Equality of functions and mods

[Jim01]

Homework Statement

Let J₃ = {0, 1, 2}, and define functions f and g from J₃ to J₃ as follows: For all x in J₃,

f(x) = (x² + x + 1) mod 3 and g(x) = (x + 2)² mod 3.

Does f = g?

Homework Equations

The Attempt at a Solution

The above is an example from the book. The section is called Equality of Functions. The procedure is given on how to solve the problem, but no explanation is given for what mod 3 means or what it is used for.

Here is the solution:

Yes, the table of values shows that f(x) = g(x) for all x in J₃.

x x² + x + 1 f(x) = (x² + x + 1)mod 3

0 1 1 mod 3 = 1
1 3 3 mod 3 = 0
2 7 7 mod 3 = 1

(x + 2)² g(x) = (x + 2)²

4 4 mod 3 = 1
9 9 mod 3 = 0
16 16 mod 3 = 1

I have poured through my elementary algebra, intermediate algebra, pre-calculus and calculus I books and can find no information on equality of functions. Is it called something else by other books? Would someone please explain this to me?

Edit: Sorry for the poor table. While it looks right when I make it, it does not format correctly when I post it.

 

[tiny-tim]

Hi Jim01!

… no explanation is given for what mod 3 means or what it is used for …

It means that they have the same remainder on division by 3.

See http://en.wikipedia.org/wiki/Modular_arithmetic" for details.

but this addition table is completely wrong (and untrue):

0 1 1 mod 3 = 1
1 3 3 mod 3 = 0
2 7 7 mod 3 = 1

it should be:

0 0 1 mod 3 = 1
1 1 1 mod 3 = 0
4 2 1 mod 3 = 1

 

[Jim01]

Hi Jim01!


It means that they have the same remainder on division by 3.

See http://en.wikipedia.org/wiki/Modular_arithmetic" for details.

Ahh. I see. They are using the word mod like it's used it in Java programming. I didn't see the connection. Thank you so much for the link. I will research it.

but this addition table is completely wrong (and untrue):

0 1 1 mod 3 = 1
1 3 3 mod 3 = 0
2 7 7 mod 3 = 1

it should be:

0 0 1 mod 3 = 1
1 1 1 mod 3 = 0
4 2 1 mod 3 = 1

Really? I double checked the book and that is what it has for that example solution. I assumed that the 0, 1, and 2 were being used in place of x. If that is the case then for x = 0, 0² + 0 + 1 = 1, for x = 1, 1² + 1 + 1 = 3, and for x = 2, 2² + 2 + 1 = 7.

I will investigate the link you gave.

 

[Mark44]

It's probably worthwhile to notice that (x + 2)² = x² + 4x + 4, and that 4 $\equiv$ 1 (mod 3).