Hey all, I got to proof
(still my backslash doesn't work, therefor no LaTeX )
I see that there is a n so that it holds for all x and for all y (n = 1)
Anyone?
Do you mean that $\displaystyle y=x^1$ for all $\displaystyle x$ and $\displaystyle y$?I see that there is a n so that it holds for all x and for all y (n = 1)
Well, here is how to show antisymmetry. Assume $\displaystyle xRy$ and $\displaystyle yRx$, i.e., $\displaystyle y=x^n$ and $\displaystyle x=y^m$ for some natural $\displaystyle m$, $\displaystyle n$. Then $\displaystyle y=y^{nm}$, so by the assumption in the problem statement $\displaystyle y=1$ or $\displaystyle nm=1$. If $\displaystyle y=1$, then $\displaystyle x=y^m=1$, so $\displaystyle x=y$. If $\displaystyle nm=1$, then $\displaystyle n=m=1$, so again $\displaystyle x=y$.
If you are having difficulties with other parts, could you describe what they are?
I forgot to add that it's a binary ordening.. Does tht make any difference to this?
If you have to pick a n out of the pos. nat. numbers so that that holds, 1 should be enough?
Therefore, x would be y, and because xRy, xRx / yRy would be true.
And if I remember right that's the definition of a reflexive ordening, or am I thinking too easy?
A reflexive (which is the same thing as non-strict, or weak) partial order is a binary relation, but I am not sure what "binary ordering" is.I forgot to add that it's a binary ordening
So that what holds?If you have to pick a n out of the pos. nat. numbers so that that holds, 1 should be enough?
Therefore, x would be y, and because xRy, xRx / yRy would be true.
Could you write precisely the statements that you have to prove and that you claim are true? What I mean in particular is that in a meaningful proposition every variable must be properly introduced. To introduce a variable one uses phrases like "Let us fix some natural number n", or "Consider x such that P(x) holds", or "let y = x^n+z^n". So every variable that you mention either in a claim that you need to prove or in a proof of that claim must be introduced in one of these ways. In your quote above, I am not sure what x and y are. Probably for some x and y what you say is true, for others it may be false.
If I were to guess, you are saying that R is reflexive for the following reason. Fix any natural number x. Then x=x^1, so xRx. Since this holds for any x, R is reflexive. This is correct, but it needs to be said more precisely.
A reflexive relation R has indeed only one property: for every x, xRx. However, a reflexive order is a relation that has three properties; they are described in the link above.And if I remember right that's the definition of a reflexive ordening, or am I thinking too easy?