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How is the relation "the smallest element is the same" reflexive?

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How is the relation “the smallest element is the same” reflexive?


Need help counting equivalence classes.Finding the smallest relation that is reflexive, transitive, and symmetricSmallest relation for reflexive, symmetry and transitivityEquivalence relation example. How is this even reflexive?Is antisymmetric the same as reflexive?Finding the smallest equivalence relation containing a specific list of ordered pairsHow is this an equivalence relation?truefalse claims in relations and equivalence relationsWhat is the least and greatest element in symmetric but not reflexive relation over $1,2,3$?How is this case a reflexive relation?













8












$begingroup$


Let $mathcalX$ be the set of all nonempty subsets of the set $1,2,3,...,10$. Define the relation $mathcalR$ on $mathcalX$ by: $forall A, B in mathcalX, A mathcalR B$ iff the smallest element of $A$ is equal to the smallest element of $B$. For example, $1,2,3 mathcalR 1,3,5,8$ because the smallest element of $1,2,3$ is $1$ which is also the smallest element of $1,3,5,8$.



Prove that $mathcalR$ is an equivalence relation on $mathcalX$.



From my understanding, the definition of reflexive is:



$$mathcalR text is reflexive iff forall x in mathcalX, x mathcalR x$$



However, for this problem, you can have the relation with these two sets:



$1$ and $1,2$



Then wouldn't this not be reflexive since $2$ is not in the first set, but is in the second set?



I'm having trouble seeing how this is reflexive. Getting confused by the definition here.










share|cite|improve this question











$endgroup$







  • 4




    $begingroup$
    Reflexive means that every element is related to itself. Thus, for reflexivity you have to consider one set only. Ok, we have that $ 1 mathcal R 1,2 $ but we have also $ 1 mathcal R 1 $ and $ 1,2 mathcal R 1,2 $
    $endgroup$
    – Mauro ALLEGRANZA
    13 hours ago






  • 6




    $begingroup$
    Note: “reflexive” does not mean that if $x$ is related to $y$, then $x=y$. It means that if $x=y$, then $x$ is related to $y$.
    $endgroup$
    – Arturo Magidin
    13 hours ago










  • $begingroup$
    So it must be reflexive because both $A$ and $B$ belong to the same set $mathcalX$?
    $endgroup$
    – qbuffer
    12 hours ago











  • $begingroup$
    @qbuffer Have a look at the updated version of my answer.
    $endgroup$
    – Haris Gusic
    12 hours ago















8












$begingroup$


Let $mathcalX$ be the set of all nonempty subsets of the set $1,2,3,...,10$. Define the relation $mathcalR$ on $mathcalX$ by: $forall A, B in mathcalX, A mathcalR B$ iff the smallest element of $A$ is equal to the smallest element of $B$. For example, $1,2,3 mathcalR 1,3,5,8$ because the smallest element of $1,2,3$ is $1$ which is also the smallest element of $1,3,5,8$.



Prove that $mathcalR$ is an equivalence relation on $mathcalX$.



From my understanding, the definition of reflexive is:



$$mathcalR text is reflexive iff forall x in mathcalX, x mathcalR x$$



However, for this problem, you can have the relation with these two sets:



$1$ and $1,2$



Then wouldn't this not be reflexive since $2$ is not in the first set, but is in the second set?



I'm having trouble seeing how this is reflexive. Getting confused by the definition here.










share|cite|improve this question











$endgroup$







  • 4




    $begingroup$
    Reflexive means that every element is related to itself. Thus, for reflexivity you have to consider one set only. Ok, we have that $ 1 mathcal R 1,2 $ but we have also $ 1 mathcal R 1 $ and $ 1,2 mathcal R 1,2 $
    $endgroup$
    – Mauro ALLEGRANZA
    13 hours ago






  • 6




    $begingroup$
    Note: “reflexive” does not mean that if $x$ is related to $y$, then $x=y$. It means that if $x=y$, then $x$ is related to $y$.
    $endgroup$
    – Arturo Magidin
    13 hours ago










