Теория графов. Проверка решений

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Здравствуйте.
Проверьте, пожалуйста решения 4-х задач по теории графов.

Задача 1. Доказать справедливость тождества для произвольных множеств А, B и C:
$% MathType!MTEF!2!1!+- % feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WGbbGaeyOkIGSaamOqaaGaayjkaiaawMcaaiabgEna0kaadoeacqGH % 9aqpdaqadaqaaiaadgeacqGHxdaTcaWGdbaacaGLOaGaayzkaaGaey % OkIG8aaeWaaeaacaWGcbGaey41aqRaam4qaaGaayjkaiaawMcaaaaa % !4A85! \[ \left( {A \cup B} \right) \times C = \left( {A \times C} \right) \cup \left( {B \times C} \right) \] $

Докажем тождество методом двух взаимных включений.
$% MathType!MTEF!2!1!+- % feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaadaqada % qaaiaadgeacqGHQicYcaWGcbaacaGLOaGaayzkaaGaey41aqRaam4q % aiabg2da9maacmaabaWaaqGaaeaadaqadaqaaiaadIhacaGGSaGaam % yEaaGaayjkaiaawMcaaaGaayjcSdGaamiEaiabgIGiolaadgeacqGH % QicYcaWGcbGaaiilaiaadMhacqGHiiIZcaWGdbaacaGL7bGaayzFaa % aabaWaaeWaaeaacaWGbbGaey41aqRaam4qaaGaayjkaiaawMcaaiab % gQIiipaabmaabaGaamOqaiabgEna0kaadoeaaiaawIcacaGLPaaacq % GH9aqpdaGadaqaamaaeiaabaWaaeWaaeaacaWG4bGaaiilaiaadMha % aiaawIcacaGLPaaaaiaawIa7aiaadIhacqGHiiIZcaWGbbGaaiilai % aadMhacqGHiiIZcaWGdbaacaGL7bGaayzFaaGaeyOkIG8aaiWaaeaa % daabcaqaamaabmaabaGaamiEaiaacYcacaWG5baacaGLOaGaayzkaa % aacaGLiWoacaWG4bGaeyicI4SaamOqaiaacYcacaWG5bGaeyicI4Sa % am4qaaGaay5Eaiaaw2haaaaaaa!7D4E! \[ \begin{array}{l} \left( {A \cup B} \right) \times C = \left\{ {\left. {\left( {x,y} \right)} \right|x \in A \cup B,y \in C} \right\} \\ \left( {A \times C} \right) \cup \left( {B \times C} \right) = \left\{ {\left. {\left( {x,y} \right)} \right|x \in A,y \in C} \right\} \cup \left\{ {\left. {\left( {x,y} \right)} \right|x \in B,y \in C} \right\} \\ \end{array} \] $

Очевидно, что $% MathType!MTEF!2!1!+- % feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WGbbGaeyOkIGSaamOqaaGaayjkaiaawMcaaiabgEna0kaadoeacqGH % gksZdaqadaqaaiaadgeacqGHxdaTcaWGdbaacaGLOaGaayzkaaGaey % OkIG8aaeWaaeaacaWGcbGaey41aqRaam4qaaGaayjkaiaawMcaaaaa % !4B80! \[ \left( {A \cup B} \right) \times C \subseteq \left( {A \times C} \right) \cup \left( {B \times C} \right) \] $ и $% MathType!MTEF!2!1!+- % feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WGbbGaey41aqRaam4qaaGaayjkaiaawMcaaiabgQIiipaabmaabaGa % amOqaiabgEna0kaadoeaaiaawIcacaGLPaaacqGHgksZdaqadaqaai % aadgeacqGHQicYcaWGdbaacaGLOaGaayzkaaGaey41aqRaam4qaaaa % !4B81! \[ \left( {A \times C} \right) \cup \left( {B \times C} \right) \subseteq \left( {A \cup C} \right) \times C \] $

Итак, тождество доказано.

Задача 2. . Доказать, что множества Х и Y равномощны, построив взаимно-однозначное соответствие между ними.
Х=[–1,1], Y=(–1,1).


