Prove that $forall x in Bbb R^+:x+0=xcdot 1=x$
Let $Bbb R^+={xinBbb R mid x>0}$ and $x,yinBbb R^+$. We define multiplication operation $(cdot)$ on $Bbb R^+$ and addition operation $(+)$ on $Bbb R$ by $$xcdot y:=inf{rcdot smid r,sinBbb Q text{ and } x<r text{ and } y<s}$$ $$x+ y:=inf{r+ smid r,sinBbb Q text{ and } x<r text{ and } y<s}$$
Prove that $$forall x in Bbb R^+:x+0=xcdot 1=x$$
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
$x+0=x$.
By definition, we have:
$x=inf {r mid rinBbb Q,x<r}=inf A$
$x+0=inf {r+s mid r,sinBbb Q,x<r,0<s}=inf B$
$bin B implies b=r+s$ for some $r,sinBbb Q$ such that $x<r,0<s$ $implies b>r$ for some $rinBbb Q$ such that $x<r$ $implies b>r$ for some $rin A$ $implies inf A le inf B implies x le x+0$.
Assume the contrary that $x<x+0$. By the fact that $Bbb Q$ is dense in $Bbb R$, there exist $pin Bbb Q$ such that $x<p<x+0$ and $rinBbb Q$ such that $x <r<p$. Let $p=r+s>r$. Then $sinBbb Q$ and $s>0$. It follows that $x+0>p$ where $p=r+s$ such that $r,sinBbb Q,x<r,0<s$. This is a contradiction.
- $xcdot 1=x$
By definition, we have:
$x=inf {r mid rinBbb Q,x<r}=inf A$
$xcdot 1=inf {rcdot s mid r,sinBbb Q,x<r,1<s}=inf B$
$bin B implies b=rcdot s$ for some $r,sinBbb Q$ such that $x<r,1<s$ $implies b>r$ for some $rinBbb Q$ such that $x<r$ $implies b>r$ for some $rin A$ $implies inf A le inf B implies x le xcdot 1$.
Assume the contrary that $x<xcdot 1$. By the fact that $Bbb Q$ is dense in $Bbb R$, there exist $pin Bbb Q$ such that $x<p<xcdot 1$ and $rinBbb Q$ such that $x <r<p$. Let $p=rcdot s>r$. Then $sinBbb Q$ and $s>1$. It follows that $xcdot 1>p$ where $p=rcdot s$ such that $r,sinBbb Q,x<r,1<s$. This is a contradiction.
proof-verification real-numbers
asked Jan 4 at 4:17
Le Anh DungLe Anh Dung
1,0251521
|
show 2 more comments
Let $Bbb R^+={xinBbb R mid x>0}$ and $x,yinBbb R^+$. We define multiplication operation $(cdot)$ on $Bbb R^+$ and addition operation $(+)$ on $Bbb R$ by $$xcdot y:=inf{rcdot smid r,sinBbb Q text{ and } x<r text{ and } y<s}$$ $$x+ y:=inf{r+ smid r,sinBbb Q text{ and } x<r text{ and } y<s}$$
Prove that $$forall x in Bbb R^+:x+0=xcdot 1=x$$
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
$x+0=x$.
By definition, we have:
$x=inf {r mid rinBbb Q,x<r}=inf A$
$x+0=inf {r+s mid r,sinBbb Q,x<r,0<s}=inf B$
$bin B implies b=r+s$ for some $r,sinBbb Q$ such that $x<r,0<s$ $implies b>r$ for some $rinBbb Q$ such that $x<r$ $implies b>r$ for some $rin A$ $implies inf A le inf B implies x le x+0$.
Assume the contrary that $x<x+0$. By the fact that $Bbb Q$ is dense in $Bbb R$, there exist $pin Bbb Q$ such that $x<p<x+0$ and $rinBbb Q$ such that $x <r<p$. Let $p=r+s>r$. Then $sinBbb Q$ and $s>0$. It follows that $x+0>p$ where $p=r+s$ such that $r,sinBbb Q,x<r,0<s$. This is a contradiction.
- $xcdot 1=x$
By definition, we have:
$x=inf {r mid rinBbb Q,x<r}=inf A$
$xcdot 1=inf {rcdot s mid r,sinBbb Q,x<r,1<s}=inf B$
$bin B implies b=rcdot s$ for some $r,sinBbb Q$ such that $x<r,1<s$ $implies b>r$ for some $rinBbb Q$ such that $x<r$ $implies b>r$ for some $rin A$ $implies inf A le inf B implies x le xcdot 1$.
