilonlt 1$
$endgroup$
– Prasun Biswas
Oct 3 '17 at 12:36
$begingroup$
I proceded as Prasun suggested me and I arrived at this point, is it correct and complete now? First part - drive.google.com/file/d/0B8vdBgWGYplSSnl6aWxYY2Fjb3c/…. And second part - drive.google.com/file/d/0B8vdBgWGYplScW9iNS1HX3hxY1k/…
$endgroup$
– Jatt-Soorma
Oct 3 '17 at 13:18
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
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');
});
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%2f2455715%2fprove-that-the-greatest-integer-function-x-is-continuous-at-all-points-except%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your argument is correct and valid. However, you still have a way to go, since...
You proved:
The function is discontinuous at $2$.
You have to prove:
- The function is discontinuous at all integer points.
- The function is continous at points that are not integers
answered Oct 3 '17 at 12:30
5xum5xum
91.4k394161
$endgroup$
$begingroup$
Is there any easy way to solve this function?
$endgroup$
– Jatt-Soorma
Oct 3 '17 at 12:34
$begingroup$
@Jatt-Soorma Depends on what you think is easy. To a graduate student of math, the statement you are trying to prove is probably obvious. To you, it is not, so you need to put some work into it. The proof that the function is discontinuous at all integer points is similar to the one you arleady made, for example.
$endgroup$
– 5xum
Oct 3 '17 at 12:36
$begingroup$
@Jatt-Soorma, Instead of only showing that it's discontinuous at $x=2$, adapt your argument for $x=k$ where $k$ is any integer. Furthermore, show that it is continuous in $[k+epsilon, k+1)$ where $kinBbb Z$ and $0lt epsilonlt 1$
$endgroup$
– Prasun Biswas
Oct 3 '17 at 12:36
$begingroup$
I proceded as Prasun suggested me and I arrived at this point, is it correct and complete now? First part - drive.google.com/file/d/0B8vdBgWGYplSSnl6aWxYY2Fjb3c/…. And second part - drive.google.com/file/d/0B8vdBgWGYplScW9iNS1HX3hxY1k/…
$endgroup$
– Jatt-Soorma
Oct 3 '17 at 13:18
add a comment |
$begingroup$
Your argument is correct and valid. However, you still have a way to go, since...
You proved:
The function is discontinuous at $2$.
You have to prove:
- The function is discontinuous at all integer points.
- The function is continous at points that are not integers
answered Oct 3 '17 at 12:30
5xum5xum
91.4k394161
$endgroup$
$begingroup$
Is there any easy way to solve this function?
$endgroup$
– Jatt-Soorma
Oct 3 '17 at 12:34
$begingroup$
@Jatt-Soorma Depends on what you think is easy. To a graduate student of math, the statement you are trying to prove is probably obvious. To you, it is not, so you need to put some work into it. The proof that the function is discontinuous at all integer points is similar to the one you arleady made, for example.
$endgroup$
– 5xum
Oct 3 '17 at 12:36
$begingroup$
@Jatt-Soorma, Instead of only showing that it's discontinuous at $x=2$, adapt your argument for $x=k$ where $k$ is any integer. Furthermore, show that it is continuous in $[k+epsilon, k+1)$ where $kinBbb Z$ and $0lt epsilonlt 1$
$endgroup$
– Prasun Biswas
Oct 3 '17 at 12:36
$begingroup$
I proceded as Prasun suggested me and I arrived at this point, is it correct and complete now? First part - drive.google.com/file/d/0B8vdBgWGYplSSnl6aWxYY2Fjb3c/…. And second part - drive.google.com/file/d/0B8vdBgWGYplScW9iNS1HX3hxY1k/…
$endgroup$
– Jatt-Soorma
Oct 3 '17 at 13:18
add a comment |
$begingroup$
Your argument is correct and valid. However, you still have a way to go, since...
You proved:
The function is discontinuous at $2$.
You have to prove:
- The function is discontinuous at all integer points.
- The function is continous at points that are not integers
answered Oct 3 '17 at 12:30
5xum5xum
91.4k394161
$endgroup$
Your argument is correct and valid. However, you still have a way to go, since...
You proved:
The function is discontinuous at $2$.
You have to prove:
- The function is discontinuous at all integer points.
- The function is continous at points that are not integers
answered Oct 3 '17 at 12:30
5xum5xum
91.4k394161
answered Oct 3 '17 at 12:30
5xum5xum
91.4k394161
answered Oct 3 '17 at 12:30
5xum5xum
91.4k394161
answered Oct 3 '17 at 12:30
5xum5xum
91.4k394161
91.4k394161
$begingroup$
Is there any easy way to solve this function?
$endgroup$
– Jatt-Soorma
Oct 3 '17 at 12:34
$begingroup$
@Jatt-Soorma Depends on what you think is easy. To a graduate student of math, the statement you are trying to prove is probably obvious. To you, it is not, so you need to put some work into it. The proof that the function is discontinuous at all integer points is similar to the one you arleady made, for example.
