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Discrete math

Which of the following functions f : {0, 1, 2, 3} ! {0, 1, 2, 3} are onto?
1) f(x) = x
2) f(x) = x^2 mod 4
3) f(x) = x^2 - x mod 4

4) f(0) = 3, f(1) = 2, f(2) = 1, f(3) = 0
5) f(0) = 1, f(1) = 2, f(2) = 1, f(3) = 2

Respuesta :

steffaniasierrag

Answer:

1) yes

2) No

3) No

4) yes

5) No

Step-by-step explanation:

1) f(0)=0, f(1)=1, f(2)=2, f(3)=3, then [tex]f[\{0, 1, 2, 3\}]=\{0, 1, 2, 3\}[/tex]

2) f(0)=0, [tex]1^2=1\equiv 1 \text{mod 4}[/tex], then f(1)=1; [tex]2^2=4\equiv 0 \text{ mod 4}[/tex], then f(2)=0; [tex]3^2=9\equiv 1 \text{mod 4}[/tex], then f(3)=1

Then [tex]f[\{0, 1, 2, 3\}]=\{0, 1, 3\}[/tex], this means that f isn't onto.

3.

  • f(0)=0;
  • [tex]1^1-1=0[/tex], then f(1)=0
  • [tex]2^2-2=2[/tex], then f(2)=2
  • [tex]3^2-3=6\equiv 2 \text{ mod 4}[/tex], then f(3)=2

Then  [tex]f[\{0, 1, 2, 3\}]=\{0, 2\}[/tex], this means that f isn't onto.

4.  [tex]f[\{0, 1, 2, 3\}]=\{0, 1, 2,3\}[/tex], then f is onto.

5. [tex]f[\{0, 1, 2, 3\}]=\{1, 2\}[/tex], this means that f isn't onto.

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