Mixing
Prove that $forall x,y,z in Bbb R^+:(xcdot y)cdot z = xcdot (ycdot z)$
$begingroup$
Let $Bbb R^+={xinBbb R mid x>0}$ and $x,yinBbb R^+$. We define the multiplication operation $(cdot)$ on $Bbb R^+$ by $$xcdot y:=inf{rcdot smid r,sinBbb Q text{ and } x<r text{ and } y<s}$$ Prove that $$forall x,y,z in Bbb R^+:(xcdot y)cdot z = xcdot (ycdot z)$$
My textbook said that the proof is straightforward, but it took me a lot of time to come up with a rigorous one. Please help me verify it. Thank you so much!
My attempt:
By definition, we have:
$xcdot y = inf{rcdot smid r,sinBbb Q, x<r, y<s}$
$ycdot z = inf{scdot tmid s,tinBbb Q, y<s, z<t}$
$(xcdot y)cdot z = inf{pcdot tmid p,tinBbb Q, xcdot y<p, z<t}=inf A$
$xcdot (ycdot z) = inf{rcdot qmid r,qinBbb Q, x<r, ycdot z<q}=inf B$
It suffices to prove that $A=B$.
Notice that $pin Bbb Q$ and $p>xcdot y iff$ $pin Bbb Q$ and $p>rcdot s$ for some $r,sinBbb Q$ such that $r>x,s>y$. Let $p=rcdot bar s>rcdot s$. Then $bar s in Bbb Q$ and $bar s>s>y$. Thus $p=rcdot bar s$ where $r,bar s in Bbb Q$ such that $r>x,bar s>y$.
Similarly, $qin Bbb Q$ and $q>ycdot z implies q=scdotbar t$ where $s,bar t in Bbb Q$ such that $s>y,bar t>z$.
$ain A implies a=pcdot t$ for some $p,tinBbb Q,p>xcdot y,t>z$ $implies a=(rcdot bar s)cdot t$ for some $r,bar s,t in Bbb Q$ such that $r>x,bar s>y,t>z$ $implies a=rcdot (bar scdot t)$ for some $r,bar s,t in Bbb Q$ such that $r>x,bar s>y,t>z$ $implies a=rcdot q$ for some $r,q=bar scdot tin Bbb Q$ such that $r>x,q>ycdot z$ $implies ain B$.
Similarly, $ain Bimplies ain A$.
Hence $A=B$ and thus $inf A=inf B$.
proof-verification real-numbers
asked Jan 3 at 17:32
Le Anh DungLe Anh Dung
1,0921521
$endgroup$
add a comment |
$begingroup$
Let $Bbb R^+={xinBbb R mid x>0}$ and $x,yinBbb R^+$. We define the multiplication operation $(cdot)$ on $Bbb R^+$ by $$xcdot y:=inf{rcdot smid r,sinBbb Q text{ and } x<r text{ and } y<s}$$ Prove that $$forall x,y,z in Bbb R^+:(xcdot y)cdot z = xcdot (ycdot z)$$
My textbook said that the proof is straightforward, but it took me a lot of time to come up with a rigorous one. Please help me verify it. Thank you so much!
My attempt:
By definition, we have:
$xcdot y = inf{rcdot smid r,sinBbb Q, x<r, y<s}$
$ycdot z = inf{scdot tmid s,tinBbb Q, y<s, z<t}$
$(xcdot y)cdot z = inf{pcdot tmid p,tinBbb Q, xcdot y<p, z<t}=inf A$
$xcdot (ycdot z) = inf{rcdot qmid r,qinBbb Q, x<r, ycdot z<q}=inf B$
It suffices to prove that $A=B$.
Notice that $pin Bbb Q$ and $p>xcdot y iff$ $pin Bbb Q$ and $p>rcdot s$ for some $r,sinBbb Q$ such that $r>x,s>y$. Let $p=rcdot bar s>rcdot s$. Then $bar s in Bbb Q$ and $bar s>s>y$. Thus $p=rcdot bar s$ where $r,bar s in Bbb Q$ such that $r>x,bar s>y$.
