Example
With Mod 2 multiplication, we ignore the carry-over from the addition. For example let's try 10101 x 1111:
Binary form: 10101 x 1111 x**4+x**2+1 x**3+x**2+x+1 Working out: 10101 x1111 -------- 10101 101010 1010100 10101000 -------- 11000011 ========
The (mod 2) multiply of 101001 and 10110 gives 1001100110:
101001 is \(x^5+x^3+1\) and 10110 gives \(x^4+x^2+x\). And so we have:
(\(x^4+x^2+x\)) \times (\(x^5+x^3+1\) )
This gives:
\(x^9+x^7+x^4+x^7+x^5+x^2+x^6+x^4+x\)
Thus gives:
\(x^9+x^6+x^5+x^2+x\)
As in mod 2, \(x^7+x^7 =0\).
Thus \(x^9+x^6+x^5+x^2+x\) is 1001100110.