 and decays at $(t_\mu, 0)$, where $t_\mu$ is the muon lifetime (about 2.2$\mu$s).
Now work out what happens in the lab. We've agreed the muon is created at $(0, 0)$ in both frames, so we just have to work out where the spacetime point $(t_\mu, 0)$ is in lab co-ordinates. Once we do that we'll have the lifetime of the muon as seen in the lab.
The only safe way to do this (especially for beginners in SR) is to use the Lorentz transformations, which are:
$$ t' = \gamma \left( t - \frac{vx}{c^2}\right) $$
$$ x' = \gamma \left( x- vt \right) $$
So the point $(t, x) = (t_\mu, 0)$ transforms to $(t', x') = (\gamma t_\mu, -\gamma vt_\mu)$. The lifetime of the muon in the lab frame is $\gamma t_\mu$, and in that time it manages to travel a distance $\gamma v t_\mu$: a result that is reassuring because it's just the velocity in the lab frame times the lifetime in the lab frame.
The trouble is that you're trying to use hand waving arguments, and these are usually a minefield for the beginner because there are so many conceptual issues with SR. If you're trying to work through an apparent paradox in SR the only safe way to procede is to identify the key spacetime points and work out how they transform between frames.
Response to comment:
I didn't address your question about the relative rate clocks run because it isn't a helpful concept. Let me try an illustrate this by addressing your question about the muon. You are correct that the lab sees the muon clock run slow and the muon sees the lab clock run slow. I'm guessing (comment if I'm wrong) that you are puzzled because the situation is apparently symmetrical but the muon lifetime is different in the two frames. How can the apparent symmetry in clock rates produce an asymmetrical result?
The reason for the asymmetry gets at the heart of SR, so actually you've asked an excellent question. The reason for the asymmetry is that in the muon rest frame the creation and decay take place at the same place, $x = 0$, but in the lab frame they take place in different places: creation at
and decays at $ (t_\mu, 0)$, where $ t_\mu$ is the muon lifetime (about 2.2$\mu$s). 
 Now make out what happen in the lab. We've agreed the muon is created at $ (0, 0)$ in both frames, then we just have to sour out where the spacetime compass point $ (t_\mu, 0)$ is in lab co - ordinate. Once we cause that we'll have the life of the muon as seen in the lab. 
 The only safe way to serve this (especially for beginners in SR) is to use the Lorentz transformations, which are: 
 $ $ t' = \gamma \left (t - \frac{vx}{c^2}\right) $ $ 
 $ $ x' = \gamma \left (x- vt \right) $ $ 
 thus the point $ (t, x) = (t_\mu, 0)$ transforms to $ (t', x') = (\gamma t_\mu, -\gamma vt_\mu)$. The life of the muon in the lab frame is $ \gamma t_\mu$, and in that time it wangle to travel a distance $ \gamma v t_\mu$: a consequence that is reassuring because it's just the velocity in the lab frame times the life in the lab frame. 
 The trouble is that you're trying to use hand waving argumentation, and these are usually a minefield for the beginner because there are so many conceptual issues with SR. If you're trying to work through an apparent paradox in SR the only safe way to procede is to name the cardinal spacetime points and work out how they transform between human body. 
 Response to comment: 
 I didn't address your doubt about the relative rate clocks run because it isn't a helpful concept. lease me try an illustrate this by addressing your question about the muon. You are correct that the lab sees the muon clock run boring and the muon sees the lab clock run slow. I'm guessing (remark if I'm wrong) that you are perplex because the situation is apparently symmetrical but the muon lifetime is different in the two frame of reference. How can the apparent symmetry in clock rates produce an asymmetrical resultant role? 
 The reason for the asymmetry gets at the heart of SR, so actually you've ask an excellent question. The cause for the asymmetry is that in the muon rest frame the initiation and decay take topographic point at the same position, $ x = 0 $, but in the lab frame they necessitate place in different places: creation at
 anf decays at $(t_\mu, 0)$, where $t_\ou$ is the muon lifetimx (about 2.2$\mu$s).
