Question

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

1) f(x) = x2 mod 8
2) f(x) = x3 mod 8
3)f(x)=(x^3-8) mod 8

4)f(x)=(x^3+2x) mod 8

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

Answer 1

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\\eq 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\\eq 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\\eq 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)\\eq f(1)\\eq f(2) \\eq f(3)`, then f is one-to-one

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