How is it that there are 'gaps' in rational numbers and yet between any two rational numbers, there exists...
$begingroup$
If there are gaps in rational numbers then lets assume we have a gap between a and b, both being rational. Then we have $frac{a+b}{2}$ which is inside the gap which essentially makes it a non-gap. What am I getting wrong?
real-analysis rational-numbers
asked Jan 7 at 11:08
Nishant Pr. DasNishant Pr. Das
234
$endgroup$
add a comment |
$begingroup$
If there are gaps in rational numbers then lets assume we have a gap between a and b, both being rational. Then we have $frac{a+b}{2}$ which is inside the gap which essentially makes it a non-gap. What am I getting wrong?
real-analysis rational-numbers
asked Jan 7 at 11:08
Nishant Pr. DasNishant Pr. Das
234
$endgroup$
1
$begingroup$
gaps are because of the existence of irrational numbers!
$endgroup$
– OmG
Jan 7 at 11:10
1
$begingroup$
How do you define a "gap"?
$endgroup$
– 5xum
Jan 7 at 11:20
$begingroup$
The rationals and its complement are both dense in the reals. That's all. The word gap is ambiguous here, e.g. If you remove one point from a line, is that a "gap"?
$endgroup$
– Ned
Jan 7 at 11:21
$begingroup$
You have shown that there is at least one real number between $a$ and $b$ that is rational. But to "fill" the gap you would have to show that every real number between $a$ and $b$ is rational.
$endgroup$
– gandalf61
Jan 7 at 11:21
$begingroup$
This is because there is no well-defined notion of the “next” rational number after each rational number. (or at least the usual ordering of the rationals doesn’t allow a well-defined next rational).
$endgroup$
– Adam Higgins
Jan 7 at 11:23
add a comment |
$begingroup$
If there are gaps in rational numbers then lets assume we have a gap between a and b, both being rational. Then we have $frac{a+b}{2}$ which is inside the gap which essentially makes it a non-gap. What am I getting wrong?
real-analysis rational-numbers
asked Jan 7 at 11:08
Nishant Pr. DasNishant Pr. Das
234
$endgroup$
If there are gaps in rational numbers then lets assume we have a gap between a and b, both being rational. Then we have $frac{a+b}{2}$ which is inside the gap which essentially makes it a non-gap. What am I getting wrong?
real-analysis rational-numbers
real-analysis rational-numbers
asked Jan 7 at 11:08
Nishant Pr. DasNishant Pr. Das
234
asked Jan 7 at 11:08
Nishant Pr. DasNishant Pr. Das
234
asked Jan 7 at 11:08
Nishant Pr. DasNishant Pr. Das
234
asked Jan 7 at 11:08
Nishant Pr. DasNishant Pr. Das
234
asked Jan 7 at 11:08
Nishant Pr. DasNishant Pr. Das
234
234
1
$begingroup$
gaps are because of the existence of irrational numbers!
$endgroup$
– OmG
Jan 7 at 11:10
1
$begingroup$
How do you define a "gap"?
$endgroup$
– 5xum
Jan 7 at 11:20
$begingroup$
The rationals and its complement are both dense in the reals. That's all. The word gap is ambiguous here, e.g. If you remove one point from a line, is that a "gap"?
$endgroup$
– Ned
Jan 7 at 11:21
$begingroup$
You have shown that there is at least one real number between $a$ and $b$ that is rational. But to "fill" the gap you would have to show that every real number between $a$ and $b$ is rational.
$endgroup$
– gandalf61
Jan 7 at 11:21
$begingroup$
This is because there is no well-defined notion of the “next” rational number after each rational number. (or at least the usual ordering of the rationals doesn’t allow a well-defined next rational).
$endgroup$
– Adam Higgins
Jan 7 at 11:23
add a comment |
1
$begingroup$
gaps are because of the existence of irrational numbers!
$endgroup$
– OmG
Jan 7 at 11:10
1
$begingroup$
How do you define a "gap"?
$endgroup$
– 5xum
Jan 7 at 11:20
$begingroup$
The rationals and its complement are both dense in the reals. That's all. The word gap is ambiguous here, e.g. If you remove one point from a line, is that a "gap"?
$endgroup$
– Ned
Jan 7 at 11:21
$begingroup$
You have shown that there is at least one real number between $a$ and $b$ that is rational. But to "fill" the gap you would have to show that every real number between $a$ and $b$ is rational.