  • $begingroup$
    So it must be reflexive because both $A$ and $B$ belong to the same set $mathcalX$?
    $endgroup$
    – qbuffer
    12 hours ago











  • $begingroup$
    @qbuffer Have a look at the updated version of my answer.
    $endgroup$
    – Haris Gusic
    12 hours ago













8












8








8





$begingroup$


Let $mathcalX$ be the set of all nonempty subsets of the set $1,2,3,...,10$. Define the relation $mathcalR$ on $mathcalX$ by: $forall A, B in mathcalX, A mathcalR B$ iff the smallest element of $A$ is equal to the smallest element of $B$. For example, $1,2,3 mathcalR 1,3,5,8$ because the smallest element of $1,2,3$ is $1$ which is also the smallest element of $1,3,5,8$.



Prove that $mathcalR$ is an equivalence relation on $mathcalX$.



From my understanding, the definition of reflexive is:



$$mathcalR text is reflexive iff forall x in mathcalX, x mathcalR x$$



However, for this problem, you can have the relation with these two sets:



$1$ and $1,2$



Then wouldn't this not be reflexive since $2$ is not in the first set, but is in the second set?



I'm having trouble seeing how this is reflexive. Getting confused by the definition here.










share|cite|improve this question











$endgroup$




Let $mathcalX$ be the set of all nonempty subsets of the set $1,2,3,...,10$. Define the relation $mathcalR$ on $mathcalX$ by: $forall A, B in mathcalX, A mathcalR B$ iff the smallest element of $A$ is equal to the smallest element of $B$. For example, $1,2,3 mathcalR 1,3,5,8$ because the smallest element of $1,2,3$ is $1$ which is also the smallest element of $1,3,5,8$.



Prove that $mathcalR$ is an equivalence relation on $mathcalX$.



From my understanding, the definition of reflexive is:



$$mathcalR text is reflexive iff forall x in mathcalX, x mathcalR x$$



However, for this problem, you can have the relation with these two sets:



$1$ and $1,2$



Then wouldn't this not be reflexive since $2$ is not in the first set, but is in the second set?



I'm having trouble seeing how this is reflexive. Getting confused by the definition here.







discrete-mathematics elementary-set-theory relations equivalence-relations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 1 hour ago









Martin Sleziak

45k10122277




45k10122277










asked 13 hours ago









qbufferqbuffer

625




625







  • 4




    $begingroup$
    Reflexive means that every element is related to itself. Thus, for reflexivity you have to consider one set only. Ok, we have that $ 1 mathcal R 1,2 $ but we have also $ 1 mathcal R 1 $ and $ 1,2 mathcal R 1,2 $
    $endgroup$
    – Mauro ALLEGRANZA
    13 hours ago






  • 6




    $begingroup$
    Note: “reflexive” does not mean that if $x$ is related to $y$, then $x=y$. It means that if $x=y$, then $x$ is related to $y$.
    $endgroup$
    – Arturo Magidin
    13 hours ago










  • $begingroup$
    So it must be reflexive because both $A$ and $B$ belong to the same set $mathcalX$?
    $endgroup$
    – qbuffer
    12 hours ago











  • $begingroup$
    @qbuffer Have a look at the updated version of my answer.
    $endgroup$
    – Haris Gusic
    12 hours ago












  • 4




    $begingroup$
    Reflexive means that every element is related to itself. Thus, for reflexivity you have to consider one set only. Ok, we have that $ 1 mathcal R 1,2 $ but we have also $ 1 mathcal R 1 $ and $ 1,2 mathcal R 1,2 $
    $endgroup$
    – Mauro ALLEGRANZA
    13 hours ago






  • 6




    $begingroup$
    Note: “reflexive” does not mean that if $x$ is related to $y$, then $x=y$. It means that if $x=y$, then $x$ is related to $y$.
    $endgroup$
    – Arturo Magidin
    13 hours ago