Assume the contrary that $x<xcdot 1$. By the fact that $Bbb Q$ is dense in $Bbb R$, there exist $pin Bbb Q$ such that $x<p<xcdot 1$ and $rinBbb Q$ such that $x <r<p$. Let $p=rcdot s>r$. Then $sinBbb Q$ and $s>1$. It follows that $xcdot 1>p$ where $p=rcdot s$ such that $r,sinBbb Q,x<r,1<s$. This is a contradiction.
proof-verification real-numbers
asked Jan 4 at 4:17
Le Anh DungLe Anh Dung
1,0251521
Here is an edit in one of your paragraphs: You write: "$b in B implies b=r+s$ for some $r,sinmathbb{Q}$ such that $x<r,0<s implies b>r$ for some $rin mathbb{Q}$ such that $x<r$ [and so $b > x= inf A$ so $inf A leq inf B$]
– Michael
Jan 4 at 5:16
Hi @Michael, I don't understand what you meant. Is the reasoning in that paragraph wrong? Could you please be more specific?
– Le Anh Dung
Jan 4 at 5:20
1
You concluded $b>r$ and $r>x$, so therefore we directly get $b>x$. That is what you want since $x$ is already defined as $inf A$. In fact the sentence would be easier to read if it was just "$b in B$ implies $b = r+s$ for some $r, s in mathbb{Q}$ such that $x<r, 0<s$. So $b=r+s>r>x = inf A$." (and since $b > inf A$ for all $b in B$ it holds that $inf B geq inf A$.) It would also help if you explicitly defined the sets $A$ and $B$, though I know what you meant.
– Michael
Jan 4 at 14:02
Hi @Michael, Thank you so much for your detailed comment! Now I got your point.
– Le Anh Dung
Jan 4 at 14:09
1
Maybe I'll do that tomorrow, or perhaps someone will post a better answer by then.
– Michael
Jan 4 at 14:34
|
show 2 more comments
Let $Bbb R^+={xinBbb R mid x>0}$ and $x,yinBbb R^+$. We define multiplication operation $(cdot)$ on $Bbb R^+$ and addition operation $(+)$ on $Bbb R$ by $$xcdot y:=inf{rcdot smid r,sinBbb Q text{ and } x<r text{ and } y<s}$$ $$x+ y:=inf{r+ smid r,sinBbb Q text{ and } x<r text{ and } y<s}$$
Prove that $$forall x in Bbb R^+:x+0=xcdot 1=x$$
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
$x+0=x$.
By definition, we have:
$x=inf {r mid rinBbb Q,x<r}=inf A$
$x+0=inf {r+s mid r,sinBbb Q,x<r,0<s}=inf B$
$bin B implies b=r+s$ for some $r,sinBbb Q$ such that $x<r,0<s$ $implies b>r$ for some $rinBbb Q$ such that $x<r$ $implies b>r$ for some $rin A$ $implies inf A le inf B implies x le x+0$.
Assume the contrary that $x<x+0$. By the fact that $Bbb Q$ is dense in $Bbb R$, there exist $pin Bbb Q$ such that $x<p<x+0$ and $rinBbb Q$ such that $x <r<p$. Let $p=r+s>r$. Then $sinBbb Q$ and $s>0$. It follows that $x+0>p$ where $p=r+s$ such that $r,sinBbb Q,x<r,0<s$. This is a contradiction.
- $xcdot 1=x$
By definition, we have:
$x=inf {r mid rinBbb Q,x<r}=inf A$
$xcdot 1=inf {rcdot s mid r,sinBbb Q,x<r,1<s}=inf B$
$bin B implies b=rcdot s$ for some $r,sinBbb Q$ such that $x<r,1<s$ $implies b>r$ for some $rinBbb Q$ such that $x<r$ $implies b>r$ for some $rin A$ $implies inf A le inf B implies x le xcdot 1$.
Assume the contrary that $x<xcdot 1$. By the fact that $Bbb Q$ is dense in $Bbb R$, there exist $pin Bbb Q$ such that $x<p<xcdot 1$ and $rinBbb Q$ such that $x <r<p$. Let $p=rcdot s>r$. Then $sinBbb Q$ and $s>1$. It follows that $xcdot 1>p$ where $p=rcdot s$ such that $r,sinBbb Q,x<r,1<s$. This is a contradiction.
proof-verification real-numbers
asked Jan 4 at 4:17
Le Anh DungLe Anh Dung
1,0251521
Let $Bbb R^+={xinBbb R mid x>0}$ and $x,yinBbb R^+$. We define multiplication operation $(cdot)$ on $Bbb R^+$ and addition operation $(+)$ on $Bbb R$ by $$xcdot y:=inf{rcdot smid r,sinBbb Q text{ and } x<r text{ and } y<s}$$ $$x+ y:=inf{r+ smid r,sinBbb Q text{ and } x<r text{ and } y<s}$$
Prove that $$forall x in Bbb R^+:x+0=xcdot 1=x$$
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
$x+0=x$.