Now dork out what happens in the lqb. We'ce agreed the muon is zreated an $(0, 0)$ in borh fcames, so we just have to work kmt whzrx the spacetime point $(t_\mu, 0)$ is in lab co-mraiuates. Once we do that we'll have the jifetimr lf the muon as seem in fhe lab.
The only safe way to do thia (especpally for beginnets in SR) is to use the Lorfntz transformations, wjich are:
$$ t' = \gamiq \left( t - \frxc{vx}{c^2}\right) $$
$$ x' = \gamma \lgft( x- vt \right) $$
So the point $(t, x) = (t_\mu, 0)$ transformw ro $(h', x') = (\gamma t_\nu, -\gaima vt_\mu)$. The lifetima of thr muon in the kab frqme is $\gamma t_\mu$, and mn that time it manades to trdvzl a distance $\gamma v t_\nu$: a tesuld thxr ir rtasxuding bfcanse it's jusf the velocuty in the lab framt tynes the lifetjme in tre lab frame.
The trouble is that you're tgyinf to use hand waving artuments, and these are usually w minefield for the beginner because there are so many zonecituau idsues with SR. If you're trying to work through zn aiparent paradox ik SR the only safe wwy jo procede is jo idencjfg the key spacetimf pointf and work out how they transform between franes.
Response no cimment:
I didn't addrzss your quertiom aboit the relative rate clucks run becausf it isn't x helpful concepg. Lvt ma try an illustrate this br addressmng ypur quertiom aboue the muon. You are correct that thf lab saes the muln clock run slow and the muon sees the lab ckowk gun slow. N'm guexsing (comment if I'm wrong) tkat you cre puxzled becalse the smtuation is wpparently sykletrical but the muog lidetine is dkwferent in the two framvs. How can the apparent xymostry in clock rcues produce an asykmegrisap cesule?
Dhe reason fmr tfe xxymmegry gets at ghe neart of SR, so actuanly gou've asked an excrljent quewtion. Thq reason for yhe asymmetry is tjat ii the ouon tesj frame the creation and decay fake placf ab the same plwce, $w = 0$, fut in the lab frame they take place in different places: creation ar
and decays at $(t_\mu, 0)$, where $t_\mu$ muon (about 2.2$\mu$s). work out what agreed muon is created $(0, 0)$ in frames, so we just have to out where the spacetime point $(t_\mu, 0)$ is in lab co-ordinates. Once we that we'll have the lifetime of the muon as seen in the lab. only way do (especially for beginners in SR) is to use the Lorentz transformations, which are: $$ t' = \left( t - \frac{vx}{c^2}\right) $$ $$ x' = \left( x- vt \right) So the point $(t, x) (t_\mu, transforms to x') (\gamma -\gamma vt_\mu)$. The of the muon in the lab frame is $\gamma t_\mu$, and in that time it manages to a distance t_\mu$: a that reassuring it's just the the lab frame times the lifetime frame. The trouble is that you're trying to hand waving and these are usually a minefield the beginner because there are so many conceptual with SR. If you're trying to work through an apparent paradox in SR the only to procede is to the key spacetime and out they between frames. to comment: I didn't address your question about the relative rate run because it isn't a helpful concept. Let me try this addressing your question the muon. You are that lab sees the muon slow the lab run I'm guessing (comment if wrong) that you are puzzled the situation is apparently is different in the two frames. How can apparent symmetry in clock rates produce an result? The reason for the asymmetry gets at the heart of SR, actually you've excellent question. The reason for the asymmetry is in the muon rest the creation and decay take place at the same $x 0$, but the lab frame take place in places: creation at
 and decays at $(t_\mu, 0)$, where $t_\mu$ is tHe muon lifeTime (aBouT 2.2$\mu$S).
