Andwhen I learneddiscretemathematics, I didn't havethisopportunity.
Sointheory, youshouldbeabletolearnthematerialmuchfasterthan I waseverableto.
So I stronglyencouragerewindingvideos, Finallyexplained.
Onceyouthinkyouunderstandthematerialofvideos, youcantrytoexplainthematerialoutloud, eithertoyourselforto a friend.
Ifnoone's aroundoryoufeelwe'retalkinghimself, thenyoucantrytoexplainitto a rubberducky.
There's a conceptinprogrammingcalledrubberduckdebugging, whichiswhereprogrammersgothroughtheircodelinebyline, tryingtoidentifybugsintheircodebyexplainingittorubberducky.
So, similarly, ifyougothroughthematerialofthevideossubjectbysubject, youwillsoonbeabletoidentifyuh, anygapsinyourunderstandingofthematerial.
Setsareusuallydenotedbycapitalletterssuchascapital, a capital, B, C, E or F, andwesaythatthesetcontainselements.
Forexample, wecouldsaythattheset A equalsthesetcontainingtheElementfive, orwecansaythatfiveisanelementof a Now I willbeinteresting.
A lotofsymbolslikethis, thisjustmeansisanelementofandit'llsaveus a lotoftimeinfuturevideosand I don't havetowriteoutisanelementofeachtime, sothat'llcomeinhandy.
Therationalnumbersaredefinedas a ratiooftoemitters, andexamplesincludetheimagesthemselvesterminatingdecimalsandrepeatingdecimalsSolet's startwithrepeatingdecimals.
Ifwelet 10 X equal.
9.999 RepeatingthenIfwedividebybothsidesby 10 then X equals 0.999 Repeating.
Andifwesubtract x from 10 x, wehavenine x equalsnotandthendividebothsides.
Byninewehave X equalsone.
Andthen, ifwemultiplyby 10 bothsidesbyTammyof 10.
X equals 10.
Sonowwehave 10.
X equals 10 but 10.
It's alsoequals 9.99 repeating.
Sowehavenow.
Andtheseareinfacttrue.
9.99 Repeatingequals 10.
Sowethinkwecannowexpress 10 X asanimager, whichis a a, um, rationalnumber.
Sothesymmetricdifferenceof A and B equalsthesetdifferenceof a and B unionwiththesetdifferenceofbeing a andfinally I willdefinetheirrationalnumberswhich I hintedtoattheendoflastvideo.
Andifwewerecalled a definitionbaitof a intersectionbeatthatequalsthesetcontainingelementsexsuchthat X isanelementof A and X isnowgonnabe.
Whichiswhytheoverlapwhere a and B overlapisshadedredweresayingallpossiblevaluesofeggs.
ThisredregionNowthesetdifferenceof A and B equalsthesetcontainingelementsexsuchthat X isanelementof A and X isalsonotanelementofbe, whichiswhyit's justthisregionhere.
Noneoftheelementsairwithinbethesetdifferenceofbeing A isthesetcontainingelementsexsuchthat X isanelementof B and X isnotanelementof a andfinally, thesymmetricdifferenceof A and B I definedastheunionof a Minus B and B minus A.
Wecansaythat B is a subsetof a becausebecontainsbecauseallelementsof B areelementsof A.
Wecanalsosaythat B is a propersubsetof a becauseallelementsof B R elementsof A andthereareelementsof a thatarenotwithintheset B.
Wecanalsosaythattheset A is a supersetofset B because A containsalloftheelementsof B andbeingevenmorespecific, wecansaythatis a propersupersetof B because a containsallelementsof B andthereareelementsof a they'renotelementsof B.
Whatwecannotsayisthatyouis a propersubsetof a Althoughthiscirclehereisbiggerthan a likeitjusttakesupmorespace.
Itdoesn't containanymoreelements.
SoalltheelementsSoyoucontainsalltheelementsof a howeveryoudoesnotcontainanyextraelements.
Soyouis a subsetof a and A isactually a subsetofyou.