Доказательство



$% MathType!MTEF!2!1!+- % feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaaeiEaaGaayjkaiaawMcaaiabg2da9maaceaaeaqabeaacaqG % 4bGaaeilaiaaykW7caaMc8UaaGPaVlaabIhacqGHGjsUcaaIWaGaai % ilamaayaaabaGaaeynaiaab6cacaqGUaGaaeOlaiaabwdaaSqaaKqz % afGaaeOBaaGccaGL44pacaqGSaGaaGPaVlaabccacaqGWaGaaeilam % aayaaabaGaae4naiaab6cacaqGUaGaaeOlaiaabEdaaSqaaKqzafGa % aeOBaaGccaGL44pacaGGSaGaaGPaVlaaykW7caqGUbGaeyicI4SaaC % OtaiaacUdaaeaacaqGTaGaaeymaiaabYcacaaMc8UaaGPaVlaaykW7 % caqG4bGaeyypa0JaaeimaiaabYcacaqG1aGaae4oaaqaaiaabgdaca % qGSaGaaGPaVlaaykW7caaMc8UaaeiEaiabg2da9iaabcdacaqGSaGa % ae4naiaabUdaaeaacaaIWaGaaiilamaayaaabaGaaeynaiaab6caca % qGUaGaaeOlaiaabwdaaSqaaKqzafGaaeOBaaGccaGL44pacaqGSaGa % aGPaVlaaykW7caaMc8UaaeiEaiabg2da9iaaicdacaGGSaWaaGbaae % aacaqG1aGaaeOlaiaab6cacaqGUaGaaeynaaWcbaqcLbuacaqGUbGa % ey4kaSIaaGymaaGccaGL44pacaqGSaGaaGPaVlaaykW7caaMc8Uaae % OBaiabgIGiolaah6eacaGG7aaabaGaaGimaiaacYcadaagaaqaaiaa % bEdacaqGUaGaaeOlaiaab6cacaqG3aaaleaajugqbiaab6gaaOGaay % jo+dGaaeilaiaaykW7caaMc8UaaGPaVlaabIhacqGH9aqpcaaIWaGa % aiilamaayaaabaGaae4naiaab6cacaqGUaGaaeOlaiaabEdaaSqaaK % qzafGaaeOBaiabgUcaRiaaigdaaOGaayjo+dGaaeilaiaaykW7caaM % c8UaaGPaVlaab6gacqGHiiIZcaWHobGaaiOlaaaacaGL7baacaWGMb % GaaiOoaiaacIcacqGHsislcaaIXaGaaiilaiaabccacaaIXaGaaiyk % aiabgsziRkaacUfacqGHsislcaaIXaGaaiilaiaabccacaaIXaGaai % yxaiaac6caaaa!CC50! \[ f\left( {\rm{x}} \right) = \left\{ \begin{array}{l} {\rm{x}}{\rm{,}}\,\,\,{\rm{x}} \ne 0,\underbrace {{\rm{5}}...{\rm{5}}}_{\rm{n}}{\rm{,}}\,{\rm{ 0}}{\rm{,}}\underbrace {{\rm{7}}...{\rm{7}}}_{\rm{n}},\,\,{\rm{n}} \in {\bf{N}}; \\ {\rm{ - 1}}{\rm{,}}\,\,\,{\rm{x}} = {\rm{0}}{\rm{,5;}} \\ {\rm{1}}{\rm{,}}\,\,\,{\rm{x}} = {\rm{0}}{\rm{,7;}} \\ 0,\underbrace {{\rm{5}}...{\rm{5}}}_{\rm{n}}{\rm{,}}\,\,\,{\rm{x}} = 0,\underbrace {{\rm{5}}...{\rm{5}}}_{{\rm{n}} + 1}{\rm{,}}\,\,\,{\rm{n}} \in {\bf{N}}; \\ 0,\underbrace {{\rm{7}}...{\rm{7}}}_{\rm{n}}{\rm{,}}\,\,\,{\rm{x}} = 0,\underbrace {{\rm{7}}...{\rm{7}}}_{{\rm{n}} + 1}{\rm{,}}\,\,\,{\rm{n}} \in {\bf{N}}. \\ \end{array} \right.f:( - 1,{\rm{ }}1) \leftrightarrow [ - 1,{\rm{ }}1]. \] $

Задача 3.