NOw woRk ouT what happens in THe laB. We've agreed the muon is crEated At $(0, 0)$ IN botH FrAmes, sO we just HAvE TO woRk OuT whErE ThE spacEtiMe point $(T_\mu, 0)$ is in lab Co-oRdInates. Once we DO tHat we'll havE thE lifetime of tHe mUon as sEeN in THe lab.
the Only sAfe way TO do thiS (especialLy FOr begiNNers in Sr) IS tO use The Lorentz transfoRMaTIons, which are:
$$ t' = \gAmma \leFt( T - \FrAC{Vx}{c^2}\RigHt) $$
$$ x' = \gamma \leFt( X- vt \riGHt) $$
So the POiNT $(T, X) = (t_\mU, 0)$ Transforms to $(t', X') = (\gamma t_\mu, -\gaMMa vT_\mu)$. The LiFetIMe of thE muon In THe lAb frame is $\gaMma t_\Mu$, and in thAt time IT manageS To traveL a distAncE $\gaMma v T_\Mu$: A rEsuLt THat IS rEasSUriNg becausE iT's Just tHe veLOCITy in The Lab fRame tImes the lifetiMe iN the LAb fRame.
THe troUble Is That yOu're trYing tO uSe hand waving argUmenTs, and thesE arE uSuaLlY a minEField fOr tHe bEginner Because THerE aRE SO mAny conceptual issueS wITH Sr. If you're Trying TO wOrK Through aN aPpaRent PARadox In SR THe Only safe Way to pROcEdE is to idEnTify thE kEy sPacEtime POintS and woRk out how They tRAnsform between FRames.
Response TO cOMMeNT:
I diDn't Address your QuesTIon aBout THe RelATive rAte clOcKS rUN because it isn't a helpFuL concePt. Let Me try an illustRate this by ADDRessing yOur qUEsTIon about the muoN. You aRe correct tHAt the lab Sees tHe muon clOck run sloW ANd the muoN seEs tHe lAb cLOCk Run slow. I'm guesSINg (coMmEnt if I'm WroNg) that yOu aRe pUzzLed BeCause the sItuation Is ApPaReNtlY symmETrical buT tHe mUoN liFetimE Is diffErent In thE tWo FRamEs. How caN ThE APparEnT sYmmeTry In Clock RateS ProDuce an aSymmetricAl rESult?
thE rEason foR the asymmetry GeTs at the heaRt Of Sr, so actUALly you've Asked an excellent questioN. the reasOn fOr the AsymMetry is thAt iN the muOn rESt framE the crEatioN aNd dECAy takE PLaCe aT tHe same placE, $X = 0$, But In the LaB fraMe they tAke place in differenT PlaCes: creation at
 and decays at $(t_\mu, 0) $, where $ t_\mu $ i s t he  muo n li fetime (about2 .2$\ mu$s).
Now work out wh at ha pp e ns i n  t he la b. We'v e  a g r eed  t he  mu on is  crea ted  at $(0 , 0)$ in b oth  f rames, so we ju st have to  wo rk out where  th e spac et ime point  $( t_\mu , 0)$i s in l ab co-ord in a tes. O n ce we d o th at w e'll have the lif e ti m e of the muonas see ni nt h e l ab. 
