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This is my attempt based on some stuff I have been seeing around:

Let $y = -x$, then $-y = -(-x)$.

Now, lets sum $y + x = (-x) + x = 0$, then we have $y + x = 0$. If we had the additive inverse of $x$, i.e $-x$, to both sides, we obtain:

$$y + x + (-x) = 0 + (-x) \\ y + 0 = 0 + (-x) \\ y = -x\\ -y = x \\ -(-x) = x$$.

Is this proof correct just using the field axioms? I think that the key point is finding that $y = -x$.

rschwieb
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4 Answers4

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The statement can be reformulated in words by "the additive inverse of $(-x)$ is $x$". So we want to show that $(-x)+x=x+(-x)=0$. But this is true by the definition of $-x$.

Another easy exercise: By the same idea, prove that $(a^{-1})^{-1}=a$ for $a\neq 0$.

Spenser
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The opposite (or additive inverse) of the element $a$ is $b$ such that $$ a+b=0=b+a $$ Such an element $b$ is unique and is denoted by $-a$.

Since the equalities are true it with $a=-x$ and $b=x$, you get that $x$ is the opposite element to $-x$, that is, $$ -(-x)=x $$

egreg
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We know the additive inverse axiom: $$ -x + x = 0 \quad\text{for all}\quad x. $$

Following Rudin's suggestion, we replace $x$ by $-x$ and get

$$ -(-x) + (-x) = 0. $$ Both expressions are equal to zero, so we can set them equal to each other.

$$ x + (-x) = -(-x) + (-x). $$

Now we note that for them to be equal $x$ must be equal to $-(-x)$, i.e., $-(-x)=x$ which was to be shown.

Joe
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0

We can start with the additive inverse $$x+(-x)=0$$ We then plug $-x$ in for $x$ to get $$(-x)+(-(-x))=0$$ Now we add these 2 values to $x$ $$(-x)+(-(-x))+x=0+x$$ Using commutative property on the left side and additive identity on the right, we get $$x+(-x)+(-(-x))=x$$ We then can use additive inverse to get $$0+(-(-x))=x$$ And finally, we use additive identity to get $$-(-x)=x$$

Harish Chandra Rajpoot
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ski Lodge
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