Ib txoj haujlwm yog bijective yog tias nws yog ob qho tib si txhaj tshuaj thiab. Ib txoj haujlwm bijective tseem hu ua bijection lossis ib-rau-ib tsab ntawv. Ib txoj haujlwm yog bijective yog thiab tsuas yog tias txhua daim duab ua tau yog mapped los ntawm ib qho kev sib cav.
Koj paub li cas yog tias txoj haujlwm yog Bijective?
A muaj nuj nqi yog hais tias yog bijective lossis bijection, yog tias muaj nuj nqi f: A → B txaus siab rau ob qho kev txhaj tshuaj (ib-rau-ib txoj haujlwm) thiab kev ua haujlwm (mus rau function) cov khoom. Nws txhais tau hais tias txhua lub ntsiab lus "b" hauv codomain B, muaj ib lub ntsiab lus "a" hauv lub npe A xws li f(a)=b.
Koj ua pov thawj li cas ib txoj haujlwm tsis yog bijective?
Kev qhia ua haujlwm tsis yog qhov tseeb peb yuav tsum show f(A)=B. Txij li qhov kev ua haujlwm tau zoo yuav tsum muaj f (A) ⊆ B, peb yuav tsum qhia B ⊆ f (A). Yog li los qhia txog kev ua haujlwm tsis yog qhov tseeb nws yog qhov txaus los nrhiav ib lub caij hauv lub codomain uas tsis yog cov duab ntawm ib lub caij ntawm lub npe.
Puas 2 x 3 yog kev ua haujlwm?
F is bijective !Yog li no 2x−3=2y−3. Peb tuaj yeem tshem tawm qhov 3 thiab faib los ntawm 2, ces peb tau x=y. … Yog li ntawd: F yog bijective!
Puas yog bijective muaj nuj nqi monotonic?
Txhua yam kev ua haujlwm tsis tu ncua los ntawm R rau R yog nruj me ntsis monotonic.
