Prove that $forall x,y,z in Bbb R^+:(xcdot y)cdot z = xcdot (ycdot z)$












1















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







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  • 1




    The proof is indeed straightforward since $(xcdot y)cdot z$ and $xcdot (ycdot z)$ are both equal to $$inf{rcdot scdot tmid r,s,tinBbb Q text{ and } x<r text{ and } y<s text{ and } z<t}$$
    – Did
    2 days ago
















1















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







share|cite|improve this question


















  • 1




    The proof is indeed straightforward since $(xcdot y)cdot z$ and $xcdot (ycdot z)$ are both equal to $$inf{rcdot scdot tmid r,s,tinBbb Q text{ and } x<r text{ and } y<s text{ and } z<t}$$
    – Did
    2 days ago














1












1








1








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







share|cite|improve this question














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






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asked 2 days ago









Le Anh Dung

1,0231521




1,0231521








  • 1




    The proof is indeed straightforward since $(xcdot y)cdot z$ and $xcdot (ycdot z)$ are both equal to $$inf{rcdot scdot tmid r,s,tinBbb Q text{ and } x<r text{ and } y<s text{ and } z<t}$$
    – Did
    2 days ago














  • 1




    The proof is indeed straightforward since $(xcdot y)cdot z$ and $xcdot (ycdot z)$ are both equal to $$inf{rcdot scdot tmid r,s,tinBbb Q text{ and } x<r text{ and } y<s text{ and } z<t}$$
    – Did
    2 days ago








1




1




The proof is indeed straightforward since $(xcdot y)cdot z$ and $xcdot (ycdot z)$ are both equal to $$inf{rcdot scdot tmid r,s,tinBbb Q text{ and } x<r text{ and } y<s text{ and } z<t}$$
– Did
2 days ago




The proof is indeed straightforward since $(xcdot y)cdot z$ and $xcdot (ycdot z)$ are both equal to $$inf{rcdot scdot tmid r,s,tinBbb Q text{ and } x<r text{ and } y<s text{ and } z<t}$$
– Did
2 days ago










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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 $xcdot (y+z) = xcdot y +xcdot z$






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    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 $xcdot (y+z) = xcdot y +xcdot z$






    share|cite|improve this answer




























      0














      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 $xcdot (y+z) = xcdot y +xcdot z$






      share|cite|improve this answer


























        0












        0








        0






        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 $xcdot (y+z) = xcdot y +xcdot z$






        share|cite|improve this answer














        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 $xcdot (y+z) = xcdot y +xcdot z$







        share|cite|improve this answer














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        share|cite|improve this answer








        edited 2 days ago

























        answered 2 days ago









        Le Anh Dung

        1,0231521




        1,0231521






























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