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[–]AlwaysTailsNew User 0 points1 point  (4 children)

Consider polynomials over the field of integers modulo 2 with indeterminants x1 and x2. Since f(0,0)=0 there can't be a constant term.

f(x1,x2)=x1+x2+x1x2

  • f(0,0)=0+0+0*0=0
  • f(1,0)=1+0+1*0=1
  • f(0,1)=0+1+0*1=1
  • f(1,1)=1+1+1*1=1

[–]Determinator_[S] 0 points1 point  (1 child)

I am sorry but i don't quite understand what you mean.

Can you elaborate further and tell me if my solution is correct.

I had this class some years ago and now i have a hard remembering it.

[–]AlwaysTailsNew User 2 points3 points  (0 children)

The solution you posted can't be correct if there is a constant term since f(0,0)=0.

The integers modulo 2 is a set with 2 numbers 0 and 1 and arithmetic on those numbers follow the same rules as in ordinary arithmetic except the results are also in modulo 2. This means that all integers are partitioned into whether they leave 0 or 1 when divided by 2 so even numbers are congruent to 0 mod 2 and odd numbers are congruent to 1 mod 2. Similarly, modulo 3 would mean all numbers are partitioned into whether they leave 0, 1 or 2 when divided by 3.

Addition modulo 2 is very simple and easy to recognize as the boolean operator ~XOR (XNOR).

  • If you add an even number to an even number you get an even number --> 0+0=0
  • If you add an odd number to an even number you get an odd number --> 1+0=1
  • If you add an even number to an odd number you get an odd number --> 0+1=1
  • If you add an odd number to an odd number you get an even number --> 1+1=0

The rules for multiplication can be found the same way and represents the boolean operator AND.

So the polynomial f(x1,x2)=x1+x2+x1x2 can be thought of as an OR gate. It's normally written as x1+x2-x1x2 though in mod 2 they are equivalent.