Даны три вещественных функции:
$% MathType!MTEF!2!1!+- % feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGMb % GaaiikaiaadIhacaGGPaGaeyypa0JaamiEamaaCaaaleqabaGaaGin % aaaakiabgUcaRiaaigdacaaI2aGaai4oaaqaaiaadEgacaGGOaGaam % iEaiaacMcacqGH9aqpcaaI1aGaamiEaiabgUcaRiaaikdacaGG7aaa % baGaamiAaiaacIcacaWG4bGaaiykaiabg2da9iaaiEdadaahaaWcbe % qaaiaadIhaaaGccqGHsislcaaIXaGaaGyoaiaacUdaaaaa!5113! \[ \begin{array}{l} f(x) = x^4 + 16; \\ g(x) = 5x + 2; \\ h(x) = 7^x - 19; \\ \end{array} \] $


1) Найти заданные композиции функций: fgh, hgf, ffh.
2) Являются ли f, g, h инъекциями, сюръекциями, биекциями на R?
3) Найти обратные функции к f, g, h. Если функции со своими областями определения обратных не имеют, то найти обратные функции к их сужениям.

1. 1) композиции функций:
fgh - $% MathType!MTEF!2!1!+- % feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacI % cacaaI1aGaeyyXICTaaiikaiaaiEdadaahaaWcbeqaaiaadIhaaaGc % cqGHsislcaaIXaGaaGyoaiaacMcacqGHRaWkcaaIYaGaaiykaiabg2 % da9iaacIcacaaI1aGaeyyXICTaaiikaiaaiEdadaahaaWcbeqaaiaa % dIhaaaGccqGHsislcaaIXaGaaGyoaiaacMcacqGHRaWkcaaIYaGaai % ykamaaCaaaleqabaGaaGinaaaakiabgUcaRiaaigdacaaI2aGaai4o % aaaa!5361! \[ f(5 \cdot (7^x - 19) + 2) = (5 \cdot (7^x - 19) + 2)^4 + 16; \] $

hgf - $% MathType!MTEF!2!1!+- % feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiAaiaacI % cacaaI1aGaeyyXICTaaiikaiaadIhadaahaaWcbeqaaiaaisdaaaGc % cqGHRaWkcaaIXaGaaGOnaiaacMcacqGHRaWkcaaIYaGaaiykaiabg2 % da9iaaiEdadaahaaWcbeqaaiaacIcacaaI1aGaeyyXICTaaiikaiaa % dIhadaahaaadbeqaaiaaisdaaaWccqGHRaWkcaaIXaGaaGOnaiaacM % cacqGHRaWkcaaIYaGaaiykaaaakiabgkHiTiaaigdacaaI5aGaai4o % aaaa!5354! \[ h(5 \cdot (x^4 + 16) + 2) = 7^{(5 \cdot (x^4 + 16) + 2)} - 19; \] $

ffh - $% MathType!MTEF!2!1!+- % feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacI % cacaGGOaGaaG4namaaCaaaleqabaGaamiEaaaakiabgkHiTiaaigda % caaI5aGaaiykamaaCaaaleqabaGaaGinaaaakiabgUcaRiaaigdaca % aI2aGaaiykaiabg2da9iaacIcacaGGOaGaaG4namaaCaaaleqabaGa % amiEaaaakiabgkHiTiaaigdacaaI5aGaaiykamaaCaaaleqabaGaaG % inaaaakiabgUcaRiaaigdacaaI2aGaaiykamaaCaaaleqabaGaaGin % aaaakiabgUcaRiaaigdacaaI2aGaai4oaaaa!50B7! \[ f((7^x - 19)^4 + 16) = ((7^x - 19)^4 + 16)^4 + 16; \] $

2. Функция f не является инъективной, т.к условие
если $% MathType!MTEF!2!1!+- % feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadAgacaGGOaGaamiEamaaBaaaleaacaaIXaaabeaakiaacMca % caaMc8UaaGPaVlaabIdbcaaMc8UaaGPaVlaadMhacqGH9aqpcaWGMb % GaaiikaiaadIhadaWgaaWcbaGaaGOmaaqabaGccaGGPaGaaiilaiaa % ykW7caaMc8UaaGPaVlaaykW7caqGcrGaaeOpeiaaykW7caaMc8UaaG % PaVlaadIhadaWgaaWcbaGaaGymaaqabaGccqGH9aqpcaWG4bWaaSba % aSqaaiaaikdaaeqaaaaa!5B29! \[ y = f(x_1 )\,\,{\rm{}}\,\,y = f(x_2 ),\,\,\,\,{\rm{}}\,\,\,x_1 = x_2 \] $ не выполняется , например f(-1) = f(1)

Функция f не является сюръективной, так как условие $% MathType!MTEF!2!1!+- % feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiaIiIaaG % PaVlaadMhacqGHiiIZcaWGzbGaaGPaVlabgoGiKiaaykW7caWG4bGa % aGPaVlabgIGiolaadIfacaaMc8UaaGPaVlaadMhacqGH9aqpcaWGMb % GaaiikaiaadIhacaGGPaaaaa!4CD7! \[ \forall \,y \in Y\,\exists \,x\, \in X\,\,y = f(x) \] $
не выполняется , например не существует x, для которых f < 0

Так как функция f не является инъективной и сюръективной, то она не является биективной.