Similarly, $qin Bbb Q$ and $q>ycdot z implies q=scdotbar t$ where $s,bar t in Bbb Q$ such that $s>y,bar t>z$.
$ain A implies a=pcdot t$ for some $p,tinBbb Q,p>xcdot y,t>z$ $implies a=(rcdot bar s)cdot t$ for some $r,bar s,t in Bbb Q$ such that $r>x,bar s>y,t>z$ $implies a=rcdot (bar scdot t)$ for some $r,bar s,t in Bbb Q$ such that $r>x,bar s>y,t>z$ $implies a=rcdot q$ for some $r,q=bar scdot tin Bbb Q$ such that $r>x,q>ycdot z$ $implies ain B$.
Similarly, $ain Bimplies ain A$.
Hence $A=B$ and thus $inf A=inf B$.
proof-verification real-numbers
asked Jan 3 at 17:32
Le Anh DungLe Anh Dung
1,0921521
$endgroup$
1
$begingroup$
The proof is indeed straightforward since $(xcdot y)cdot z$ and $xcdot (ycdot z)$ are both equal to $$inf{rcdot scdot tmid r,s,tinBbb Q text{ and } x<r text{ and } y<s text{ and } z<t}$$
$endgroup$
– Did
Jan 3 at 17:55
add a comment |
$begingroup$
Let $Bbb R^+={xinBbb R mid x>0}$ and $x,yinBbb R^+$. We define the multiplication operation $(cdot)$ on $Bbb R^+$ by $$xcdot y:=inf{rcdot smid r,sinBbb Q text{ and } x<r text{ and } y<s}$$ Prove that $$forall x,y,z in Bbb R^+:(xcdot y)cdot z = xcdot (ycdot z)$$
My textbook said that the proof is straightforward, but it took me a lot of time to come up with a rigorous one. Please help me verify it. Thank you so much!
My attempt:
By definition, we have:
$xcdot y = inf{rcdot smid r,sinBbb Q, x<r, y<s}$
$ycdot z = inf{scdot tmid s,tinBbb Q, y<s, z<t}$
$(xcdot y)cdot z = inf{pcdot tmid p,tinBbb Q, xcdot y<p, z<t}=inf A$
$xcdot (ycdot z) = inf{rcdot qmid r,qinBbb Q, x<r, ycdot z<q}=inf B$
It suffices to prove that $A=B$.
Notice that $pin Bbb Q$ and $p>xcdot y iff$ $pin Bbb Q$ and $p>rcdot s$ for some $r,sinBbb Q$ such that $r>x,s>y$. Let $p=rcdot bar s>rcdot s$. Then $bar s in Bbb Q$ and $bar s>s>y$. Thus $p=rcdot bar s$ where $r,bar s in Bbb Q$ such that $r>x,bar s>y$.
Similarly, $qin Bbb Q$ and $q>ycdot z implies q=scdotbar t$ where $s,bar t in Bbb Q$ such that $s>y,bar t>z$.
$ain A implies a=pcdot t$ for some $p,tinBbb Q,p>xcdot y,t>z$ $implies a=(rcdot bar s)cdot t$ for some $r,bar s,t in Bbb Q$ such that $r>x,bar s>y,t>z$ $implies a=rcdot (bar scdot t)$ for some $r,bar s,t in Bbb Q$ such that $r>x,bar s>y,t>z$ $implies a=rcdot q$ for some $r,q=bar scdot tin Bbb Q$ such that $r>x,q>ycdot z$ $implies ain B$.
Similarly, $ain Bimplies ain A$.
