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Changing commutative to associative when discussing properties of addition per issue #95

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Minh T. Nguyen 2014-08-12 08:36:11 -07:00
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commit 0283588c75
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@ -53,7 +53,7 @@ In some cases (i.e., if P~1~ and P~2~ have the same x values but different y val
If one of P~1~ is the "point at infinity", then the sum P~1~ + P~2~ = P~2~. Similary, if P~2~ is the point at infinity, then P~1~ + P~2~ = P~1~. This shows how the point at infinity plays the roll of 0. If one of P~1~ is the "point at infinity", then the sum P~1~ + P~2~ = P~2~. Similary, if P~2~ is the point at infinity, then P~1~ + P~2~ = P~1~. This shows how the point at infinity plays the roll of 0.
It turns out that + is commutative, which means that `(A+B)+C = A+(B+C)`. That means we can write A+B+C without parentheses without any ambiguity. It turns out that + is associative, which means that `(A+B)+C = A+(B+C)`. That means we can write A+B+C without parentheses without any ambiguity.
Now that we have defined addition, we can define multiplication in the standard way that extends addition. For a point P on the elliptic curve, if k is a whole number, then kP = P + P + P + ... + P (k times). Note that k is sometimes confusingly called an "exponent" in this case. (It would make a lot more sense to call it this if we used an operator that looked like multiplication rather than "+".) Now that we have defined addition, we can define multiplication in the standard way that extends addition. For a point P on the elliptic curve, if k is a whole number, then kP = P + P + P + ... + P (k times). Note that k is sometimes confusingly called an "exponent" in this case. (It would make a lot more sense to call it this if we used an operator that looked like multiplication rather than "+".)