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From Stephen Abbott's - understanding analysis there is a section in the text which says:

"The finite set $\{0,1,2,3,4\}$ is a field when addition and multiplication are computed modulo 5."

I wasn't so familiar with the term "modulo" so quick googling yielded this Is the idea that an addition or multiplication which in theory yields a number greater than 4, is supposed to "wrap around" the set and begin from zero again?

so $4+1 = 0 $ ?

Otherwise, I do not see how this satisfies the properties of a field, namely having an additive identity and multiplicative inverse.

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elbarto
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    Yes, $4+1 = 0,, 3\cdot 3 = 4$ etc. – Daniel Fischer Feb 16 '15 at 13:42
  • The additive identity is 0. – steedsnisps Feb 16 '15 at 13:45
  • @wowlolbrommer Sorry - meant additive inverse, which I am thinking is satisfied when the set loops around itself after traversing 4, is this correct? – elbarto Feb 16 '15 at 13:50
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    I don't understand what you mean. 4 and 1 are each others' additive inverses (4+1=5=0 modulo 5), and 2 and 3 also (2+3=5=0 modulo 5). – steedsnisps Feb 16 '15 at 13:52
  • Other things to prove: associative, commutative, distributive laws. – GEdgar Feb 16 '15 at 14:09
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    It turns out that the set of numbers ${0,\dots,p-1}$ with everything computed modulo $p$ is a field if and only if $p$ is prime. On the other hand, consider $\Bbb Z/6\Bbb Z$ (a notation for ${0,\dots,5}$ modulo $6$ — another common notation is $\Bbb Z_6$). This is not a field, because $2$ has no multiplicative inverse. (If it did, then we would have $3=(2^{-1}\cdot2)\cdot3=2^{-1}\cdot(2\cdot3)=2^{-1}\cdot0=0$. Remember that $2\cdot3=0$ modulo $6$.) – Akiva Weinberger Jul 22 '16 at 14:17

2 Answers2

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You can think of modular arithmetic either as the numbers wrapping around, or what's more typical is to think of the numbers as remainders when divided by 5 (in this case). You can write any integer $m$ as $m=5n+r$ then $m\equiv r \text{ mod } 5$. 0 is the additive identity. 1 is the multiplicative identity. Both operations are commutative and associative. To see that every non-zero element has a multiplicative inverse you could just write out the multiplication table... The additive inverse is just the negative (as usual). So $-2 = -1*5+3$ and hence $-2 \equiv 3\text{ mod }5$. So the additive inverse of 2 is 3 (because $2+3=5\equiv 0$).

TravisJ
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Multiplicative inverses: $$ 1\cdot 1 \equiv 1 \pmod{5} \\ 2\cdot 3 \equiv 1 \pmod{5} \\ 3\cdot 2 \equiv 1 \pmod{5} \\ 4\cdot 4 \equiv 1 \pmod{5} $$

Generally: addition mod $m$ is a field if and only if $m$ is prime.

GEdgar
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