Функция g является инъективной, т.к условие
если $% MathType!MTEF!2!1!+- % feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadAgacaGGOaGaamiEamaaBaaaleaacaaIXaaabeaakiaacMca % caaMc8UaaGPaVlaabIdbcaaMc8UaaGPaVlaadMhacqGH9aqpcaWGMb % GaaiikaiaadIhadaWgaaWcbaGaaGOmaaqabaGccaGGPaGaaiilaiaa % ykW7caaMc8UaaGPaVlaaykW7caqGcrGaaeOpeiaaykW7caaMc8UaaG % PaVlaadIhadaWgaaWcbaGaaGymaaqabaGccqGH9aqpcaWG4bWaaSba % aSqaaiaaikdaaeqaaaaa!5B29! \[ y = f(x_1 )\,\,{\rm{}}\,\,y = f(x_2 ),\,\,\,\,{\rm{}}\,\,\,x_1 = x_2 \] $ выполняется

Функция g является сюръективной, так как условие $% MathType!MTEF!2!1!+- % feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiaIiIaaG % PaVlaadMhacqGHiiIZcaWGzbGaaGPaVlabgoGiKiaaykW7caWG4bGa % aGPaVlabgIGiolaadIfacaaMc8UaaGPaVlaadMhacqGH9aqpcaWGMb % GaaiikaiaadIhacaGGPaaaaa!4CD7! \[ \forall \,y \in Y\,\exists \,x\, \in X\,\,y = f(x) \] $ выполняется

Так как функция g является инъективной и сюръективной, то она является биективной (т.е. взаимнооднозначной)


Функция h является инъективной, т.к условие
если $% MathType!MTEF!2!1!+- % feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadAgacaGGOaGaamiEamaaBaaaleaacaaIXaaabeaakiaacMca % caaMc8UaaGPaVlaabIdbcaaMc8UaaGPaVlaadMhacqGH9aqpcaWGMb % GaaiikaiaadIhadaWgaaWcbaGaaGOmaaqabaGccaGGPaGaaiilaiaa % ykW7caaMc8UaaGPaVlaaykW7caqGcrGaaeOpeiaaykW7caaMc8UaaG % PaVlaadIhadaWgaaWcbaGaaGymaaqabaGccqGH9aqpcaWG4bWaaSba % aSqaaiaaikdaaeqaaaaa!5B29! \[ y = f(x_1 )\,\,{\rm{}}\,\,y = f(x_2 ),\,\,\,\,{\rm{}}\,\,\,x_1 = x_2 \] $ выполняется

Функция h не является сюръективной, так как условие
$% MathType!MTEF!2!1!+- % feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiaIiIaaG % PaVlaadMhacqGHiiIZcaWGzbGaaGPaVlabgoGiKiaaykW7caWG4bGa % aGPaVlabgIGiolaadIfacaaMc8UaaGPaVlaadMhacqGH9aqpcaWGMb % GaaiikaiaadIhacaGGPaaaaa!4CD7! \[ \forall \,y \in Y\,\exists \,x\, \in X\,\,y = f(x) \] $ не выполняется , например не существует x, для которых h < -19

Так как функция h не является сюръективной, то она не является биективной

3. Найдем обратные функции:

$% MathType!MTEF!2!1!+- % feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacI % cacaWG4bGaaiykaiabg2da9iaadIhadaahaaWcbeqaaiaaisdaaaGc % cqGHRaWkcaaIXaGaaGOnaiaacUdaaaa!3F42! \[ f(x) = x^4 + 16; \] $ $% MathType!MTEF!2!1!+- % feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaaCa % aaleqabaGaeyOeI0IaaGymaaaakiaacIcacaWG4bGaaiykaiabg2da % 9maabmaabaGaamiEaiabgkHiTiaaigdacaaI2aaacaGLOaGaayzkaa % WaaWbaaSqabeaadaWcaaqaaiaaigdaaeaacaaI0aaaaaaakiaacUda % aaa!4380! \[ f^{ - 1} (x) = \left( {x - 16} \right)^{\frac{1}{4}} ; \] $