By definition, we have:
$x=inf {r mid rinBbb Q,x<r}=inf A$
$x+0=inf {r+s mid r,sinBbb Q,x<r,0<s}=inf B$
$bin B implies b=r+s$ for some $r,sinBbb Q$ such that $x<r,0<s$ $implies b>r$ for some $rinBbb Q$ such that $x<r$ $implies b>r$ for some $rin A$ $implies inf A le inf B implies x le x+0$.
Assume the contrary that $x<x+0$. By the fact that $Bbb Q$ is dense in $Bbb R$, there exist $pin Bbb Q$ such that $x<p<x+0$ and $rinBbb Q$ such that $x <r<p$. Let $p=r+s>r$. Then $sinBbb Q$ and $s>0$. It follows that $x+0>p$ where $p=r+s$ such that $r,sinBbb Q,x<r,0<s$. This is a contradiction.
- $xcdot 1=x$
By definition, we have:
$x=inf {r mid rinBbb Q,x<r}=inf A$
$xcdot 1=inf {rcdot s mid r,sinBbb Q,x<r,1<s}=inf B$
$bin B implies b=rcdot s$ for some $r,sinBbb Q$ such that $x<r,1<s$ $implies b>r$ for some $rinBbb Q$ such that $x<r$ $implies b>r$ for some $rin A$ $implies inf A le inf B implies x le xcdot 1$.
Assume the contrary that $x<xcdot 1$. By the fact that $Bbb Q$ is dense in $Bbb R$, there exist $pin Bbb Q$ such that $x<p<xcdot 1$ and $rinBbb Q$ such that $x <r<p$. Let $p=rcdot s>r$. Then $sinBbb Q$ and $s>1$. It follows that $xcdot 1>p$ where $p=rcdot s$ such that $r,sinBbb Q,x<r,1<s$. This is a contradiction.
proof-verification real-numbers
proof-verification real-numbers
asked Jan 4 at 4:17
Le Anh DungLe Anh Dung
1,0251521
asked Jan 4 at 4:17
Le Anh DungLe Anh Dung
1,0251521
edited Jan 4 at 4:29
Le Anh Dung
asked Jan 4 at 4:17
Le Anh DungLe Anh Dung
1,0251521
asked Jan 4 at 4:17
Le Anh DungLe Anh Dung
1,0251521
asked Jan 4 at 4:17
Le Anh DungLe Anh Dung
1,0251521
1,0251521
Here is an edit in one of your paragraphs: You write: "$b in B implies b=r+s$ for some $r,sinmathbb{Q}$ such that $x<r,0<s implies b>r$ for some $rin mathbb{Q}$ such that $x<r$ [and so $b > x= inf A$ so $inf A leq inf B$]
– Michael
Jan 4 at 5:16
Hi @Michael, I don't understand what you meant. Is the reasoning in that paragraph wrong? Could you please be more specific?
– Le Anh Dung
Jan 4 at 5:20
1
You concluded $b>r$ and $r>x$, so therefore we directly get $b>x$. That is what you want since $x$ is already defined as $inf A$. In fact the sentence would be easier to read if it was just "$b in B$ implies $b = r+s$ for some $r, s in mathbb{Q}$ such that $x<r, 0<s$. So $b=r+s>r>x = inf A$." (and since $b > inf A$ for all $b in B$ it holds that $inf B geq inf A$.) It would also help if you explicitly defined the sets $A$ and $B$, though I know what you meant.
– Michael
Jan 4 at 14:02
Hi @Michael, Thank you so much for your detailed comment! Now I got your point.
– Le Anh Dung
Jan 4 at 14:09
1
Maybe I'll do that tomorrow, or perhaps someone will post a better answer by then.
– Michael
Jan 4 at 14:34
|
show 2 more comments
Here is an edit in one of your paragraphs: You write: "$b in B implies b=r+s$ for some $r,sinmathbb{Q}$ such that $x<r,0<s implies b>r$ for some $rin mathbb{Q}$ such that $x<r$ [and so $b > x= inf A$ so $inf A leq inf B$]
– Michael
Jan 4 at 5:16
Hi @Michael, I don't understand what you meant. Is the reasoning in that paragraph wrong? Could you please be more specific?
– Le Anh Dung
Jan 4 at 5:20
1
You concluded $b>r$ and $r>x$, so therefore we directly get $b>x$. That is what you want since $x$ is already defined as $inf A$. In fact the sentence would be easier to read if it was just "$b in B$ implies $b = r+s$ for some $r, s in mathbb{Q}$ such that $x<r, 0<s$. So $b=r+s>r>x = inf A$." (and since $b > inf A$ for all $b in B$ it holds that $inf B geq inf A$.) It would also help if you explicitly defined the sets $A$ and $B$, though I know what you meant.