$endgroup$
– gandalf61
Jan 7 at 11:21
$begingroup$
This is because there is no well-defined notion of the “next” rational number after each rational number. (or at least the usual ordering of the rationals doesn’t allow a well-defined next rational).
$endgroup$
– Adam Higgins
Jan 7 at 11:23
1
1
$begingroup$
gaps are because of the existence of irrational numbers!
$endgroup$
– OmG
Jan 7 at 11:10
$begingroup$
gaps are because of the existence of irrational numbers!
$endgroup$
– OmG
Jan 7 at 11:10
1
1
$begingroup$
How do you define a "gap"?
$endgroup$
– 5xum
Jan 7 at 11:20
$begingroup$
How do you define a "gap"?
$endgroup$
– 5xum
Jan 7 at 11:20
$begingroup$
The rationals and its complement are both dense in the reals. That's all. The word gap is ambiguous here, e.g. If you remove one point from a line, is that a "gap"?
$endgroup$
– Ned
Jan 7 at 11:21
$begingroup$
The rationals and its complement are both dense in the reals. That's all. The word gap is ambiguous here, e.g. If you remove one point from a line, is that a "gap"?
$endgroup$
– Ned
Jan 7 at 11:21
$begingroup$
You have shown that there is at least one real number between $a$ and $b$ that is rational. But to "fill" the gap you would have to show that every real number between $a$ and $b$ is rational.
$endgroup$
– gandalf61
Jan 7 at 11:21
$begingroup$
You have shown that there is at least one real number between $a$ and $b$ that is rational. But to "fill" the gap you would have to show that every real number between $a$ and $b$ is rational.
$endgroup$
– gandalf61
Jan 7 at 11:21
$begingroup$
This is because there is no well-defined notion of the “next” rational number after each rational number. (or at least the usual ordering of the rationals doesn’t allow a well-defined next rational).
$endgroup$
– Adam Higgins
Jan 7 at 11:23
$begingroup$
This is because there is no well-defined notion of the “next” rational number after each rational number. (or at least the usual ordering of the rationals doesn’t allow a well-defined next rational).
$endgroup$
– Adam Higgins
Jan 7 at 11:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A gap in $Bbb Q$ means there exist non-empty sets $A, B$ with $Bbb Q=Acup B,$ such that (i) every $ain A$ is less than every $bin B,$ and (ii) $A$ has no largest member and $B$ has no smallest member. It does NOT mean that there are rationals $x, y$ with $x<y$ such that no rational is between $x$ and $y$. No such $x,y$ exist but if they did, the pair $(x,y)$ would be called a jump.
Example: No $xin Bbb Q$ satisfies $x^2=2.$ Let $B={bin Bbb Q: 0<bland b^2>2}$ and let $A=Bbb Q setminus B.$ If $bin B$ then $b>frac {1}{2}(b+frac {2}{b})in B,$ so $B$ has no least member. This implies that ${ain A:a>0}={frac {2}{b}: bin B}$ has no largest member. So $A$ has no largest member.
answered Jan 7 at 11:43
DanielWainfleetDanielWainfleet
35.1k31648
$endgroup$
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%2f3064887%2fhow-is-it-that-there-are-gaps-in-rational-numbers-and-yet-between-any-two-rati%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
$begingroup$
A gap in $Bbb Q$ means there exist non-empty sets $A, B$ with $Bbb Q=Acup B,$ such that (i) every $ain A$ is less than every $bin B,$ and (ii) $A$ has no largest member and $B$ has no smallest member. It does NOT mean that there are rationals $x, y$ with $x<y$ such that no rational is between $x$ and $y$. No such $x,y$ exist but if they did, the pair $(x,y)$ would be called a jump.
Example: No $xin Bbb Q$ satisfies $x^2=2.$ Let $B={bin Bbb Q: 0<bland b^2>2}$ and let $A=Bbb Q setminus B.$ If $bin B$ then $b>frac {1}{2}(b+frac {2}{b})in B,$ so $B$ has no least member. This implies that ${ain A:a>0}={frac {2}{b}: bin B}$ has no largest member. So $A$ has no largest member.
answered Jan 7 at 11:43
DanielWainfleetDanielWainfleet
35.1k31648
$endgroup$
add a comment |
$begingroup$
A gap in $Bbb Q$ means there exist non-empty sets $A, B$ with $Bbb Q=Acup B,$ such that (i) every $ain A$ is less than every $bin B,$ and (ii) $A$ has no largest member and $B$ has no smallest member. It does NOT mean that there are rationals $x, y$ with $x<y$ such that no rational is between $x$ and $y$. No such $x,y$ exist but if they did, the pair $(x,y)$ would be called a jump.
