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# <math>\,\!F_2 = \{ 0, 1 \}</math> |
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# <math>\,\!F_2 = \{ 0, 1 \}</math> |
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# <math>\,\!F_7 = \{ 0, 1,2,3,4,5,6 \}</math> |
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# <math>\,\!F_7 = \{ 0, 1,2,3,4,5,6 \}</math> |
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# <math>\,\!F_6 = \{ 0, 1,2,3,4,5 \}</math> is not a field (counterexample) |
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# <math>\,\!F_6 = \{ 0, 1,2,3,4,5 \}</math> is not a field because not every element has a multiplicative inverse. |
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#: Let <math>a = 2.</math> |
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#: Then <math>a \cdot 0 = 0, a \cdot 1 = 2, a \cdot 3 = 0, a \cdot 4 = 2, a \cdot 5 = 4</math> |
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#: Therefore F4 fails; there is '''no''' number ''b'' in ''F''<sub>6</sub> s.t. ''a · b'' = 1 |
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{| |
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! + !! 0 !! 1 |
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! × !! 0 !! 1 |
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! + !! 0 !! 1 !! 2 !! 3 !! 4 !! 5 !! 6 |
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! × !! 0 !! 1 !! 2 !! 3 !! 4 !! 5 !! 6 |
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The real numbers A set
with two binary operators and two special elements
s.t.
![{\displaystyle F1.\quad \forall a,b\in \mathbb {R} ,a+b=b+a{\mbox{ and }}a\cdot b=b\cdot a}](https://wikimedia.org/api/rest_v1/media/math/render/svg/89e2084a598d3aee71bf052e635228ff28680dde)
![{\displaystyle F2.\quad \forall a,b,c,(a+b)+c=a+(b+c){\mbox{ and }}(a\cdot b)\cdot c=a\cdot (b\cdot c)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f7a5a7306b8734d61df7726f4e4b61779dc48022)
![{\displaystyle {\mbox{(So for any real numbers }}a_{1},a_{2},...,a_{n},{\mbox{ one can sum them in any order and achieve the same result.}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7834e3c93dc288e9cbddb9a82ee0359267eb3b8e)
![{\displaystyle F3.\quad \forall a,a+0=a{\mbox{ and }}a\cdot 0=0{\mbox{ and }}a\cdot 1=a}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f2b9eafa64d8b239c92290bc5821a91929dfc6b4)
![{\displaystyle F4.\quad \forall a,\exists b,a+b=0{\mbox{ and }}\forall a\neq 0,\exists b,a\cdot b=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c28670c5b19bcc977e870f9b1eb5dadf80219396)
![{\displaystyle {\mbox{So }}a+(-a)=0{\mbox{ and }}a\cdot a^{-1}=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/001dc0d80a90e25b173645ead18e028aa45db4dd)
![{\displaystyle {\mbox{(So }}(a+b)\cdot (a-b)=a^{2}-b^{2})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4e4dbaa0a51225955499a55d55a1a3b46579353a)
![{\displaystyle \forall a,\exists x,x\cdot x=a{\mbox{ or }}a+x\cdot x=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2d71a04dc0d2158e1688436802e87c744d5a4a1c)
- Note: or means inclusive or in math.
![{\displaystyle F5.\quad (a+b)\cdot c=a\cdot c+b\cdot c}](https://wikimedia.org/api/rest_v1/media/math/render/svg/21604a4fa96fe8e1edba621cc64a185f452fedea)
Definition: A field is a set F with two binary operators
: F×F → F,
: F×F → F and two elements
s.t.
![{\displaystyle F1\quad {\mbox{Commutativity }}a+b=b+a{\mbox{ and }}a\cdot b=b\cdot a\forall a,b\in F}](https://wikimedia.org/api/rest_v1/media/math/render/svg/93874137d44d34260a9f9fcbe7fe7d523dda1a86)
![{\displaystyle F2\quad {\mbox{Associativity }}(a+b)+c=a+(b+c){\mbox{ and }}(a\cdot b)\cdot c=a\cdot (b\cdot c)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f2fe1a974b35f76aec8419cbe96228452778168e)
![{\displaystyle F3\quad a+0=a,a\cdot 1=a}](https://wikimedia.org/api/rest_v1/media/math/render/svg/038718cdb35f50ca32b3f80148bdb18d771fece8)
![{\displaystyle F4\quad \forall a,\exists b,a+b=0{\mbox{ and }}\forall a\neq 0,\exists b,a\cdot b=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9b3fec383fede7d5334150f44d93f4a502330a3c)
![{\displaystyle F5\quad {\mbox{Distributivity }}(a+b)\cdot c=a\cdot c+b\cdot c}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e0aa3a67ba8c2a33052b2319934b954ab4c640f1)
Examples
![{\displaystyle F=\mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/153faf37401acde5841386b70649e8616aabdea0)
![{\displaystyle F=\mathbb {Q} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/e178128ef93304af241f3706aacc15908eb9bfb3)
![{\displaystyle i={\sqrt {-1}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/370c8cebe9634fbfc84c29ea61680b0ad4a1ae0d)
![{\displaystyle \,\!(a+bi)+(c+di)=(a+c)+(b+d)i}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b426dc2b841f290fb62b08b29a7f78c4a0a0daa7)
![{\displaystyle \,\!0=0+0i,1=1+0i}](https://wikimedia.org/api/rest_v1/media/math/render/svg/293b5598eb3af046e8bed474b5c5fb8155fd8d05)
![{\displaystyle \,\!F_{2}=\{0,1\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d33ac4cd92e1a50e706b4713830b0e0af70b5258)
![{\displaystyle \,\!F_{7}=\{0,1,2,3,4,5,6\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2ca480aae37fc5ace8a4d80c553b95e7e904a1aa)
is not a field because not every element has a multiplicative inverse.
- Let
![{\displaystyle a=2.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/393128c6a62f3c6c70a643fd904ff6851ed32a34)
- Then
![{\displaystyle a\cdot 0=0,a\cdot 1=2,a\cdot 3=0,a\cdot 4=2,a\cdot 5=4}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8d0bc7f3f2fcd00c32dab006cbc4c0e9213fda2a)
- Therefore F4 fails; there is no number b in F6 s.t. a · b = 1
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Ex. 5
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Ex. 5
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Theorem:
for
is a field iff (if and only if)
is a prime number
Tedious Theorem
"cancellation property"
![{\displaystyle a\cdot b=c\cdot b,(b\neq 0)\Rightarrow a=c}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e2ced49d865d0432d4f4d17f3e4da7be8d9ad620)
...