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I'm trying to prove theorem $1.4$ with the following:

enter image description here

$$a=a$$

$$a-a=\stackrel{0}{\overline{a-a}}\tag{Ax.5}$$

$$a-a=0\tag{Inverse def.}$$

$$a+(-a)=0\tag{Thm 1.3}$$

Here I thought about packing the $a$ with a minus sign and then it would yield a new minus sign on Its inverse due to the inverse definition.

$$(-a)+(-(-a))=0$$

Adding $a$ to both sides.

$$\stackrel{0}{\overline{a+(-a)}}+(-(-a))=a$$

$$-(-a)=a$$

Is my proof correct?

Red Banana
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    I see you removed the overbrace notation edit I made? I made the change because I thought the overline notation makes the quantities look like fractions – graydad Oct 12 '15 at 21:24
  • What do those overlines mean? – fleablood Oct 12 '15 at 21:29
  • Oh, I get it. I thought they were fractions. Overbrace would be better. – fleablood Oct 12 '15 at 21:30
  • Your proof is correct and good. But all the initial stuff isn't nesc. If you started with "(-a)+(-(-a))=0" that'd be fine. You can forget everything before. – fleablood Oct 12 '15 at 21:33

5 Answers5

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Better wording:

By theorem 1.2. For every a there is a unique -a such that a + (-a) = 0. Likewise for (-a) there is a unique -(-a) such that (-a) + (-(-a)) = 0.

Therefore:

(-a) + (-(-a)) = 0

a + (-a) + (-(-a)) = a+ 0 = a

By associativity:

[a + (-a)] + (-(-a)) = a so

0 + (-(-a)) = (-(-a)) = a.

fleablood
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    It is probably time for you to lear to format your answers. Take a look at this http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Giovanni Oct 12 '15 at 21:47
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Your proof is correct, but it looks like you're trying to be as precise as possible, so I'll write out a proof you might find helpful. Let $a$ be a real number. By axiom $5$, we there exists a real number $b$, satisfying $a+b=0$.

Since $b$ is a real number, it also has an inverse, which we may call $c$. The aim is to show $c=a$. Since $a+b=0$, we have $$(a+b)+c=0+c=c$$ And then by axiom $2$ (associativity) we deduce that $$a+(b+c)=c$$ But since $c$ was the inverse of $b$, $b+c=0$, so $a+0=c \implies a=c$. Hopefully that's helpful.

preferred_anon
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What is "packing with a minus sign"? This step seems a little obscure. You are acting like there is an operation which is "putting minus in front", even though (in this stage of axioms) $-a$ is a symbol in itself.

For proving $-(-a)=a$, we have to prove (by definition) that $-a+a=0$. But this is trivial due to commutativity and the characterization of $-a$.

Also, more leisurely:

Proof: We have that $a+(-a)=0$. Now look again. $\blacksquare$

Aloizio Macedo
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  • If I have $a$ and it's inverse is $-a$, then It seems natural to expect the inverse of $-a$ to be $-(-a)$. That's what I meant with "packing with a minus sign". – Red Banana Oct 12 '15 at 21:37
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    What is a minus sign? The "minus sign" exists only representatively because we chose to represent the inverse of $a$ as $-a$. If we denoted the inverse of $a$ as $A$, what would be "packing with a minus sign?" Making $A$ bigger? Then the inverse of $A$ would be a twice bigger $A$? – Aloizio Macedo Oct 12 '15 at 21:39
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    Good point. I'll think about it. – Red Banana Oct 12 '15 at 21:43
  • I don't get it. Suppose the scenario you proposed: $a+A=0$. How would you do from here? – Red Banana Oct 12 '15 at 21:53
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    In the proof I presented? If so, it is exactly the same thing: $A$ being the inverse of $a$ implies $a+A=0$. Now note that, due to commutativity, $A+a=0$. But this means that $a$ is the inverse of $A$ (since, adding it to $A$, we get $0$). It follows then that the inverse of the inverse of $a$ is $a$ itself. EDIT: Sorry for the incoherence of notation. What I call "inverse of" is what your text calls "negative of". – Aloizio Macedo Oct 12 '15 at 21:55
  • Is it too problematic to work on a representation? I wrote some ideas on a paper that was intended to teach mathematics. My goal was to bring up the idea of equivalent rewritings, loosely based on this. So these naive rewritings became my media for doing proofs in mathematics and for the first time I had a deeper access to mathematics. I guess I was following something like the formalist school. – Red Banana Oct 12 '15 at 22:29
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The proof uses a bunch of manipulations that are generally legitimate for later proofs--but notice that in this particular axiomatization of algebra, the subtraction operation and negative sign are not introduced until Theorem 1.2. If you look closely at Theorem 1.2, you may notice that it provides a definition of $-a$ that is, in essence,

$$a + (-a) = 0.$$

That is, Theorem 1.2 is actually the source of the statement you justified by Theorem 1.3, and it supports that statement directly.

The rest of the proof looks OK. I'm not sure what "packing" means, but since the statement $a + (-a) = 0$ is true for any real number $a$, it is certainly true for $-a$, which also is a real number, and you can write $-a$ instead of $a$ to obtain $(-a) + (-(-a)) = 0.$

David K
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I'm not exactly clear on how you go from $a+(-a)=0$ to $(-a)+(-(-a))=0$ via "packing". But starting with $$(-a)+(-(-a)) = 0$$ Then by theorem $1.2$ you have $(-a(a)) = x$ as the unique element such that $c+x = b$ where $b=0$ and $c = (-a).$ However you can just as well satisfy the same values of $c,b$ if $x = a$. Hence by uniqueness $a = -(-a)$.

graydad
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