Hence $A=B$ and thus $inf A=inf B$.
proof-verification real-numbers
asked Jan 3 at 17:32
Le Anh DungLe Anh Dung
1,0921521
$endgroup$
Let $Bbb R^+={xinBbb R mid x>0}$ and $x,yinBbb R^+$. We define the multiplication operation $(cdot)$ on $Bbb R^+$ by $$xcdot y:=inf{rcdot smid r,sinBbb Q text{ and } x<r text{ and } y<s}$$ Prove that $$forall x,y,z in Bbb R^+:(xcdot y)cdot z = xcdot (ycdot z)$$
My textbook said that the proof is straightforward, but it took me a lot of time to come up with a rigorous one. Please help me verify it. Thank you so much!
My attempt:
By definition, we have:
$xcdot y = inf{rcdot smid r,sinBbb Q, x<r, y<s}$
$ycdot z = inf{scdot tmid s,tinBbb Q, y<s, z<t}$
$(xcdot y)cdot z = inf{pcdot tmid p,tinBbb Q, xcdot y<p, z<t}=inf A$
$xcdot (ycdot z) = inf{rcdot qmid r,qinBbb Q, x<r, ycdot z<q}=inf B$
It suffices to prove that $A=B$.
Notice that $pin Bbb Q$ and $p>xcdot y iff$ $pin Bbb Q$ and $p>rcdot s$ for some $r,sinBbb Q$ such that $r>x,s>y$. Let $p=rcdot bar s>rcdot s$. Then $bar s in Bbb Q$ and $bar s>s>y$. Thus $p=rcdot bar s$ where $r,bar s in Bbb Q$ such that $r>x,bar s>y$.
Similarly, $qin Bbb Q$ and $q>ycdot z implies q=scdotbar t$ where $s,bar t in Bbb Q$ such that $s>y,bar t>z$.
$ain A implies a=pcdot t$ for some $p,tinBbb Q,p>xcdot y,t>z$ $implies a=(rcdot bar s)cdot t$ for some $r,bar s,t in Bbb Q$ such that $r>x,bar s>y,t>z$ $implies a=rcdot (bar scdot t)$ for some $r,bar s,t in Bbb Q$ such that $r>x,bar s>y,t>z$ $implies a=rcdot q$ for some $r,q=bar scdot tin Bbb Q$ such that $r>x,q>ycdot z$ $implies ain B$.
Similarly, $ain Bimplies ain A$.
Hence $A=B$ and thus $inf A=inf B$.
proof-verification real-numbers
proof-verification real-numbers
asked Jan 3 at 17:32
Le Anh DungLe Anh Dung
1,0921521
asked Jan 3 at 17:32
Le Anh DungLe Anh Dung
1,0921521
asked Jan 3 at 17:32
Le Anh DungLe Anh Dung
1,0921521
asked Jan 3 at 17:32
Le Anh DungLe Anh Dung
1,0921521
asked Jan 3 at 17:32
Le Anh DungLe Anh Dung
1,0921521
1,0921521
1
$begingroup$
The proof is indeed straightforward since $(xcdot y)cdot z$ and $xcdot (ycdot z)$ are both equal to $$inf{rcdot scdot tmid r,s,tinBbb Q text{ and } x<r text{ and } y<s text{ and } z<t}$$
$endgroup$
– Did
Jan 3 at 17:55
add a comment |
1
$begingroup$
The proof is indeed straightforward since $(xcdot y)cdot z$ and $xcdot (ycdot z)$ are both equal to $$inf{rcdot scdot tmid r,s,tinBbb Q text{ and } x<r text{ and } y<s text{ and } z<t}$$
$endgroup$
– Did
Jan 3 at 17:55
1
1
$begingroup$
The proof is indeed straightforward since $(xcdot y)cdot z$ and $xcdot (ycdot z)$ are both equal to $$inf{rcdot scdot tmid r,s,tinBbb Q text{ and } x<r text{ and } y<s text{ and } z<t}$$
$endgroup$
– Did
Jan 3 at 17:55
$begingroup$
The proof is indeed straightforward since $(xcdot y)cdot z$ and $xcdot (ycdot z)$ are both equal to $$inf{rcdot scdot tmid r,s,tinBbb Q text{ and } x<r text{ and } y<s text{ and } z<t}$$
$endgroup$
– Did
Jan 3 at 17:55
add a comment |
1 Answer
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$begingroup$
We define the addition operation $(+)$ on $Bbb R$ by $$x+ y:=inf{r+ smid r,sinBbb Q text{ and } x<r text{ and } y<s}$$
I would like to present a proof that $forall x,y,z in Bbb R^+:xcdot (y+z) = xcdot y +xcdot z$.
