text: # Modular arithmetic
  
  • Feb 28th 2010, 09:03 AM
  integral
  Modular arithmetic
  $\displaystyle \textrm{I\; am\; trying\; to\; learn\; modular\; arithmetic,\; and\; this\; right \; here\; stumped\; me.}$
  $\displaystyle 10\cdot(-13)+27\cdot 6\equiv-122\mathrm{mod}2$
  $\displaystyle \textrm{I\; thought\; It\; would\; go\; like\; this:}$
  $\displaystyle 10\cdot(-13)+27\cdot 6\; \; \; \textrm{Can be reduced to in mod}2$
  $\displaystyle 0\cdot(1)+1\cdot 0=0\mathrm{mod}2$
  
  $\displaystyle \textrm{Please explain this to me someone}$
  • Feb 28th 2010, 10:45 AM
  Bacterius
  Hi Integral,
  
  Suppose you have :
  $\displaystyle 10 \times (-13) + 27 \times 6$
  
  This can be written as :
  
  $\displaystyle (2 \times 5 + 0) \times (2 \times (-6) + 1) + (2 \times 13 + 1) \times (2 \times 3 + 0)$
  
  Note that modulo two, adding two doesn't change the residue, and multiplying by two makes the residue zero. So this simplifies to :
  
  $\displaystyle 0 \times 1 + 1 \times 0$ (mod 2)
  
  And therefore the whole expression is equivalent to zero modulo 2 (it is even). These are just modular arithmetic theorems that need to be learnt (that adding the modulus to terms doesn't change anything, multiplying by the modulus makes the term zero, that the residue of a sum of terms is the same as the sum of the residues of each term, ... then you got the powers and Fermat's/Euler's Theorem ...)
  
  :)