Question

Which of the following functions f : {0, 1, 2, 3} ! {0, 1, . . . , 7} are one-to-one?

Answer 1

Answer:

1. No.

$f(0)=0\\1^2=1; \text{then } f(1)=1\\2^2=4\equiv 4 \text{ mod 8}; \text{then } f(2)=4\\3^2=9\equiv 1 \text{mod 8 }; \text{then } f(3)=1$

Since f(1)=f(3) and $1\neq 3$ then f isn't one-to-one.

2. No

$f(0)=0\\1^3=1\equiv 1\text{ mod 8}; \text{then } f(1)=1\\2^3=8\equiv 0\text{ mod 8}; \text{then } f(2)=0\\3^3=27\equiv 3 \text{ mod 8}; \text{then } f(3)=3$

Since f(0)=f(2) and $0\neq 2$ then f isn't one-to-one.

3. No

$0^3-8=-8\equiv 0\text{ mod 8}; \text{then } f(0)=0\\1^3-8=-7\equiv 1\text{ mod 8}; \text{then } f(1)=1\\2^3-8=0\equiv 0 \text{ mod 8}; \text{then } f(2)=0\\3^3-8=27-8=19\equiv 3 \text{ mod 8}; \text{then } f(3)=3$

Since f(0)=f(2) and $0\neq 2$ then f isn't one-to-one.

4. Yes

$0^3+2*0=0; \text{then } f(0)=0\\1^3+2*1=3\equiv 3\text{ mod 8}; \text{then } f(1)=3\\2^3+2*2=8+4=12\equiv 4 \text{ mod 8}; \text{then } f(2)=4\\3^3+2*3=27+6=33\equiv 1\text{ mod 8}; \text{then } f(3)=1$

Since $f(0)\neq f(1)\neq f(2) \neq f(3)$, then f is one-to-one

5. Since f(1)=f(3) and $1\neq 3$ then, f isn't one-to-one