So I wanttobeginbytalkingabouttheuniversalset, whichiscommonlyreferredtoastheuniversenowthereason I usethisfancylookingyouhereisthat I wasdefiningtheuniverse, whichistosay, themaximumboundariesofmyset.
Now, if I don't definethis, if I don't definemyuniversalset, thenwejustassumeittobetherealnumbers.
Soif I justerasethis, thenyouknowthiswilljustbetherealnumbers, whichiswe'vetalkedaboutinpreviousvideos.
Sotheotherthing I wanttotalkabouthiscompliments.
Soif I sayif I'm lookingforthecomplimentof B and I'llput a moreformaldefinitionofif I'm lookingforthecomplimentof B, then, um, itiseverythingthatisnotwithin.
Don't letitconcernyouthatthereareelementsoutsideofouruniversethatjustmeansthatthey're I mean, thatjustmeansthey'renotintheuniversethatwehaveinquestionrightnow.
Likewise, thissethereitcontainselements 31 andactswhile a contains 31 Next.
Sothisistheset A andfinally, thissaidcontainsElementsoneandthree.
Thissetcontainshelmetsoneandthree.
Sothisistheset.
See?
Nowwhatcouldbesaidaboutthesesets?
Well, wehaveWecansaythatseais a subsetoftheuniversebecauseallelementsofseaareelementsoftheuniverse.
Furthermore, wecansaythattoset B isnot a subsetoftheuniversebecausetheirelementsof B they'renotwithintheuniverse.
Also, wecansaythattheuniverseisnot a subset.
Areyousupersetof a Becausethere's elementsof a thatarenotwithinarenotcontainedbytheuniverse.
Andfinally, wecansaythattheuniverseis a propersupersetofseabecausetheuniversecontainsallelementsofseaexcept a LLAndthereareelementswithintheuniversethatarenotwithinthesensi.
Soifwewanttobemorespecific, wecouldactuallysaythatseais a propersubsetoftheuniverse.
Sointhisvideo, I havedefinedtheuniversetobeatheimagersonethroughfive.
Andit's notagraphicallybythisblueline.
Andifwerecallthedefinitionofthecomplimentswithherwithregardsto a wehave a complimentequalsthesetcontainingelementseggssuchthat X isanelementoftheuniverse, and X isnotanelementof a Sointhatcase, wecanlookatalltheimagesonethroughfiveandifitcontainsanyelementfrom a wethrowitout.
Sointhiscase, ifwelookattheimageofone, weknowthatthat's notgonnabetherebecausethat's a weelookattheaterjertothat'llbeanotherset.
Welookattheenergyorthreethatwon't beinoursetbecausethat's alsoan A andthenfourandfivearewithintheuniverse, andthey'renotwithanegg.
Andlastly, wehave C complimentto 45 arewithintheuniverse, andthey'renotwithin.
Seebecausewehadtoexcludeoneinthree.
I wanttobeginthisvideobyreviewingourdefinitionsoftheunionandintersection.
Afterthat, I'llbeintroducingtoalgebraiclawsforsetssothedefinitionof a unionbeequalssetcontainingelementsExsuchthat X isanelementof A or X isanelementof B.
Also, a intersection B isdefinedassecondtakingelements.
Exsuchthat X isanelementof A and X isanelementof B.
Sowhen I introducethecomponentslaw, itshouldbeclearthat a union A equals A because a U.
E.
T.
A isdefinedasthesetcontainingelements.
ExSubset X isanelement A or X isanelementof a whichshouldyieldareyoushouldseewhythatequalstheset.
A also a intersection, A equalstheSECcontainingelementsactssuchthat X isanelementof A and X isanelementof a whichitshouldalsobeclearwhythatistheset A.
Nextwehaveouridentitylaws.
Soidentitylawsarelawswherewehave a givensetandwe'reperforminganoperationonthatsetwith a givenelement.
So a unionnosetequals a andbydefinition, thatmeansthat a unionnosetequalsthesecondTaneyElementsexsuchthat X isanelementof A or X isanelementoftheNelson.