Example: No $xin Bbb Q$ satisfies $x^2=2.$ Let $B={bin Bbb Q: 0<bland b^2>2}$ and let $A=Bbb Q setminus B.$ If $bin B$ then $b>frac {1}{2}(b+frac {2}{b})in B,$ so $B$ has no least member. This implies that ${ain A:a>0}={frac {2}{b}: bin B}$ has no largest member. So $A$ has no largest member.
answered Jan 7 at 11:43
DanielWainfleetDanielWainfleet
35.1k31648
$endgroup$
add a comment |
$begingroup$
A gap in $Bbb Q$ means there exist non-empty sets $A, B$ with $Bbb Q=Acup B,$ such that (i) every $ain A$ is less than every $bin B,$ and (ii) $A$ has no largest member and $B$ has no smallest member. It does NOT mean that there are rationals $x, y$ with $x<y$ such that no rational is between $x$ and $y$. No such $x,y$ exist but if they did, the pair $(x,y)$ would be called a jump.
Example: No $xin Bbb Q$ satisfies $x^2=2.$ Let $B={bin Bbb Q: 0<bland b^2>2}$ and let $A=Bbb Q setminus B.$ If $bin B$ then $b>frac {1}{2}(b+frac {2}{b})in B,$ so $B$ has no least member. This implies that ${ain A:a>0}={frac {2}{b}: bin B}$ has no largest member. So $A$ has no largest member.
answered Jan 7 at 11:43
DanielWainfleetDanielWainfleet
35.1k31648
$endgroup$
A gap in $Bbb Q$ means there exist non-empty sets $A, B$ with $Bbb Q=Acup B,$ such that (i) every $ain A$ is less than every $bin B,$ and (ii) $A$ has no largest member and $B$ has no smallest member. It does NOT mean that there are rationals $x, y$ with $x<y$ such that no rational is between $x$ and $y$. No such $x,y$ exist but if they did, the pair $(x,y)$ would be called a jump.
Example: No $xin Bbb Q$ satisfies $x^2=2.$ Let $B={bin Bbb Q: 0<bland b^2>2}$ and let $A=Bbb Q setminus B.$ If $bin B$ then $b>frac {1}{2}(b+frac {2}{b})in B,$ so $B$ has no least member. This implies that ${ain A:a>0}={frac {2}{b}: bin B}$ has no largest member. So $A$ has no largest member.
answered Jan 7 at 11:43
DanielWainfleetDanielWainfleet
35.1k31648
answered Jan 7 at 11:43
DanielWainfleetDanielWainfleet
35.1k31648
answered Jan 7 at 11:43
DanielWainfleetDanielWainfleet
35.1k31648
answered Jan 7 at 11:43
DanielWainfleetDanielWainfleet
35.1k31648
35.1k31648
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');
});
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%2f3064887%2fhow-is-it-that-there-are-gaps-in-rational-numbers-and-yet-between-any-two-rati%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
1
$begingroup$
gaps are because of the existence of irrational numbers!
$endgroup$
– OmG
Jan 7 at 11:10
1
$begingroup$
How do you define a "gap"?
$endgroup$
– 5xum
Jan 7 at 11:20
$begingroup$
The rationals and its complement are both dense in the reals. That's all. The word gap is ambiguous here, e.g. If you remove one point from a line, is that a "gap"?
$endgroup$
– Ned
Jan 7 at 11:21
$begingroup$
You have shown that there is at least one real number between $a$ and $b$ that is rational. But to "fill" the gap you would have to show that every real number between $a$ and $b$ is rational.
$endgroup$
– gandalf61
Jan 7 at 11:21
$begingroup$
This is because there is no well-defined notion of the “next” rational number after each rational number. (or at least the usual ordering of the rationals doesn’t allow a well-defined next rational).
$endgroup$
– Adam Higgins
Jan 7 at 11:23