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Membuktikan Fungsi Injektif, Surjektif, dan Bijektif

Kita akan menerapkan kaidah-kaidah yang telah kita pelajari sebelumnya (dalam tulisan Cara Membuktikan dalam Matematika) untuk membuktikan apakah suatu fungsi termasuk fungsi injektif, surjektif, atau bijektif. Pertama, kita harus memahami definisi-definisi dari sifat-sifat fungsi tersebut. Saya berikan dua referensi berikut ini dari blog Matematika Ku Bisa. Supaya dalam tulisan ini, kita cepat langsung ke contoh soal yang pemaca inginkan.

  1. Definisi Fungsi dan Fungsi-Fungsi Khusus (Fungsi Injektif, Surjektif, dan Bijektif)

Tapi bagi Anda yang tidak mau membacanya, saya berikan definisi dalam tulisan ini saja.

Definisi Fungsi

Secara intuisi, suatu fungsi dari himpunan A ke himpunan B adalah relasi (aturan korespondensi) yang memasangkan masing-masing unsur x di A secara tunggal dengan unsur f(x) di B.

Definisi di atas mungkin tidak jelas, dikarenakan ketidakjelasan frase “aturan korespondensi”. Untuk mengatasi hal ini, kita akan mendefenisikan fungsi dengan menggunakan himpunan. Dengan definisi tersebut, dapat saja kita kehilangan kandungan intuitif definisi sebelumnya, tetapi kita akan mendapat kejelasan. Ide dasar pendefinisian ini adalah memikirkan gambar dari suatu fungsi, yaitu suatu korelasi pasangan berurut. Bila kita perhatikan tidak setiap koleksi pasangan berurut merupakan gambar suatu fungsi, karena sekali unsur pertama dalam pasangan berurut diambil, unsur keduanya ditentukan secara tunggal. Berikut ini diberikan definisi tersebut.

Misalkan A dan B himpunan. Suatu fungsi dari A ke B adalah himpunan pasangan berurut f di $A \times B$ sedemikian hingga untuk masing-masing a $\in$ A terdapat b $\in$ B yang tunggal, yaitu jika (a,b) dan (a,b’) $\in$ f maka b=b’. Himpunan A dari unsur-unsur pertama dari f disebut daerah asal (domain) dari f, dan dituliskan D(f). Sedangkan unsur-unsur di B yang menjadi unsur kedua dari f disebut “range” dari f dan dituliskan dengan R(f).

Notasi $f: \ A \rightarrow B$ menunjukkan bahwa f suatu fungsi dari A ke B; akan sering kita katakan bahwa f suatu pemetaan dari A ke dalam B atau f memetakan A ke dalam B. Bila (a,b) suatu unsur di f, sering ditulis dengan $b=f(a)$ daripada (a,b) $\in$ f. Dalam hal ini b merupakan nilai f di titik a, atau peta a terhadap f.

Fungsi Injektif. Suatu fungsi $f: \ A \rightarrow B$ dikatakan injektif (satu-satu) jika dan hanya jika $x_1 \neq x_2$ mengakibatkan $f(x_1) \neq f(x_2)$. Bila f satu-satu, kita katakan f injeksi. Karena implikasi $x_1 \neq x_2 \Rightarrow f(x_1) \neq f(x_2)$ ekivalen dengan implikasi $f(x_1)=f(x_2) \Rightarrow x_1=x_2$ maka secara ekivalen f fungsi injektif jika dan hanya jika $f(x_1)=f(x_2)$ mengakibatkan $x_1=x_2$.

Fungsi Surjektif. Suatu fungsi $f: \ A \rightarrow B$ dikatakan surjektif (atau memetakan A pada B) bila f(A)=B yang berarti range f adalah semua anggota himpunan B sehingga secara ekivalen $f: \ A \rightarrow B$ dikatakan surjektif bila untuk setiap $y \in B$ terdapat $x \in A$ sehingga f(x)=y. Bila f surjektif, kita sebut f suatu surjeksi.

Fungsi Bijektif. Suatu fungsi $f: \ A \rightarrow B$ dikatakan bijektif (satu-satu dan pada) bila bersifat injektif dan surjektif. Bila f bijektif, kita sebut f suatu bijeksi.

