1
$\begingroup$

My course book for mathematical structures uses a different set of axioms than Peano in order to define the natural numbers, and the operations on those:

We take a set $\mathbb{N}$, two elements $0,1 \in \mathbb{N}$, and two functions $+: \mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}$, and $\cdot: \mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}$ such that the following statements hold $\forall a,b,c \in \mathbb{N}$:

  1. $a + b = b+a$
  2. $a + (b+c) = (a+b) + c$
  3. $a + 0 = 0 + a = a$
  4. $a + b = a + c \implies b = c$
  5. $a \cdot b = b\cdot a$
  6. $a \cdot(b\cdot c) = (a\cdot b)\cdot c$
  7. $a \cdot1 = 1\cdot a = a$
  8. $a \cdot(b + c) = a \cdot b + a \cdot c$
  9. $0 \neq 1$
  10. $1 + a \neq 0$
  11. If $A \subset \mathbb{N}$ such that $0 \in A$ and $a \in A \implies a+1 \in A$, then $A = \mathbb{N}$.

Their claim is now, that this uniquely defines the natural numbers, and I believe that. I did, however, notice that the addition axioms 1-3 were repeated as 5-7 for multiplication, but the 4th axiom was not. I have checked an older (by 10 years) version of the course book, and this axiom was there at the time, meaning it was for some reason removed somewhere between then and now. This seems to imply that "$a \cdot b = a\cdot c \implies b = c$" can be constructed from the other axioms, but I don't see how that would be the case. The only similar statement would be 4, but it is not clear to me how a multiplication can be written as only additions, which would be necessary to use that axiom. Does anyone have any hint on why this should be the case?

I have tried googling this question, but not knowing what the property this axiom describes is called in English didn't help (I only know it's "schrapwet" in dutch).

$\endgroup$
5
  • 4
    $\begingroup$ It isn’t true that $a\cdot b=a\cdot c$ implies that $b=c$: what if $a=0$? $\endgroup$ Dec 28, 2020 at 22:06
  • $\begingroup$ Good point, but would it then be possible if we assumed $a \neq 0$? $\endgroup$
    – Xander L
    Dec 28, 2020 at 22:09
  • $\begingroup$ Yes, one could then include it, though if Yves is right, it would be redundant. $\endgroup$ Dec 28, 2020 at 22:26
  • $\begingroup$ Which book are you referring to? $\endgroup$
    – Shaun
    Dec 28, 2020 at 23:36
  • $\begingroup$ It's actually a syllabus written by my teachers at uni. The 2013/14 version (in dutch) can be found online, with the axioms at page 22. The following link will download the dutch file directly: google.com/… $\endgroup$
    – Xander L
    Dec 28, 2020 at 23:43

1 Answer 1

3
$\begingroup$

If I am right, the property $a\ne0,a\cdot b=a\cdot c\implies b=c$ can be proven by induction, using 8., 4., 5. and 11.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.