  • $begingroup$
    So it must be reflexive because both $A$ and $B$ belong to the same set $mathcalX$?
    $endgroup$
    – qbuffer
    12 hours ago











  • $begingroup$
    @qbuffer Have a look at the updated version of my answer.
    $endgroup$
    – Haris Gusic
    12 hours ago







4




4




$begingroup$
Reflexive means that every element is related to itself. Thus, for reflexivity you have to consider one set only. Ok, we have that $ 1 mathcal R 1,2 $ but we have also $ 1 mathcal R 1 $ and $ 1,2 mathcal R 1,2 $
$endgroup$
– Mauro ALLEGRANZA
13 hours ago




$begingroup$
Reflexive means that every element is related to itself. Thus, for reflexivity you have to consider one set only. Ok, we have that $ 1 mathcal R 1,2 $ but we have also $ 1 mathcal R 1 $ and $ 1,2 mathcal R 1,2 $
$endgroup$
– Mauro ALLEGRANZA
13 hours ago




6




6




$begingroup$
Note: “reflexive” does not mean that if $x$ is related to $y$, then $x=y$. It means that if $x=y$, then $x$ is related to $y$.
$endgroup$
– Arturo Magidin
13 hours ago




$begingroup$
Note: “reflexive” does not mean that if $x$ is related to $y$, then $x=y$. It means that if $x=y$, then $x$ is related to $y$.
$endgroup$
– Arturo Magidin
13 hours ago












$begingroup$
So it must be reflexive because both $A$ and $B$ belong to the same set $mathcalX$?
$endgroup$
– qbuffer
12 hours ago





$begingroup$
So it must be reflexive because both $A$ and $B$ belong to the same set $mathcalX$?
$endgroup$
– qbuffer
12 hours ago













$begingroup$
@qbuffer Have a look at the updated version of my answer.
$endgroup$
– Haris Gusic
12 hours ago




$begingroup$
@qbuffer Have a look at the updated version of my answer.
$endgroup$
– Haris Gusic
12 hours ago










2 Answers
2






active

oldest

votes


















8












$begingroup$

Why are you testing reflexivity by looking at two different elements of $mathcalX$? The definition of reflexivity says that a relation is reflexive iff each element of $mathcal X$ is in relation with itself.



To check whether $mathcal R$ is reflexive, just take one element of $mathcal X$, let's call it $x$. Then check whether $x$ is in relation with $x$. Because $x=x$, the smallest element of $x$ is equal to the smallest element of $x$. Thus, by definition of $mathcal R$, $x$ is in relation with $x$. Now, prove that this is true for all $x in mathcal X$. Of course, this is true because $min(x) = min(x)$ is always true, which is intuitive. In other words, $x mathcalR x$ for all $x in mathcal X$, which is exactly what you needed to prove that $mathcal R$ is reflexive.



You must understand that the definition of reflexivity says nothing about whether different elements (say $x,y$, $xneq y$) can be in the relation $mathcal R$. The fact that $1mathcal R 1,2$ does not contradict the fact that $1,2mathcal R 1,2$ as well.






share|cite|improve this answer











$endgroup$




















    4












    $begingroup$

    A binary relation $R$ over a set $mathcalX$ is reflexive if every element of $mathcalX$ is related to itself. The more formal definition has already been given by you, i.e. $$mathcalR text is reflexive iff forall x in mathcalX, x mathcalR x$$



    Note here that you've picked two different elements of the set to make your comparison when you should be comparing an element with itself. Also make sure you understand that an element may be related to other elements as well, reflexivity does not forbid that. It just says that every element must be related to itself.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      8












      $begingroup$

      Why are you testing reflexivity by looking at two different elements of $mathcalX$? The definition of reflexivity says that a relation is reflexive iff each element of $mathcal X$ is in relation with itself.