$endgroup$
– 5xum
Oct 3 '17 at 12:36
$begingroup$
@Jatt-Soorma, Instead of only showing that it's discontinuous at $x=2$, adapt your argument for $x=k$ where $k$ is any integer. Furthermore, show that it is continuous in $[k+epsilon, k+1)$ where $kinBbb Z$ and $0lt epsilonlt 1$
$endgroup$
– Prasun Biswas
Oct 3 '17 at 12:36
$begingroup$
I proceded as Prasun suggested me and I arrived at this point, is it correct and complete now? First part - drive.google.com/file/d/0B8vdBgWGYplSSnl6aWxYY2Fjb3c/…. And second part - drive.google.com/file/d/0B8vdBgWGYplScW9iNS1HX3hxY1k/…
$endgroup$
– Jatt-Soorma
Oct 3 '17 at 13:18
add a comment |
$begingroup$
Is there any easy way to solve this function?
$endgroup$
– Jatt-Soorma
Oct 3 '17 at 12:34
$begingroup$
@Jatt-Soorma Depends on what you think is easy. To a graduate student of math, the statement you are trying to prove is probably obvious. To you, it is not, so you need to put some work into it. The proof that the function is discontinuous at all integer points is similar to the one you arleady made, for example.
$endgroup$
– 5xum
Oct 3 '17 at 12:36
$begingroup$
@Jatt-Soorma, Instead of only showing that it's discontinuous at $x=2$, adapt your argument for $x=k$ where $k$ is any integer. Furthermore, show that it is continuous in $[k+epsilon, k+1)$ where $kinBbb Z$ and $0lt epsilonlt 1$
$endgroup$
– Prasun Biswas
Oct 3 '17 at 12:36
$begingroup$
I proceded as Prasun suggested me and I arrived at this point, is it correct and complete now? First part - drive.google.com/file/d/0B8vdBgWGYplSSnl6aWxYY2Fjb3c/…. And second part - drive.google.com/file/d/0B8vdBgWGYplScW9iNS1HX3hxY1k/…
$endgroup$
– Jatt-Soorma
Oct 3 '17 at 13:18
$begingroup$
Is there any easy way to solve this function?
$endgroup$
– Jatt-Soorma
Oct 3 '17 at 12:34
$begingroup$
Is there any easy way to solve this function?
$endgroup$
– Jatt-Soorma
Oct 3 '17 at 12:34
$begingroup$
@Jatt-Soorma Depends on what you think is easy. To a graduate student of math, the statement you are trying to prove is probably obvious. To you, it is not, so you need to put some work into it. The proof that the function is discontinuous at all integer points is similar to the one you arleady made, for example.
$endgroup$
– 5xum
Oct 3 '17 at 12:36
$begingroup$
@Jatt-Soorma Depends on what you think is easy. To a graduate student of math, the statement you are trying to prove is probably obvious. To you, it is not, so you need to put some work into it. The proof that the function is discontinuous at all integer points is similar to the one you arleady made, for example.
$endgroup$
– 5xum
Oct 3 '17 at 12:36
$begingroup$
@Jatt-Soorma, Instead of only showing that it's discontinuous at $x=2$, adapt your argument for $x=k$ where $k$ is any integer. Furthermore, show that it is continuous in $[k+epsilon, k+1)$ where $kinBbb Z$ and $0lt epsilonlt 1$
$endgroup$
– Prasun Biswas
Oct 3 '17 at 12:36
$begingroup$
@Jatt-Soorma, Instead of only showing that it's discontinuous at $x=2$, adapt your argument for $x=k$ where $k$ is any integer. Furthermore, show that it is continuous in $[k+epsilon, k+1)$ where $kinBbb Z$ and $0lt epsilonlt 1$
$endgroup$
– Prasun Biswas
Oct 3 '17 at 12:36
$begingroup$
I proceded as Prasun suggested me and I arrived at this point, is it correct and complete now? First part - drive.google.com/file/d/0B8vdBgWGYplSSnl6aWxYY2Fjb3c/…. And second part - drive.google.com/file/d/0B8vdBgWGYplScW9iNS1HX3hxY1k/…
$endgroup$
– Jatt-Soorma
Oct 3 '17 at 13:18
$begingroup$
I proceded as Prasun suggested me and I arrived at this point, is it correct and complete now? First part - drive.google.com/file/d/0B8vdBgWGYplSSnl6aWxYY2Fjb3c/…. And second part - drive.google.com/file/d/0B8vdBgWGYplScW9iNS1HX3hxY1k/…
$endgroup$
– Jatt-Soorma
Oct 3 '17 at 13:18
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');
});
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%2f2455715%2fprove-that-the-greatest-integer-function-x-is-continuous-at-all-points-except%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');
});
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');
});
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');
});
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
$begingroup$
Your answer presentation is not correct. Because in these type of question answer should not be explained with help of an example but it should be explained mathematically...
$endgroup$
– Akshit Bansal
Jun 3 '18 at 4:42