Why does $<^{-1}$ not equal $>$?

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Let us investigate the powers of $<$:

  • ${<^1} = \{(0,1);(0,2);(0,3);...;(1;2);...\}$
  • ${<^2} = \{(0,2);(0,3);(0,4);...;(1;3);...\}$
  • ${<^3} = \{(0,3);(0,4);(0,5);...;(1;4);...\}$
  • ...
  • ${<^N} = \{(0,N);(0,N+1);(0,N+2);...;(1;N+1);...\}$
  • ${<^0} = \{(0,0);(0,1);(0,2);...;(1;1);...\} = {\le}$
  • $<^{-1} = \{(0,-1);(0,0);(0,1);...;(1;0);...\}\ne {>}$

But why?

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You got it wrong. So, you're looking at a relation $<\,\, \in \mathbb{N}_0 \times \mathbb{N}_0$. By definition,

$$<^n \,\,:= \{(a_0, a_n) \mid \exists a_0, a_1, \ldots, a_n \in \mathbb{N}_0 : a_0 < a_1 < \ldots < a_n \}$$

for positive $n$, but

$$<^0 \,\,:= \{(a, a) \mid a \in \mathbb{N_0}\} = \,\,\,=$$

and

$$<^{-1} \,\,:= \{(b, a) \mid (a, b) \in \,\,<\} = \{(b, a) \mid a < b\} = \{(b, a) \mid b > a\} =\,\,>.$$

Note that these are entirely different definitions than positive powers of relations. So you can't just extrapolate for negative numbers.