      To check whether $mathcal R$ is reflexive, just take one element of $mathcal X$, let's call it $x$. Then check whether $x$ is in relation with $x$. Because $x=x$, the smallest element of $x$ is equal to the smallest element of $x$. Thus, by definition of $mathcal R$, $x$ is in relation with $x$. Now, prove that this is true for all $x in mathcal X$. Of course, this is true because $min(x) = min(x)$ is always true, which is intuitive. In other words, $x mathcalR x$ for all $x in mathcal X$, which is exactly what you needed to prove that $mathcal R$ is reflexive.



      You must understand that the definition of reflexivity says nothing about whether different elements (say $x,y$, $xneq y$) can be in the relation $mathcal R$. The fact that $1mathcal R 1,2$ does not contradict the fact that $1,2mathcal R 1,2$ as well.






      share|cite|improve this answer











      $endgroup$

















        8












        $begingroup$

        Why are you testing reflexivity by looking at two different elements of $mathcalX$? The definition of reflexivity says that a relation is reflexive iff each element of $mathcal X$ is in relation with itself.



        To check whether $mathcal R$ is reflexive, just take one element of $mathcal X$, let's call it $x$. Then check whether $x$ is in relation with $x$. Because $x=x$, the smallest element of $x$ is equal to the smallest element of $x$. Thus, by definition of $mathcal R$, $x$ is in relation with $x$. Now, prove that this is true for all $x in mathcal X$. Of course, this is true because $min(x) = min(x)$ is always true, which is intuitive. In other words, $x mathcalR x$ for all $x in mathcal X$, which is exactly what you needed to prove that $mathcal R$ is reflexive.



        You must understand that the definition of reflexivity says nothing about whether different elements (say $x,y$, $xneq y$) can be in the relation $mathcal R$. The fact that $1mathcal R 1,2$ does not contradict the fact that $1,2mathcal R 1,2$ as well.






        share|cite|improve this answer











        $endgroup$















          8












          8








          8





          $begingroup$

          Why are you testing reflexivity by looking at two different elements of $mathcalX$? The definition of reflexivity says that a relation is reflexive iff each element of $mathcal X$ is in relation with itself.



          To check whether $mathcal R$ is reflexive, just take one element of $mathcal X$, let's call it $x$. Then check whether $x$ is in relation with $x$. Because $x=x$, the smallest element of $x$ is equal to the smallest element of $x$. Thus, by definition of $mathcal R$, $x$ is in relation with $x$. Now, prove that this is true for all $x in mathcal X$. Of course, this is true because $min(x) = min(x)$ is always true, which is intuitive. In other words, $x mathcalR x$ for all $x in mathcal X$, which is exactly what you needed to prove that $mathcal R$ is reflexive.



          You must understand that the definition of reflexivity says nothing about whether different elements (say $x,y$, $xneq y$) can be in the relation $mathcal R$. The fact that $1mathcal R 1,2$ does not contradict the fact that $1,2mathcal R 1,2$ as well.






          share|cite|improve this answer











          $endgroup$



          Why are you testing reflexivity by looking at two different elements of $mathcalX$? The definition of reflexivity says that a relation is reflexive iff each element of $mathcal X$ is in relation with itself.



          To check whether $mathcal R$ is reflexive, just take one element of $mathcal X$, let's call it $x$. Then check whether $x$ is in relation with $x$. Because $x=x$, the smallest element of $x$ is equal to the smallest element of $x$. Thus, by definition of $mathcal R$, $x$ is in relation with $x$. Now, prove that this is true for all $x in mathcal X$. Of course, this is true because $min(x) = min(x)$ is always true, which is intuitive. In other words, $x mathcalR x$ for all $x in mathcal X$, which is exactly what you needed to prove that $mathcal R$ is reflexive.