Soit's reallysayingthat a unionthatNellsaidisequaltothesetcontainingelementsexsuchthat X isanelementof a ormoresimplyequals a.
I wanttobeginthisvideobyreviewingthedefinitionofcompliments.
Sothedefinitionof a complementof a setfor, forinstance, thecomplimentof a equals.
Thesetcontainingelementsexsuchthat X isnotanelementof A I willnowbegoingovertonewalgebraiclawsforsets.
Thefirstarethelawofcompliments.
A lotofcomplimentsstatesthatifwetaketheunionof a setanditscomplement, itequals a universe.
Orifyoutakethecomplimentofthenullsetequalstheuniverseforthemoretheatersectionof a set.
Andit's complimentisthatNelson?
Andthatshouldbeclearbecauseyouhavetheset A andyouareintersectingitwitheverythingthatisnotinside a There's clearlynotgonnabeanyoverlap.
Soyou'regonnahaveanemptyset.
Thenextoneisthelawofevolution.
Soevolutionstatesthatifwehave a, uh, statehere, let's cutwillcallitSTzero, andwefeedthatstatetoe a functioninthiscase, we'regonnafeedittothecomplimentfunction.
Beintersection, Seeisequaltotheset a intersection, beintersection, see?
Andifyoulookatthedefinitionsofintersection, youcanactuallygothroughandfigureoutwhythisis, and I'm prettysureyou'llbeabletodothatbasedonthevideosthatyou'vealreadywatched.
So a intersection B isthesamethingisbeinghersection A andthesamethingis a union B isequaltobeunion, eh?
Sowelookattheactualdefinition.
Wehave a section B isdefinedasthesetcontainingelementsexsuchthat X isanelementof A and X isanelementof B, andthat's goingtoequalthissecondtrainingelementsexsuchthat X isanelementof B andexisanelementof.
Soifyoujustlookatthese, theconditionsforthesefortheseptalimitationhere, it's essentiallyjust a argumentofgrammaratthispoint.
Andifwetaketheintersectionofthatsetwith a, we'releftwiththesetcontainingelementsoneandthree, whichisgoodbecausenowwe'veshownthatthisisthat.
Now, ifwelookat a union, beintersection, See, that's whatequal a B intersection a union.
See?
Soyouanalyze, beintersectionSeethatequalsSingletonthreeandthen a B we'rehereequals 13 X betagammain a union C equalsthesetcontainelements 13 and X.
Sowhenwetaketheintersectionofthesetwosetswereleftwiththesetcontainingmoments 13 and X Andwhenwetaketheunionoftheintersection, seewith a we'releftwiththesecondtrainingelements 13 x, whichisexactlywhatwewanted.
TheMorganslawwasfoundedbyAugustustoMorgan, anditstatesthatthecomplimentof a unionistheintersectionofthecompliments.
Oritsaysthatthecomplimentoftheintersectionis a unionofthecompliments.
Sowe'regonnaprovethatthecomplimentoftheUnionof A and B isisinfactequaltothe A complimentof a intersectedwiththecomplimentofBeSoTodothis, wehavetoshowthatthecomplimentof a U.
B is a subsetof a compliment.
Intersectionbecomplimentandviceversa.
Solet's getstarted.
Supposethat X isanelementof a union.
Theconflict.
Ifthat's thecase, thenbydefinition X isnotanelementof a human.
Be.
Andif X isnotanelementof a baby, thatis, If X isnotanelementoftheregion, a Unit B, whichistheredhere, thensurely X isnotanelementof A and X isnotanelementof B.
Andifthat's thecase, thenbydefinition, X isnotanelementof a compliment, and X isnotanelementofbecompliment.
Therefore, X isnotanelementof a complimentIntersectionbecalm.
Soifwegototheothersidehere, westartbysaying, Let's let's supposethat X isanelementof a compliment.
Intersected.
Becompliment.
Then, bydefinition, X isanelementof a complimentandexeselementofbecompliment.