My attempt:
First, we have some useful observations:
$min Bbb Q$ and $m > xcdot y$ $iff m=r cdot s$ for some $r,sin Bbb Q$ such that $r>x,s>y$.
$pinBbb Q$ and $p>y+z$ $iff p=s+t$ for some $s,tin Bbb Q$ such that $s>y,t>z$.
By definition, we have:
$xcdot (y+z) = inf {rcdot p mid r,pinBbb Q,x<r,y+z<p}=inf {rcdot(s+t) mid r,s,tinBbb Q, x<r,y<s,z<t}=inf {rcdot s+rcdot t mid r,s,tinBbb Q, x<r,y<s,z<t}.$
$xcdot y+xcdot z = inf {m+n mid m,ninBbb Q,xcdot y<m,xcdot z<n}=inf {rcdot s+rcdot t mid r,s,tinBbb Q, x<r,y<s,z<t}.$
Hence $xcdot (y+z) = xcdot y +xcdot z$
answered Jan 4 at 3:34
Le Anh DungLe Anh Dung
1,0921521
$endgroup$
add a comment |
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$begingroup$
We define the addition operation $(+)$ on $Bbb R$ by $$x+ y:=inf{r+ smid r,sinBbb Q text{ and } x<r text{ and } y<s}$$
I would like to present a proof that $forall x,y,z in Bbb R^+:xcdot (y+z) = xcdot y +xcdot z$.
My attempt:
First, we have some useful observations:
$min Bbb Q$ and $m > xcdot y$ $iff m=r cdot s$ for some $r,sin Bbb Q$ such that $r>x,s>y$.
$pinBbb Q$ and $p>y+z$ $iff p=s+t$ for some $s,tin Bbb Q$ such that $s>y,t>z$.
By definition, we have:
$xcdot (y+z) = inf {rcdot p mid r,pinBbb Q,x<r,y+z<p}=inf {rcdot(s+t) mid r,s,tinBbb Q, x<r,y<s,z<t}=inf {rcdot s+rcdot t mid r,s,tinBbb Q, x<r,y<s,z<t}.$
$xcdot y+xcdot z = inf {m+n mid m,ninBbb Q,xcdot y<m,xcdot z<n}=inf {rcdot s+rcdot t mid r,s,tinBbb Q, x<r,y<s,z<t}.$
Hence $xcdot (y+z) = xcdot y +xcdot z$
answered Jan 4 at 3:34
Le Anh DungLe Anh Dung
1,0921521
$endgroup$
add a comment |
$begingroup$
We define the addition operation $(+)$ on $Bbb R$ by $$x+ y:=inf{r+ smid r,sinBbb Q text{ and } x<r text{ and } y<s}$$
I would like to present a proof that $forall x,y,z in Bbb R^+:xcdot (y+z) = xcdot y +xcdot z$.
My attempt:
First, we have some useful observations:
$min Bbb Q$ and $m > xcdot y$ $iff m=r cdot s$ for some $r,sin Bbb Q$ such that $r>x,s>y$.
$pinBbb Q$ and $p>y+z$ $iff p=s+t$ for some $s,tin Bbb Q$ such that $s>y,t>z$.
