prove in a field with four elements, F = {0,1,a,b}, that 1 + 1 = 0.
The given four elements are F = {0, 1, a, b}.
Consider ab belongs to F and then there are four possibilities for ab.
If ab = 0, then a = 0 or b = 0 which is contradiction as fields do not have ‘zero divisors’.
If ab = a, then b = a/a = 1 which is contradiction same as ab ≠ b. So, ab must be equal to 1. In other words b = a^–1.
Clearly a^2 ≠ 0, if a^2 = 1, then a = a^–1 = b which is not allowed.
If a^2 = a, then a = 1 which is not allowed.
That is, a^2 = a will be true.
Finally if a + 1 = 0, then a = –1 = 1 will be tr...
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