MATH SOLVE

3 months ago

Q:
# Which of the following functions f : {0, 1, 2, 3} ! {0, 1, . . . , 7} are one-to-one?1) f(x) = x2 mod 82) f(x) = x3 mod 83)f(x)=(x^3-8) mod 84)f(x)=(x^3+2x) mod 85) f(0) = 3, f(1) = 1, f(2) = 4, f(3) = 1

Accepted Solution

A:

Answer:1. No.[tex]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[/tex]Since f(1)=f(3) and [tex]1\neq 3[/tex] then f isn't one-to-one.2. No[tex]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[/tex]Since f(0)=f(2) and [tex]0\neq 2[/tex] then f isn't one-to-one.3. No[tex]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\\[/tex]Since f(0)=f(2) and [tex]0\neq 2[/tex] then f isn't one-to-one.4. Yes[tex]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[/tex]Since [tex]f(0)\neq f(1)\neq f(2) \neq f(3)[/tex], then f is one-to-one5. Since f(1)=f(3) and [tex]1\neq 3[/tex] then, f isn't one-to-one