Bydefinitionofthecompliments, X isnotanelementof A and X isnotanelementof B andwiththesamelogic, X isnotanelementof a youneedbebecause X isnotanelementofthisregion, and X isnot a momentofthisregion.
Sosurelyit's notanelementofthetworegionscombinesandbydefinitionofcompliments, X isanelementof a unionbeaccomplishedinthisvideo.
Soifwecheckoutthesenewthingthisnewuniversalsethereitisequaltothesecondtrainingimagerssuchthat X isgreaterthanequalzeroand X islessthan I thought.
Ormoresimply, theuniverseequalsthesetcontainingimagersfrom 0 to 5.
Soifwedecomposeourfirstequation, a Intersection B equalsSingleton.
Three.
Thecomplimentof a intersection B isgoingtobeimagesfrom 05 Excluding a intersectionbeatso 012 foreignfucknowthecomplimentof a visitto 2 to 0245 because a containsthemanager's oneandthree.
I wanttobeginwithinexcerptfromTheBookofProofbyRichardHammond.
Thisis a freetextbook, andit's excellent.
Youcanfinditonline, anditcoversallfoundationalmathematicstoincludediscretemathematics, and I I cannotrecommenditenough.
SotheexcerptisLogicis a systematicwayofthinkingthatallowsustodeducenewinformationfromoldinformationandtoparsethemeaningofsentences.
Sowhatthisissayingisthatessentiallywithwithoutlogic, wewouldn't beabletodeduceormovefrompoint A topoint B andormakeclaimsfrompoint A topoint B and 100% affirmative.
Soifyouremember, a P and Q P isdefinedasrainfallfromthisguyand Q isGhanais a countryinAsia, thenpeeor Q arebothbeingtoo.
Youhavetwopossiblevalues, trueorfalse, theirprimitivepropositionssoWejustputiton a tableTrueandfallsandthentrueorfalse.
But I circledthesebecause I knowthat P istrue.
And I knowthat Q wasfalse.
Butwhen I analyze a conjunctionordisjunction, I havetoaddtwomoreroadstothetruetable.
Andthat's becauseifyoulike, ifyouimaginethisasstartingon a pathhere, youcanchoosetogoeithertrueorfalseWhatitsay I tooktrueWhen I gettotheendofthatpath, I couldbend, choosetrueorfalseagain.
So I'm stuckwithfourdifferentpossibilitieshere, the 1st 1 beingtrue, truetruefaultsandthenfallstrueandinfalsefalls.
Sowhen I actuallygotomytablehere, I havetrue, truetruefallsinyetAndthen, uh, when I stickinmyconjunction I havetrueandtrueequalstrue, trueandfalseegosfalse, trueandfalseandtruevehiclesfalseandfalseandfalseeaglespotsPeorquetruearetruelegalistrue, trueorfalseequalstruefalseortrueeaglestrueandfalsefalseeaglespotsandtheseairbythedefinitionsoftheconjunctionandthedestruction.
Ifyourecallit, impotenceis a whenyoucantake a propositionandapply a binaryoperatortothatpropositionoverandoverandwillneverchangethevalueoftheoriginalproposition.
Forinstance, P isthelogicalequivalent.
That's whatthesemeanthelogicalequivalentof P, or P, whichistheologicalequivalentof P and P.
Weknowthatbasedonthepathproblemthatwedidwhen I firstintroducedtruthtables, weknowthatthere's fourpossiblecasesforboththedisjunctionandtheconjunctionof P and Q.
Sowhenwecommutepeeing Q andwehave Q and P, it's notsurprisingthatwegettheexactsameanswer, whichisgood, becausenowwe'veshownthat P and Q islogicallyequivalentto Q and P.
Nowwe'regoingtotakethis a stepfurther, andwe'retohaveeightpropositionsareeightpossibleoutcomesforourtourfalsepathbecausewe'redealingwiththreepropositions.
Andtheconverseissimplyqueimplies P inverses, not p implies, not Q andnottoyouimpliesnot P islot, whichiscontrapositiveislogicallyequivalenttotheoriginalconditionalstatement.