– Michael
Jan 4 at 14:02
Hi @Michael, Thank you so much for your detailed comment! Now I got your point.
– Le Anh Dung
Jan 4 at 14:09
1
Maybe I'll do that tomorrow, or perhaps someone will post a better answer by then.
– Michael
Jan 4 at 14:34
Here is an edit in one of your paragraphs: You write: "$b in B implies b=r+s$ for some $r,sinmathbb{Q}$ such that $x<r,0<s implies b>r$ for some $rin mathbb{Q}$ such that $x<r$ [and so $b > x= inf A$ so $inf A leq inf B$]
– Michael
Jan 4 at 5:16
Here is an edit in one of your paragraphs: You write: "$b in B implies b=r+s$ for some $r,sinmathbb{Q}$ such that $x<r,0<s implies b>r$ for some $rin mathbb{Q}$ such that $x<r$ [and so $b > x= inf A$ so $inf A leq inf B$]
– Michael
Jan 4 at 5:16
Hi @Michael, I don't understand what you meant. Is the reasoning in that paragraph wrong? Could you please be more specific?
– Le Anh Dung
Jan 4 at 5:20
Hi @Michael, I don't understand what you meant. Is the reasoning in that paragraph wrong? Could you please be more specific?
– Le Anh Dung
Jan 4 at 5:20
1
1
You concluded $b>r$ and $r>x$, so therefore we directly get $b>x$. That is what you want since $x$ is already defined as $inf A$. In fact the sentence would be easier to read if it was just "$b in B$ implies $b = r+s$ for some $r, s in mathbb{Q}$ such that $x<r, 0<s$. So $b=r+s>r>x = inf A$." (and since $b > inf A$ for all $b in B$ it holds that $inf B geq inf A$.) It would also help if you explicitly defined the sets $A$ and $B$, though I know what you meant.
– Michael
Jan 4 at 14:02
You concluded $b>r$ and $r>x$, so therefore we directly get $b>x$. That is what you want since $x$ is already defined as $inf A$. In fact the sentence would be easier to read if it was just "$b in B$ implies $b = r+s$ for some $r, s in mathbb{Q}$ such that $x<r, 0<s$. So $b=r+s>r>x = inf A$." (and since $b > inf A$ for all $b in B$ it holds that $inf B geq inf A$.) It would also help if you explicitly defined the sets $A$ and $B$, though I know what you meant.
– Michael
Jan 4 at 14:02
Hi @Michael, Thank you so much for your detailed comment! Now I got your point.
– Le Anh Dung
Jan 4 at 14:09
Hi @Michael, Thank you so much for your detailed comment! Now I got your point.
– Le Anh Dung
Jan 4 at 14:09
1
1
Maybe I'll do that tomorrow, or perhaps someone will post a better answer by then.
– Michael
Jan 4 at 14:34
Maybe I'll do that tomorrow, or perhaps someone will post a better answer by then.
– Michael
Jan 4 at 14:34
|
show 2 more comments
0
active
oldest
votes
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%2f3061318%2fprove-that-forall-x-in-bbb-rx0-x-cdot-1-x%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3061318%2fprove-that-forall-x-in-bbb-rx0-x-cdot-1-x%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
Here is an edit in one of your paragraphs: You write: "$b in B implies b=r+s$ for some $r,sinmathbb{Q}$ such that $x<r,0<s implies b>r$ for some $rin mathbb{Q}$ such that $x<r$ [and so $b > x= inf A$ so $inf A leq inf B$]
– Michael
Jan 4 at 5:16
Hi @Michael, I don't understand what you meant. Is the reasoning in that paragraph wrong? Could you please be more specific?
– Le Anh Dung
Jan 4 at 5:20
1
You concluded $b>r$ and $r>x$, so therefore we directly get $b>x$. That is what you want since $x$ is already defined as $inf A$. In fact the sentence would be easier to read if it was just "$b in B$ implies $b = r+s$ for some $r, s in mathbb{Q}$ such that $x<r, 0<s$. So $b=r+s>r>x = inf A$." (and since $b > inf A$ for all $b in B$ it holds that $inf B geq inf A$.) It would also help if you explicitly defined the sets $A$ and $B$, though I know what you meant.
– Michael
Jan 4 at 14:02
Hi @Michael, Thank you so much for your detailed comment! Now I got your point.
– Le Anh Dung
Jan 4 at 14:09
1
Maybe I'll do that tomorrow, or perhaps someone will post a better answer by then.
– Michael
Jan 4 at 14:34