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Yesterday my Honors Calculus professor introduced four basic postulates regarding (real) numbers and the operation $+$:

(P1) $(a+b)+c=a+(b+c), \forall a,b,c.$

(P2) $\exists 0:a+0=0+a=a, \forall a.$

(P3) $\forall a,\exists (-a): a+(-a)=(-a)+a=0.$

(P4) $a+b=b+a, \forall a,b.$

And of course, we can write $a + (-b) = a-b$. Then he proposed a challenge, which was to prove that $$a-b=b-a\iff a=b$$ using only these four basic properties. The $(\Leftarrow )$ is extremely easy and we can prove using only (P3), but I'm struggling to prove $(\Rightarrow )$ and I'm starting to think that it is not possible at all.

My question is how to prove $(\Rightarrow )$, or how to prove that proving $(\Rightarrow )$ isn't possible, using only (P1), (P2), (P3), (P4)?

Community
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Deathkamp Drone
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2 Answers2

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In fact, it is impossible to prove that result using only the information provided.

To show that this is impossible, we can build a system that obeys the postulates, but does not satisfy the provided statement. In particular, we can consider the following system:

  • The only numbers are $0$ and $1$
  • $1+0=0+1=1$
  • $0+0=1+1=0$ (so, $a = -a$ for $a = 0,1$)

Now, show that $a = 0$ and $b = 1$ satisfy $(a-b) = (b-a)$ but $a \neq b$.

What you can say (once you allow multiplication by integers) is that $$ a - b = b - a \iff 2(a-b) = (a-b) + (a-b) = 0 $$ in our system, however, multiplying anything by $2$ makes it zero.

Ben Grossmann
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  • To prove that this system is associative, we just need to verify that it works for the 8 different ways of summing 3 elements from ${0,1}$, right? – Deathkamp Drone Mar 30 '16 at 21:30
  • @DeathkampDrone that certainly works. However, I think you can make things easier using the fact that the system is commutative. – Ben Grossmann Mar 30 '16 at 21:57
  • Are you saying that commutativity implies in associativity? Or are you saying that I can use commutative property to reduce the number of cases that I need to verify in order to prove associativity? – Deathkamp Drone Mar 30 '16 at 22:08
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    I meant the second, since I'm not sure about the first. – Ben Grossmann Mar 31 '16 at 03:01
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You are right, you can't only from those axioms, and here is why.

Consider $A=\mathbb{Z}/2\mathbb{Z}$. Then $1-0=1=0-1$. But $0 \neq 1$.

But you can show this is true if $2 \neq 0$ and your ring is a integral domain, where $2:=1+1$. In fact,

$a-b=b-a \implies a=b-a+b \implies 0=b-a+b-a \implies 0=2b-2a \implies 0=2(b-a) \implies b-a=0 \implies b=a. $

Aloizio Macedo
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  • Good points but you made it a bit hard for this level. :-) – Mikasa Mar 30 '16 at 20:48
  • I don't know what is an integral domain, and I have only a vague idea of what is $\mathbb{Z}/2\mathbb{Z}$, so I don't really understand this proof. – Deathkamp Drone Mar 30 '16 at 21:07
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    @DeathkampDrone Let me try to clear up: you can see $\mathbb{Z}/2\mathbb{Z}$ as a shorthand for the structure that Omnomnomnom described. And an integral domain is simply a space where $ab=0 \implies a=0$ or $b=0$. – Aloizio Macedo Mar 30 '16 at 21:09