By definition, we have:
$xcdot (y+z) = inf {rcdot p mid r,pinBbb Q,x<r,y+z<p}=inf {rcdot(s+t) mid r,s,tinBbb Q, x<r,y<s,z<t}=inf {rcdot s+rcdot t mid r,s,tinBbb Q, x<r,y<s,z<t}.$
$xcdot y+xcdot z = inf {m+n mid m,ninBbb Q,xcdot y<m,xcdot z<n}=inf {rcdot s+rcdot t mid r,s,tinBbb Q, x<r,y<s,z<t}.$
Hence $xcdot (y+z) = xcdot y +xcdot z$
answered Jan 4 at 3:34
Le Anh DungLe Anh Dung
1,0921521
$endgroup$
add a comment |
$begingroup$
We define the addition operation $(+)$ on $Bbb R$ by $$x+ y:=inf{r+ smid r,sinBbb Q text{ and } x<r text{ and } y<s}$$
I would like to present a proof that $forall x,y,z in Bbb R^+:xcdot (y+z) = xcdot y +xcdot z$.
My attempt:
First, we have some useful observations:
$min Bbb Q$ and $m > xcdot y$ $iff m=r cdot s$ for some $r,sin Bbb Q$ such that $r>x,s>y$.
$pinBbb Q$ and $p>y+z$ $iff p=s+t$ for some $s,tin Bbb Q$ such that $s>y,t>z$.
By definition, we have:
$xcdot (y+z) = inf {rcdot p mid r,pinBbb Q,x<r,y+z<p}=inf {rcdot(s+t) mid r,s,tinBbb Q, x<r,y<s,z<t}=inf {rcdot s+rcdot t mid r,s,tinBbb Q, x<r,y<s,z<t}.$
$xcdot y+xcdot z = inf {m+n mid m,ninBbb Q,xcdot y<m,xcdot z<n}=inf {rcdot s+rcdot t mid r,s,tinBbb Q, x<r,y<s,z<t}.$
Hence $xcdot (y+z) = xcdot y +xcdot z$
answered Jan 4 at 3:34
Le Anh DungLe Anh Dung
1,0921521
$endgroup$
We define the addition operation $(+)$ on $Bbb R$ by $$x+ y:=inf{r+ smid r,sinBbb Q text{ and } x<r text{ and } y<s}$$
I would like to present a proof that $forall x,y,z in Bbb R^+:xcdot (y+z) = xcdot y +xcdot z$.
My attempt:
First, we have some useful observations:
$min Bbb Q$ and $m > xcdot y$ $iff m=r cdot s$ for some $r,sin Bbb Q$ such that $r>x,s>y$.
$pinBbb Q$ and $p>y+z$ $iff p=s+t$ for some $s,tin Bbb Q$ such that $s>y,t>z$.
By definition, we have:
$xcdot (y+z) = inf {rcdot p mid r,pinBbb Q,x<r,y+z<p}=inf {rcdot(s+t) mid r,s,tinBbb Q, x<r,y<s,z<t}=inf {rcdot s+rcdot t mid r,s,tinBbb Q, x<r,y<s,z<t}.$
$xcdot y+xcdot z = inf {m+n mid m,ninBbb Q,xcdot y<m,xcdot z<n}=inf {rcdot s+rcdot t mid r,s,tinBbb Q, x<r,y<s,z<t}.$
Hence $xcdot (y+z) = xcdot y +xcdot z$
answered Jan 4 at 3:34
Le Anh DungLe Anh Dung
1,0921521
edited Jan 4 at 4:30
answered Jan 4 at 3:34
Le Anh DungLe Anh Dung
1,0921521
answered Jan 4 at 3:34
Le Anh DungLe Anh Dung
1,0921521
answered Jan 4 at 3:34
Le Anh DungLe Anh Dung
1,0921521
1,0921521
add a comment |
add a comment |
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The proof is indeed straightforward since $(xcdot y)cdot z$ and $xcdot (ycdot z)$ are both equal to $$inf{rcdot scdot tmid r,s,tinBbb Q text{ and } x<r text{ and } y<s text{ and } z<t}$$
$endgroup$
– Did
Jan 3 at 17:55