          You must understand that the definition of reflexivity says nothing about whether different elements (say $x,y$, $xneq y$) can be in the relation $mathcal R$. The fact that $1mathcal R 1,2$ does not contradict the fact that $1,2mathcal R 1,2$ as well.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 12 hours ago

























          answered 13 hours ago









          Haris GusicHaris Gusic

          3,331526




          3,331526





















              4












              $begingroup$

              A binary relation $R$ over a set $mathcalX$ is reflexive if every element of $mathcalX$ is related to itself. The more formal definition has already been given by you, i.e. $$mathcalR text is reflexive iff forall x in mathcalX, x mathcalR x$$



              Note here that you've picked two different elements of the set to make your comparison when you should be comparing an element with itself. Also make sure you understand that an element may be related to other elements as well, reflexivity does not forbid that. It just says that every element must be related to itself.






              share|cite|improve this answer









              $endgroup$

















                4












                $begingroup$

                A binary relation $R$ over a set $mathcalX$ is reflexive if every element of $mathcalX$ is related to itself. The more formal definition has already been given by you, i.e. $$mathcalR text is reflexive iff forall x in mathcalX, x mathcalR x$$



                Note here that you've picked two different elements of the set to make your comparison when you should be comparing an element with itself. Also make sure you understand that an element may be related to other elements as well, reflexivity does not forbid that. It just says that every element must be related to itself.






                share|cite|improve this answer









                $endgroup$















                  4












                  4








                  4





                  $begingroup$

                  A binary relation $R$ over a set $mathcalX$ is reflexive if every element of $mathcalX$ is related to itself. The more formal definition has already been given by you, i.e. $$mathcalR text is reflexive iff forall x in mathcalX, x mathcalR x$$



                  Note here that you've picked two different elements of the set to make your comparison when you should be comparing an element with itself. Also make sure you understand that an element may be related to other elements as well, reflexivity does not forbid that. It just says that every element must be related to itself.






                  share|cite|improve this answer









                  $endgroup$



                  A binary relation $R$ over a set $mathcalX$ is reflexive if every element of $mathcalX$ is related to itself. The more formal definition has already been given by you, i.e. $$mathcalR text is reflexive iff forall x in mathcalX, x mathcalR x$$



                  Note here that you've picked two different elements of the set to make your comparison when you should be comparing an element with itself. Also make sure you understand that an element may be related to other elements as well, reflexivity does not forbid that. It just says that every element must be related to itself.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 13 hours ago