Kita tuliskan definisi masing-masing definisi di atas dengan menggunakan simbol-simbol logika sebagai berikut, yang akan digunakan secara praktis dalam membuktikan hal yang diminta (apakah f(x) suatu fungsi, fungsi injektif, fungsi surjektif, atau fungsi bijektif):

Misalkan $f: \ X \rightarrow Y$

  • f adalah fungsi jika dan hanya jika  $\forall x_1, \ x_2 \in X$, $x_1=x_2 \Rightarrow f(x_1)=f(x_2)$.
  • f adalah fungsi injektif jika dan hanya jika  $\forall x_1, \ x_2 \in X$, $f(x_1)=f(x_2)  \Rightarrow x_1=x_2$.
  • f adalah fungsi surjektif jika dan hanya jika  $\forall y \in Y$, $\exists x$ sedemikian sehingga y=f(x).

Contoh soal: Selidiki $f(x)=x^2-1$

  • Pertama-tama, kita tunjukkan dulu apakah implikasi $x_1=x_2 \Rightarrow f(x_1)=f(x_2)$ bernilai benar.


$\begin{align} f(x_1) &=f(x_2) \\ (x_1)^2 – 1 &= (x_2)^2 – 1 \\ x_1^2 &= x_2^2 \end{align}$

Karena $x_1=x_2$, mengakibatkan $f(x_1)=f(x_2)$ maka terbukti $f(x)=x^2-1$ adalah fungsi.

  • Kedua, kita tunjukkan apakah implikasi $f(x_1)=f(x_2)  \Rightarrow x_1=x_2$ bernilai benar.

Berdasarkan yang pertama, dari $f(x_1)=f(x_2)$ diperoleh $x_1^2=x_2^2$. Dari $x_1^2=x_2^2$, tidak dicapai kesimpulan bahwa $x_1=x_2$, kecuali jika $x_1$ dan $x_2$ memiliki tanda yang sama. Sebagai contoh $x_1=2$ dan $x_2=-2$, $(2)^2 =(-2)^2 $ tetapi $2 \neq –2$.

Jadi, $f(x)=x^2-1$ bukan fungsi injektif.

Namun, apabila domain dari fungsi tersebut kita batasi, $(- \infty,0]$ atau $[0, \infty)$, maka fungsi terseut akan merupakan fungsi injektif pada domain $(- \infty,0]$ atau $[0, \infty)$.

  • Ketiga, kita tunjukkan apakah $\forall y \in Y$, $\exists x$ sedemikian sehingga y=f(x) bernilai benar.

Karena dalam penulisan suatu fungsi, misalkan tidak dituliskan doaminnya, berarti domain fungsinya domain alami yaitu domain dimana saja fungsi tersebut terdefinisi. Dalam soal ini, f(x) terdefinisi pada bilangan real sehingga domainnya adalah x bilangan real. Kodomain dari fungsi tersebut adalah bilangan real juga. Jadi, untuk membuktikan bahwa f(x) tersebut surjeksi maka harus ditunjukkan untuk semua $y \in R$ terdapat $x \in R$ sehingga $y=x^2-1$.


$x= \pm \sqrt{y+1}$.

Ambil sebarang $y \in R$. Kemungkinannya $y>0$, $y=0$, dan $y<0$. Untuk y tak-negatif, terdapat $x= \pm \sqrt{y+1}$ sehingga $y=x^2-1$. Tapi, untuk $y<0$ tidak terdapat x pada bilangan real.

Jadi, $f(x)=x^2-1$ juga bukan fungsi surjektif.

Nampak, jika kita batasi kodomainnya untuk bilangan real yang tak negatif maka fungsi tersebut akan merupakan fungsi surjektif.

  • Terakhir, karena $f(x)=x^2-1$ bukan fungsi injektif dan/atau fungsi surjektif maka $f(x)=x^2-1$ bukan fungsi bijektif. Namun, jika kita batasi domain dan kodomainnya maka akan menjadi fungsi bijektif. $f(x)=x^2-1$ merupakan fungsi bijektif apabila domainnya $(- \infty,0]$ atau $[0, \infty)$, dan kodomainnya adalah $[0, \infty)$.

Referensi: Buku Analisis Real karya Bartle dan Sherbert (edisi 4) dan terjemahannya.

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