$% MathType!MTEF!2!1!+- % feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4zaiaacI % cacaWG4bGaaiykaiabg2da9iaaiwdacaWG4bGaey4kaSIaaGOmaiaa % cUdaaaa!3E4E! \[ g(x) = 5x + 2; \] $ $% MathType!MTEF!2!1!+- % feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4zamaaCa % aaleqabaGaeyOeI0IaaGymaaaakiaacIcacaWG4bGaaiykaiabg2da % 9maalaaabaWaaeWaaeaacaWG4bGaeyOeI0IaaGOmaaGaayjkaiaawM % caaaqaaiaaiwdaaaGaai4oaaaa!41D1! \[ g^{ - 1} (x) = \frac{{\left( {x - 2} \right)}}{5}; \] $

$% MathType!MTEF!2!1!+- % feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiAaiaacI % cacaWG4bGaaiykaiabg2da9iaaiEdadaahaaWcbeqaaiaadIhaaaGc % cqGHsislcaaIXaGaaGyoaiaacUdaaaa!3F55! \[ h(x) = 7^x - 19; \] $ $% MathType!MTEF!2!1!+- % feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiAamaaCa % aaleqabaGaeyOeI0IaaGymaaaakiaacIcacaWG4bGaaiykaiabg2da % 9iGacYgacaGGVbGaai4zamaaBaaaleaacaaI3aaabeaakiaacIcaca % WG4bGaey4kaSIaaGymaiaaiMdacaGGPaaaaa!4492! \[ h^{ - 1} (x) = \log _7 (x + 19) \] $

Для 1-ой функции сужение x>=16
Для 3-ей функции сужение x > -19

Задача 4.

Является ли антисимметричным бинарное отношение $% MathType!MTEF!2!1!+- % feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOuamaaCa % aaleqabaGaeyOeI0IaaGymaaaaaaa!3899! \[ R^{ - 1} \] $ если отношение R антисимметрично? В случае отрицательного ответа необходимо привести конкретный пример.

Решение:
Обратное отношение:
Отношение называется обратным к отношению Г(обозначение $% MathType!MTEF!2!1!+- % feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeu4KdC0aaW % baaSqabeaacqGHsislcaaIXaaaaOGaaiilaiabfo5ahnaaCaaaleqa % baGaeyOeI0IaaGymaaaakiabgAOinlaadIfadaahaaWcbeqaaiaaik % daaaGccaWG-qGaamiqeiaadIdbcqqHtoWrcqGHgksZcaWGybWaaWba % aSqabeaacaaIYaaaaaaa!487D! \[ \Gamma ^{ - 1} ,\Gamma ^{ - 1} \subseteq X^2 \Gamma \subseteq X^2 \] $), если $% MathType!MTEF!2!1!+- % feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEamaaBa % aaleaacaWGQbaabeaakiabfo5ahnaaCaaaleqabaGaeyOeI0IaaGym % aaaakiaadIhadaWgaaWcbaGaamyAaaqabaaaaa!3D6D! \[ x_j \Gamma ^{ - 1} x_i \] $ тогда и только тогда, когда $% MathType!MTEF!2!1!+- % feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEamaaBa % aaleaacaWGPbaabeaakiabfo5ahjaadIhadaWgaaWcbaGaamOAaaqa % baaaaa!3B8E! \[ x_i \Gamma x_j \] $ для всех $% MathType!MTEF!2!1!+- % feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaadI % hadaWgaaWcbaGaamyAaaqabaGccaGGSaGaamiEamaaBaaaleaacaWG % QbaabeaakiaacMcacqGHiiIZcqqHtoWraaa!3F25! \[ (x_i ,x_j ) \in \Gamma \] $

Антисимметричность: хГу и уГх
Значит х=у;

Меняя местами x и y по определению обратного отношения для
антисимметричности получим:
yГx и xГy
Значит y=x;
т.е. свойство антисимметричности сохранилось.
Примером могут быть отношения
<= (антисимметрично) и обратное >= (тоже антисимметрично)

[PAV]

Не вполне ясно, почему тема называется "про графы"...

По поводу первой задачи может последовать вопрос - а докажите-ка это "очевидно". Мне кажется, что более наглядно рассуждать так: взять конкретный элемент из множества из левой части и показать, что он принадлежит правой части. А затем наоборот.

-- Пн дек 14, 2009 22:25:19 --

Второе правильно, вполне наглядное построение.

-- Пн дек 14, 2009 22:27:47 --

Остальные долго проверять, может кто-то еще посмотрит.