                  s0ulr3aper07s0ulr3aper07

                  658112




                  658112



























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                      یوتیوب محتویات پیشینه[ویرایش] فناوری‌های ویدئویی[ویرایش] شوخی‌های آوریل[ویرایش] سانسور و فیلترینگ[ویرایش] آمار و ارقامی از یوتیوب[ویرایش] تأثیر اجتماعی[ویرایش] سیاست اجتماعی[ویرایش] نمودارها[ویرایش] یادداشت‌ها[ویرایش] پانویس[ویرایش] پیوند به بیرون[ویرایش] منوی ناوبریبررسی شده‌استYouTube.com[بروزرسانی]"Youtube.com Site Info""زبان‌های یوتیوب""Surprise! There's a third YouTube co-founder"سایت یوتیوب برای چندمین بار در ایران فیلتر شدنسخهٔ اصلیسالار کمانگر جوان آمریکایی ایرانی الاصل مدیر سایت یوتیوب شدنسخهٔ اصلیVideo websites pop up, invite postingsthe originalthe originalYouTube: Overnight success has sparked a backlashthe original"Me at the zoo"YouTube serves up 100 million videos a day onlinethe originalcomScore Releases May 2010 U.S. Online Video Rankingsthe originalYouTube hits 4 billion daily video viewsthe originalYouTube users uploading two days of video every minutethe originalEric Schmidt, Princeton Colloquium on Public & Int'l Affairsthe original«Streaming Dreams»نسخهٔ اصلیAlexa Traffic Rank for YouTube (three month average)the originalHelp! YouTube is killing my business!the originalUtube sues YouTubethe originalGoogle closes $A2b YouTube dealthe originalFlash moves on to smart phonesthe originalYouTube HTML5 Video Playerنسخهٔ اصلیYouTube HTML5 Video Playerthe originalGoogle tries freeing Web video with WebMthe originalVideo length for uploadingthe originalYouTube caps video lengths to reduce infringementthe originalAccount Types: Longer videosthe originalYouTube bumps video limit to 15 minutesthe originalUploading large files and resumable uploadingthe originalVideo Formats: File formatsthe originalGetting Started: File formatsthe originalThe quest for a new video codec in Flash 8the originalAdobe Flash Video File Format Specification Version 10.1the originalYouTube Mobile goes livethe originalYouTube videos go HD with a simple hackthe originalYouTube now supports 4k-resolution videosthe originalYouTube to get high-def 1080p playerthe original«Approximate YouTube Bitrates»نسخهٔ اصلی«Bigger and Better: Encoding for YouTube 720p HD»نسخهٔ اصلی«YouTube's 1080p – Failure Depends on How You Look At It»نسخهٔ اصلیYouTube in 3Dthe originalYouTube in 3D?the originalYouTube 3D Videosthe originalYouTube adds a dimension, 3D goggles not includedthe originalYouTube Adds Stereoscopic 3D Video Support (And 3D Vision Support, Too)the original«Sharing YouTube Videos»نسخهٔ اصلی«Downloading videos from YouTube is not supported, except for one instance when it is permitted.»نسخهٔ اصلی«Terms of Use, 5.B»نسخهٔ اصلی«Some YouTube videos get download option»نسخهٔ اصلی«YouTube looks out for content owners, disables video ripping»«Downloading videos from YouTube is not supported, except for one instance when it is permitted.»نسخهٔ اصلی«YouTube Hopes To Boost Revenue With Video Downloads»نسخهٔ اصلی«YouTube Mobile»نسخهٔ اصلی«YouTube Live on Apple TV Today; Coming to iPhone on June 29»نسخهٔ اصلی«Goodbye Flash: YouTube mobile goes HTML5 on iPhone and Android»نسخهٔ اصلی«YouTube Mobile Goes HTML5, Video Quality Beats Native Apps Hands Down»نسخهٔ اصلی«TiVo Getting YouTube Streaming Today»نسخهٔ اصلی«YouTube video comes to Wii and PlayStation 3 game consoles»نسخهٔ اصلی«Coming Up Next... YouTube on Your TV»نسخهٔ اصلی«Experience YouTube XL on the Big Screen»نسخهٔ اصلی«Xbox Live Getting Live TV, YouTube & Bing Voice Search»نسخهٔ اصلی«YouTube content locations»نسخهٔ اصلی«April fools: YouTube turns the world up-side-down»نسخهٔ اصلی«YouTube goes back to 1911 for April Fools' Day»نسخهٔ اصلی«Simon Cowell's bromance, the self-driving Nascar and Hungry Hippos for iPad... the best April Fools' gags»نسخهٔ اصلی"YouTube Announces It Will Shut Down""YouTube Adds Darude 'Sandstorm' Button To Its Videos For April Fools' Day"«Censorship fears rise as Iran blocks access to top websites»نسخهٔ اصلی«China 'blocks YouTube video