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[Ch-01] Case when k is equal to the order of the set #180

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[Ch-01] Case when k is equal to the order of the set #180

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2 changes: 1 addition & 1 deletion ch01.asciidoc
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Expand Up @@ -509,7 +509,7 @@ include::code-ch01/answers.py[tag=exercise5,indent=0]
[NOTE]
====
The answer to Exercise 5 is why we choose to use finite fields with a _prime_ number of elements.
No matter what _k_ you choose, as long as it's greater than 0, multiplying the entire set by _k_ will result in the same set as you started with.
No matter what _k_ you choose, as long as it's greater than 0 (and different from the order of the set), multiplying the entire set by _k_ will result in the same set as you started with.

Intuitively, the fact that we have a prime order results in every element of a finite field being equivalent.
If the order of the set was a composite number, multiplying the set by one of the divisors would result in a smaller set.
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