Prove that $forall x in Bbb R^+:x+0=xcdot 1=x$
$begingroup$
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
1,0701521
$endgroup$
|
show 2 more comments
$begingroup$
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
1,0701521
$endgroup$
$begingroup$
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 $inf A leq inf B$]
$endgroup$
– Michael
Jan 4 at 5:16
$begingroup$
Hi @Michael, I don't understand what you meant. Is the reasoning in that paragraph wrong? Could you please be more specific?
$endgroup$
– Le Anh Dung
Jan 4 at 5:20
1
$begingroup$
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.
$endgroup$
– Michael
Jan 4 at 14:02
$begingroup$
Hi @Michael, Thank you so much for your detailed comment! Now I got your point.
$endgroup$
– Le Anh Dung
Jan 4 at 14:09
1
$begingroup$
Maybe I'll do that tomorrow, or perhaps someone will post a better answer by then.
$endgroup$
– Michael
Jan 4 at 14:34
|
show 2 more comments
$begingroup$
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
1,0701521
$endgroup$
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
1,0701521
asked Jan 4 at 4:17
Le Anh DungLe Anh Dung
1,0701521
edited Jan 4 at 4:29
Le Anh Dung
asked Jan 4 at 4:17
Le Anh DungLe Anh Dung
1,0701521
asked Jan 4 at 4:17
Le Anh DungLe Anh Dung
1,0701521
asked Jan 4 at 4:17
Le Anh DungLe Anh Dung
1,0701521
1,0701521
$begingroup$
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 $inf A leq inf B$]
$endgroup$
– Michael
Jan 4 at 5:16
$begingroup$
Hi @Michael, I don't understand what you meant. Is the reasoning in that paragraph wrong? Could you please be more specific?
$endgroup$
– Le Anh Dung
Jan 4 at 5:20
1
$begingroup$
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.
$endgroup$
– Michael
Jan 4 at 14:02
$begingroup$
Hi @Michael, Thank you so much for your detailed comment! Now I got your point.
$endgroup$
– Le Anh Dung
Jan 4 at 14:09
1
$begingroup$
Maybe I'll do that tomorrow, or perhaps someone will post a better answer by then.
$endgroup$
– Michael
Jan 4 at 14:34
|
show 2 more comments
$begingroup$
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 $inf A leq inf B$]
$endgroup$
– Michael
Jan 4 at 5:16
$begingroup$
Hi @Michael, I don't understand what you meant. Is the reasoning in that paragraph wrong? Could you please be more specific?
$endgroup$
– Le Anh Dung
Jan 4 at 5:20
1
$begingroup$
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.
$endgroup$
– Michael
Jan 4 at 14:02
$begingroup$
Hi @Michael, Thank you so much for your detailed comment! Now I got your point.
$endgroup$
– Le Anh Dung
Jan 4 at 14:09
1
$begingroup$
Maybe I'll do that tomorrow, or perhaps someone will post a better answer by then.
$endgroup$
– Michael
Jan 4 at 14:34
$begingroup$
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 $inf A leq inf B$]
$endgroup$
– Michael
Jan 4 at 5:16
$begingroup$
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 $inf A leq inf B$]
$endgroup$
– Michael
Jan 4 at 5:16
$begingroup$
Hi @Michael, I don't understand what you meant. Is the reasoning in that paragraph wrong? Could you please be more specific?
$endgroup$
– Le Anh Dung
Jan 4 at 5:20
$begingroup$
Hi @Michael, I don't understand what you meant. Is the reasoning in that paragraph wrong? Could you please be more specific?
$endgroup$
– Le Anh Dung
Jan 4 at 5:20
1
1
$begingroup$
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.
$endgroup$
– Michael
Jan 4 at 14:02
$begingroup$
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.
$endgroup$
– Michael
Jan 4 at 14:02
$begingroup$
Hi @Michael, Thank you so much for your detailed comment! Now I got your point.
$endgroup$
– Le Anh Dung
Jan 4 at 14:09
$begingroup$
Hi @Michael, Thank you so much for your detailed comment! Now I got your point.
$endgroup$
– Le Anh Dung
Jan 4 at 14:09
1
1
$begingroup$
Maybe I'll do that tomorrow, or perhaps someone will post a better answer by then.
$endgroup$
– Michael
Jan 4 at 14:34
$begingroup$
Maybe I'll do that tomorrow, or perhaps someone will post a better answer by then.
$endgroup$
– Michael
Jan 4 at 14:34
|
show 2 more comments
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$begingroup$
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 $inf A leq inf B$]
$endgroup$
– Michael
Jan 4 at 5:16
$begingroup$
Hi @Michael, I don't understand what you meant. Is the reasoning in that paragraph wrong? Could you please be more specific?
$endgroup$
– Le Anh Dung
Jan 4 at 5:20
1
$begingroup$
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.
$endgroup$
– Michael
Jan 4 at 14:02
$begingroup$
Hi @Michael, Thank you so much for your detailed comment! Now I got your point.
$endgroup$
– Le Anh Dung
Jan 4 at 14:09
1
$begingroup$
Maybe I'll do that tomorrow, or perhaps someone will post a better answer by then.
$endgroup$
– Michael
Jan 4 at 14:34