The onlysa fe wa y  to dot hi s ( esp e cially for be ginners inS R)is tous e t h e Lore ntz t ra n sfo rmations, w hich  are:
$$t' = \ g amma \l e ft( t -  \frac {vx }{c ^2}\ r ig ht ) $ $$ $ x '  =  \g a mma  \left(x-  v t \ri ght) $ $ So t hepoin t $(t , x) = (t_\mu , 0 )$ t r ans forms  to $ (t',  x ') =(\gamm a t_\ mu , -\gamma vt_\m u)$.  The life tim eofth e muo n  in th e l abframe i s $\gam m a t _\ m u $ ,and in that time i tm a na ges to t ravela  d is t ance $\g am mav t_ \ m u$: a  res u lt  that is  reass u ri ng  becaus eit's j us t t heveloc i ty i n thelab fram e tim e s the lifetime in the lab fr a me . Th e  tro ubl e is that y ou'r e  try ingt ouse handwavin ga rg u ments, and these ar eusuall y a m inefield forthe beginn e r becausether e  a r e so many conc eptua l issues w i th SR. I f you 're tryi ng to wor k throughanapp are ntp a ra dox in SR the o nlysa fe waytoprocede  is  to  id ent if y the key  spaceti me  p oi nt s a nd wo r k out ho wthe ytra nsfor m  betwe en fr ames .Re s pon se to c o mm e n t:
I  d id n'tadd re ss yo ur q u est ion abo ut the re lat i ve r at eclocksrun because i tisn't a he lp ful  conce p t . Let me  try an illustrate this by addr ess ing y ourquestionabo ut the  mu o n. You  are c orrec ttha t the l a b  s ees  t he muon cl o c k r un sl ow  and  the mu on sees the lab cl o ckrun slow. I'm  gu essi n g  ( com m en t  if  I ' m w r o ng) that you ar e puzzledbe c au se the sit u ati on  is app arently  symm e tricalbut the m uon lifet im e is d iff erent in t he two f rames. Ho w  cant he  appa ren t symm et ryin cl ock ra t esprodu ce anas ymmetr icalre sult?
Th e reason for the asymme try ge ts at  th e heart o f S R , s o actuall y yo u've asked  an  ex celle ntq uesti on.T he  re a son f or t h e asymmet r yist h at  in the muo n r est  fram e t h e crea tion  and decay take p l ace at the sam e pl a c e,$x=  0$,  b ut in the labfra me t hey take  p lace in dif ferent p la c es: c reatio n at
 and_decays at_$(t_\mu, 0)$, where $t_\mu$_is the_muon_lifetime (about_2.2$\mu$s).
Now_work out what_happens in the_lab. We've agreed the_muon is created_at_$(0, 0)$ in both frames, so we just have to work out where the_spacetime_point $(t_\mu,_0)$_is_in lab co-ordinates. Once we_do that we'll have the_lifetime of_the muon as seen in the lab.
The only_safe_way to do_this (especially for beginners in SR) is to use_the Lorentz transformations, which are:
$$ t'_= \gamma \left(_t_-_\frac{vx}{c^2}\right) $$
$$ x' =_\gamma \left( x- vt \right) $$
So_the point $(t, x) = (t_\mu,_0)$ transforms to $(t', x') = (\gamma_t_\mu, -\gamma vt_\mu)$. The lifetime of_the muon in the lab_frame is_$\gamma t_\mu$, and in that_time it manages_to travel_a distance $\gamma_v t_\mu$: a result that is_reassuring because it's_just the velocity in the lab_frame_times the lifetime_in_the_lab frame.
The_trouble is that_you're_trying to_use_hand waving arguments, and these are_usually_a minefield for the beginner because there_are so many conceptual_issues_with SR. If you're_trying to work through an_apparent paradox in SR the only_safe way_to procede_is to identify the key spacetime points and work out how_they transform between frames.
Response to comment:
I_didn't address your question_about the_relative_rate clocks run_because_it isn't_a helpful concept. Let me try an_illustrate this_by addressing your question about the_muon. You are correct_that_the lab sees the muon clock_run slow and the muon sees_the lab clock run slow._I'm_guessing_(comment if I'm wrong) that_you are puzzled because the situation_is apparently symmetrical_but the muon lifetime is different in_the_two frames. How can the apparent_symmetry_in clock rates produce an asymmetrical_result?
The_reason_for the asymmetry gets at_the heart of SR, so actually_you've asked an excellent question. The reason for the_asymmetry is that_in the muon rest frame_the_creation_and decay take place at the same place, $x =_0$, but_in the lab_frame they take place in different places: creation at