Question

Discrete math

Answer 1

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 $f[\{0, 1, 2, 3\}]=\{0, 1, 2, 3\}$

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

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

3.

• f(0)=0;

• $1^1-1=0$, then f(1)=0
• $2^2-2=2$, then f(2)=2
• $3^2-3=6\equiv 2 \text{ mod 4}$, then f(3)=2

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

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

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