site'»نسخهٔ اصلی«YouTube shut down in Morocco»نسخهٔ اصلی«Thailand blocks access to YouTube»نسخهٔ اصلی«Ban on YouTube lifted after deal»نسخهٔ اصلی«Google's Gatekeepers»نسخهٔ اصلی«Turkey goes into battle with Google»نسخهٔ اصلی«Turkey lifts two-year ban on YouTube»نسخهٔ اصلیسانسور در ترکیه به یوتیوب رسیدلغو فیلترینگ یوتیوب در ترکیه«Pakistan blocks YouTube website»نسخهٔ اصلی«Pakistan lifts the ban on YouTube»نسخهٔ اصلی«Pakistan blocks access to YouTube in internet crackdown»نسخهٔ اصلی«Watchdog urges Libya to stop blocking websites»نسخهٔ اصلی«YouTube»نسخهٔ اصلی«Due to abuses of religion, customs Emirates, YouTube is blocked in the UAE»نسخهٔ اصلی«Google Conquered The Web - An Ultimate Winner»نسخهٔ اصلی«100 million videos are viewed daily on YouTube»نسخهٔ اصلی«Harry and Charlie Davies-Carr: Web gets taste for biting baby»نسخهٔ اصلی«Meet YouTube's 224 million girl, Natalie Tran»نسخهٔ اصلی«YouTube to Double Down on Its 'Channel' Experiment»نسخهٔ اصلی«13 Some Media Companies Choose to Profit From Pirated YouTube Clips»نسخهٔ اصلی«Irate HK man unlikely Web hero»نسخهٔ اصلی«Web Guitar Wizard Revealed at Last»نسخهٔ اصلی«Charlie bit my finger – again!»نسخهٔ اصلی«Lowered Expectations: Web Redefines 'Quality'»نسخهٔ اصلی«YouTube's 50 Greatest Viral Videos»نسخهٔ اصلیYouTube Community Guidelinesthe original«Why did my YouTube account get closed down?»نسخهٔ اصلی«Why do I have a sanction on my account?»نسخهٔ اصلی«Is YouTube's three-strike rule fair to users?»نسخهٔ اصلی«Viacom will sue YouTube for $1bn»نسخهٔ اصلی«Mediaset Files EUR500 Million Suit Vs Google's YouTube»نسخهٔ اصلی«Premier League to take action against YouTube»نسخهٔ اصلی«YouTube law fight 'threatens net'»نسخهٔ اصلی«Google must divulge YouTube log»نسخهٔ اصلی«Google Told to Turn Over User Data of YouTube»نسخهٔ اصلی«US judge tosses out Viacom copyright suit against YouTube»نسخهٔ اصلی«Google and Viacom: YouTube copyright lawsuit back on»نسخهٔ اصلی«Woman can sue over YouTube clip de-posting»نسخهٔ اصلی«YouTube loses court battle over music clips»نسخهٔ اصلیYouTube to Test Software To Ease Licensing Fightsthe original«Press Statistics»نسخهٔ اصلی«Testing YouTube's Audio Content ID System»نسخهٔ اصلی«Content ID disputes»نسخهٔ اصلیYouTube Community Guidelinesthe originalYouTube criticized in Germany over anti-Semitic Nazi videosthe originalFury as YouTube carries sick Hillsboro video insultthe originalYouTube attacked by MPs over sex and violence footagethe originalAl-Awlaki's YouTube Videos Targeted by Rep. Weinerthe originalYouTube Withdraws Cleric's Videosthe originalYouTube is letting users decide on terrorism-related videosthe original«Time's Person of the Year: You»نسخهٔ اصلی«Our top 10 funniest YouTube comments – what are yours?»نسخهٔ اصلی«YouTube's worst comments blocked by filter»نسخهٔ اصلی«Site Info YouTube»نسخهٔ اصلیوبگاه YouTubeوبگاه موبایل YouTubeوووووو

                      Magento 2 - Auto login with specific URL Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Customer can't login - Page refreshes but nothing happensCustom Login page redirectURL to login with redirect URL after completionCustomer login is case sensitiveLogin with phone number or email address - Magento 1.9Magento 2: Set Customer Account Confirmation StatusCustomer auto connect from URLHow to call customer login form in the custom module action magento 2?Change of customer login error message magento2Referrer URL in modal login form

                      Rest API with Magento using PHP with example. Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?How to update product using magento client library for PHP?Oauth Error while extending Magento Rest APINot showing my custom api in wsdl(url) and web service list?Using Magento API(REST) via IXMLHTTPRequest COM ObjectHow to login in Magento website using REST APIREST api call for Guest userMagento API calling using HTML and javascriptUse API rest media management by storeView code (admin)Magento REST API Example ErrorsHow to log all rest api calls in magento2?How to update product using magento client library for PHP?