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Based on the axioms for a mathematical field, the wiki article states that 0·a = 0 and (-1)·a = -a are consequences of the axioms, but doesn't show how they are derived. There was a similar question asked before, but I'm not sure about the accepted answer.

https://en.wikipedia.org/wiki/Field_(mathematics)#Elementary_consequences_of_the_definition

Also, it would seem logical that if a ≠ b, and if c ≠ 0, then c·a ≠ c·b, a uniqueness property that should hold true for a finite field (unordered) that I'm wondering if it can be derived from the axioms (perhaps something like induction?) .

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  • $\begingroup$ These are in fact standard easy arguments (your guess in the last sentence is correct). I'm sure someone will step up and provide them. Here's a hint for the first one: consider $(0 + 0) \times a$. I'm curious: what prompts your curiousity? A course you're taking? $\endgroup$ – Ethan Bolker Apr 22 '17 at 0:12
  • $\begingroup$ @EthanBolker - a question came up at another forum. An "addition" table was given for a finite field with 4 numbers (0,1,2,3), where the "addition" turns out to be exclusive or, and the problem was asking to produce the multiplication table, based on the axioms rather than knowledge of GF(4). That prompted me to wonder how the "consequences" as noted in the wiki article are derived, and what operations are considered acceptable as derivations for field math. $\endgroup$ – rcgldr Apr 22 '17 at 0:42
  • $\begingroup$ @EthanBolker - I'm aware that algebra works with finite finite fields (I've worked with RS ECC), but never thought about the derivations based on the axioms. For that other forum question, after taking into account 0·a = a·0 = 0 and 1·a = a·1 = a, only 4 products in the multiplication table (2·2, 2·3, 3·2, 3·3) needed to be determined, so this could be done by trial and error, where an axiom would fail if the wrong set of 4 products was chosen. Noting that every number has a multiplicative inverse, meant that two of those products = 1 => 2·3 = 3·2 = 1, which simplified the problem. $\endgroup$ – rcgldr Apr 22 '17 at 1:06
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For $0 \times a$: $$ 0 \times a = (0+0)\times a = 0\times a + 0 \times a . $$ Now whatever $0 \times a$ is, it has an additive inverse, so you can subtract it from each side of that equation to conclude that $0 \times a = 0$.

For $(-1) \times a$: $$ 0 \times a = (1 + (-1)) \times a = 1 \times a + (-1) \times a = a + (-1) \times a $$ but $a$ has a unique additive inverse $-a$.

For your third question.

If $c \ne 0$ then it has a multiplicative inverse $d$. Then $$ ca = cb \implies ca - cb = c(a-b) = 0 \implies dc(a-b) = 0 \implies a-b = 0 \implies a = b. $$

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  • $\begingroup$ Shouldn't the third part somewhere include a - b ≠ 0 or perhaps a - b = e ≠ 0 ? $\endgroup$ – rcgldr Apr 22 '17 at 0:26
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    $\begingroup$ @rcgldr No. I proved your claim by reasoning with the contrapositive form. If $a \ne b$ then the last entry in my chain of correct implications is false, so the first entry must be false. $\endgroup$ – Ethan Bolker Apr 22 '17 at 0:30
  • $\begingroup$ OK, thanks for the explanation. I was also considering let a-b = e ≠ 0, then ca - cb = ce ≠ 0, since c ≠ 0 and e ≠ 0. $\endgroup$ – rcgldr Apr 22 '17 at 0:35
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I presume that by "-a" you mean the additive inverse of a so that you want to prove that -1 times a is the additive inverse of a. Again, that follows from the distributive law. (1+ (-1))a= 0a= 0. But, by the distributive law, (1+ (-1)))a= 1a+ (-1)a= a+ (-1)a= 0 also. "a+ (-1)a= 0" is precisely what is meant by "additive inverse of a".

The statement "if a ≠ b, and if c ≠ 0, then c·a ≠ c·b" is most easily proved by proving the "contrapositive". For any statement "if p then q", the contrapositive is the statement "if not q then not p" and it is easy to show in general that a statement is true if and only if its contrapositive is true.

The contrapositive of "if a ≠ b, and if c ≠ 0, then c·a ≠ c·b" is "if ca= cb then a= b". To prove that add the additive inverse of (subtract) cb from both sides: ca- cb= 0. By the distributive law, c(a- b)= 0. Since c is not 0 we must have a- b= 0 so a= b.

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  • $\begingroup$ Should the contrapositive be stated as $$\text{If $ca=cb$ the either $a=b$ or $c=0$}?$$ $\endgroup$ – Juniven Apr 22 '17 at 0:37
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$(-1)a=-a $ means $a+(-1)a=0$ as additive inverses are unique.

$a+(-1)a=1*a+(-1)a $ (existence of multiplicative identity.

$=(1+(-1))a $ (distributive property)

$=0*a $. Remains to show $0*a=0$ for all $a $

$0*a= (0+0)*a $ (Associativity and definition of additive property.

$=0*a+0*a $.

$0a=0a+0a $

$0a+(-0a)=0a+0a+(-0a) $ (existence of additive inverse and acknowledgement that addition is a closed binary opperation)

$0=0a+0=0a $(additive inverse and implied associtivity.)

So $a+(-1)a=0$. So $(-1)a=-a$ (and $a= -(-1 (a)) $.)

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