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I was wondering whether the statement $n>n−1$, where $n$ is a real number, is considered an axiom. I know that the real numbers are built upon axiomatic systems such as the field axioms and order axioms, but I am unsure if this specific inequality is directly derived from these axioms or if it is taken as an axiom itself.

The usually proof I see is that $n>n-1$ so $0>-1$ and this is true so $n>n-1$ , but here you are assuming that $n>n-1$ for the case $n = 0$. So is it a axiom or is it a theorem.

Pawan J
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1 Answers1

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The axioms are usually that i) if $a > b$ then $a+c > b+c$ and ii) if $a > b$ and $c > 0$ then $ac > bc$ (and, of course, iii) exactly one and only one of the following is always true: either $a < b; a=b;$ or $a > b$.)

From this we prove that if $a \ne 0$ then $a^2 > 0$. Because: If $a < 0$ then $a-a < 0 -a$ so $0 < (-a)$ so $0\cdot (-a) < (-a)\cdot (-a)$ so $0 < a^2$. And if $a > 0$ then $a^2 =a\times a > a\times 0 = 0$.

And therefore $1 = 1^2 > 0$.

And therefore $-1 < 0$ and $-1 + n < 0 + n$ and $n-1 < n$.

But, no, $n> n-1$ is not axiom. Not even $1 > 0$ is an axiom. What is an axiom is if $a < b$ then $n+a < n+b$ and what can be proven is that $-1 < 0$ and what can then be proven by application is $n-1 < n$.

fleablood
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  • How did you get -1<0 from 1>0? – Pawan J Oct 11 '24 at 16:03
  • $1> 0$ so $1+(-1) > 0 +(-1)$ so $0 > -1$. – fleablood Oct 11 '24 at 16:04
  • I did leave out of the proof the following: $(-a)^2 = a^2$ and $0\cdot a = 0$. – fleablood Oct 11 '24 at 16:07
  • So what is the proof of $(-a)^2 = a^2$ ? – Pawan J Oct 11 '24 at 16:10
  • First we prove $-(-a)=a$. Then $0\cdot a = 0$. Then $(-a)b =-(ab)$. From there we have $(-a)^2= (-a)(-a)=-(a(-a))=-(-(aa))=-(-(a^2)) = a^2$. To prove these: If $c+d=0$ then by definition $d=-c$. So we have $a + (-a)=0$ and $(-a) + a=0$. So by definition $a=-(-a)$. To prove $0\cdot a=0$ we have $0+0=0$ so $a\cdot 0 = a\cdot (0 + 0)=a\cdot 0 + a\cdot 0$ (distribution is an axiom). Then we add $-(a\cdot 0)$ to each side to get $a\cdot 0 - (a\cdot 0) = a\cdot 0 + a\cdot 0 - (a\cdot 0)$ so $0 -a\cdot 0$. To be continued.... – fleablood Oct 11 '24 at 16:34
  • So to prove $(-a)b = -(ab)$ consider that $ab + (-a)b = (a+(-a))b = 0\cdot b = 0$. As we have $ab+(-a)b$ that means, by definition that $(-a)b = -(ab)$. So from all that we have $(-a)^2 =(-a)(-a)= -(a(-a)) =-(-(aa)) = (aa) =a^2$. – fleablood Oct 11 '24 at 16:36