Deductions using Field axioms

Question

Deduce $\alpha(-\beta)=(-\alpha)(\beta)=-(\alpha\beta)$

I'm struggling with where to start and which axioms will be useful.

The axioms:

Axioms for a field $\mathbb{K}$ are with respect to the additive operation $+ : \mathbb{K} \times \mathbb{K} \rightarrow \mathbb{K}$:

1) $\forall x,y,z\in \mathbb{K}: (x+y)+z = x+(y+z)$

2) $\forall x,y \in \mathbb{K}: x+y = y+x$

3) $\exists 0 \in \mathbb{K} $such that$ : x+0=x$

4) $\forall x \in \mathbb{K} \exists y \in \mathbb{K}: x+y=0 \rightarrow y = -x$

for the multiplicative operation:

1) $\forall x,y,z \in \mathbb{K}: (xy)z = x(yz)$

2) $\forall x,y \in \mathbb{K}: xy=yx$

3) $\exists 1 \in K : \forall x \in \mathbb{K}: x1=x$

4) $\forall x\neq 0 \in \mathbb{K} \exists y \in \mathbb{K}: xy=1 \rightarrow y = x^{-1}= \frac{1}{x}$

and the distributivity property

$(x+y)z=xz+yz$

Answer 1

You'll have to first prove a lemma using the axioms, namely that $x \cdot 0 = 0 \cdot x = 0$ for all $x$.

Once you've done that, then $$\alpha\beta + \alpha(-\beta) = \alpha(\beta + (-\beta)) = \alpha \cdot 0 = 0 $$ and so $\alpha(-\beta)$ is equal to the additive inverse of $\alpha\beta$ which is $-\alpha\beta$.

Similarly for $(-\alpha)\beta$.

Answer 2

Hint:

Use this fact: For all $\alpha$, $-\alpha = (-1)\cdot \alpha$.

Proof of the fact: $\alpha+(-1)\cdot \alpha = 1\cdot \alpha + (-1)\cdot\alpha = (1-1)\cdot \alpha = 0\cdot \alpha = 0$.

The last equality holds because $0\cdot \alpha = (0+0)\cdot \alpha = 0\cdot\alpha + 0 \cdot \alpha$ and so by removing $0\cdot \alpha$ from both sides we get $0\cdot\alpha = 0$.

From this fact you can deduce all of the equalities. I'll do one of them and leave the rest for you

Example: $\alpha \cdot (-\beta) = (-\alpha)\beta$

Proof: $\alpha\cdot (-\beta) = \alpha\cdot ((-1)\cdot \beta) = (\alpha \cdot (-1))\cdot \beta = ((-1)\cdot \alpha)\cdot \beta = (-\alpha)\cdot \beta$.

Answer 3

Attempt at solving equality $(\alpha)(-\beta)=-(\alpha\beta)$

$\alpha(-\beta)=\alpha((-1)(\beta)=(\alpha(-1))(\beta)=((-1)\alpha)\beta)=(-1)((\alpha)(\beta))=(-1)(\alpha\beta)=-(\alpha\beta)$

Answer 4

$α(−β)=(−α)(β)=−(αβ)$

First

$α(−β) = α(−1)(β)$

(Proof in above comment)

Then

$∀x,y,z∈K:(xy)z=x(yz)$, thus

$(α)(−1)(β) = (α)(β)(-1)$

$∀x,y∈K:xy=yx$, thus

$(α)(β)(-1) = (-1)(α)(β)$

$(-1)(α)(β) = -(αβ)$ (Maybe make that distinction earlier on perhaps)

:)