I How can -a be equal to (-1)*a?

[transmini]

I will say that this question is coming from a lack of explanation in a classroom, however this particular proof is not homework and is just explanation over a proof that was discussed briefly in class, so I didn't think it belong in the homework section. I'm also not certain it belongs in the calculus section, as it seems more of an algebra concept, but it was in a calculus course.

So we were doing another proof which relied on this one, however I couldn't catch what was being said and the "proof" done for this statement really seemed like it was done only for the special case of a=1.

All of the proofs are supposed to be done using solely the following 10 field axioms:
1) Commutativity of Addition and Multiplication
2) Associativity of Addition and Multiplication
3) Additive and Multiplicative Identities
4) Additive and Multiplicative Inverses
5) Distributive Property
6) Nontrivial assumption of 1 $\neq$ 0

This is fairly easy to do, provided you can add something to both sides of the equation:
$$a*0 = 0$$
$$a*(1+(-1))=0$$
$$a*(1)+a*(-1)=0$$
$$a+(-1)*a=0$$
$$-a+a+(-1)*a=-a$$
$$0+(-1)*a=-a$$
$$(-1)*a=-a$$

which is all well and good, however we were told "we don't know how to add to both sides just yet" and as such can not perform line 5, making this an invalid method. Does anyone know how to do this using absolutely nothing but the axioms listed above?

 

[andrewkirk]

There is no need to take the proof beyond the fourth line: $a+-1\cdot a=0$.

That line tells us that $-1\cdot a$ is the additive inverse of $a$ (because they add to zero) and by definition that is written as $-a$.

By the way, the first line in the proof $a.0=0$ is not an axiom. It is a theorem that needs to be proven.

 

[transmini]

There is no need to take the proof beyond the fourth line: $a+-1\cdot a=0$.

That line tells us that $-1\cdot a$ is the additive inverse of $a$ (because they add to zero) and by definition that is written as $-a$.

By the way, the first line in the proof $a.0=0$ is not an axiom. It is a theorem that needs to be proven.

I know the first line isn't an axiom, but it was something previously stated, and I made it made starting point. However, that answer makes a lot of sense. I'm not sure why this is what was mentioned in class, rather than some make a=1 and some other stuff that didn't quite make sense. Thanks for the insight, that definitely makes things more clear.

 

[transmini]

There is no need to take the proof beyond the fourth line: $a+-1\cdot a=0$.

That line tells us that $-1\cdot a$ is the additive inverse of $a$ (because they add to zero) and by definition that is written as $-a$.

By the way, the first line in the proof $a.0=0$ is not an axiom. It is a theorem that needs to be proven.

Though I guess technically we were trying to prove a*0=0 using the at then unproven -a=(-1)*a, so how could you start the proof without beginning at a*0=0? Since you mentioned it not being an axiom.

 

[andrewkirk]

I would use it, but first prove it:
$$0\cdot a=(0+0)\cdot a = 0\cdot a + 0\cdot a$$
then add $-0\cdot a$ to both sides.

The qualm over not using adding to both sides is unnecessary, because the justification for that is based solely on fundamental logic, not on any mathematical axioms or even on the definition of '+', viz:

1. $a=a$ [Logical axiom of equality]
2. $c+d = c+d$ [substituting $c+d$ for all occurrences of $a$ in line 1, using the logical axiom of specification]
3. $b=d$ [hypothesis to open conditional proof]
4. $c+b=c+d$ [substituting $b$ for the first $d$ in line 2, using the logical axiom of substitution together with line 3]
5. $(b=d)\to (c+b=c+d)$ [conclusion from closing conditional proof]

The proof works just as well if we replace '+' by $\times$ or in fact any binary function.

 

[transmini]

I would use it, but first prove it:
$$0\cdot a=(0+0)\cdot a = 0\cdot a + 0\cdot a$$
then add $-0\cdot a$ to both sides.

The qualm over not using adding to both sides is unnecessary, because the justification for that is based solely on fundamental logic, not on any mathematical axioms or even on the definition of '+', viz:

1. $a=a$ [Logical axiom of equality]
2. $c+d = c+d$ [substituting $c+d$ for all occurrences of $a$ in line 1, using the logical axiom of specification]
3. $b=d$ [hypothesis to open conditional proof]
4. $c+b=c+d$ [substituting $b$ for the first $d$ in line 2, using the logical axiom of substitution together with line 3]
5. $(b=d)\to (c+b=c+d)$ [conclusion from closing conditional proof]

The proof works just as well if we replace '+' by $\times$ or in fact any binary function.

Whether it justified on fundamental logic or not, we were still told that we couldn't use it, and that's where I was running into problems.

 

[I like Serena]

then add $-0\cdot a$ to both sides

Shouldn't that be $-(0\cdot a)$?
That is, the additive inverse of whatever the as yet unknown $0\cdot a$ is?
And not the as yet unknown additive inverse of 0 times a?

(As a challenge to the OP, can we prove the trivial theorem $-0=0$ from the axioms?)