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a) $X$ is positive b) $X=0$ c) $-X$ is positive

We finished building integers and now, we are building rational numbers. This question is on my text book and i was trying to exercise. I dont know where to start. I know that integers are equivalence classes in $\mathbb N\times\mathbb N$. And I know rule of being positive in integers.

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    Why can't you type $X$ or $X = 0$ or $-X$ or $\Bbb{N}\times\Bbb{N}$ on your phone? MathJax only needs ASCII characters. E.g., $\$$\Bbb{N}\times\Bbb{N}$\$$ gives $\Bbb{N}\times\Bbb{N}$. – 2017-02-19
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    Oh and as for the mathematics: if $(m, n)$ is a pair of natural numbers, what can you say about the relative magnitudes of $m$ and $n$? Can you think of three cases that might relate to your three cases for the sign of an integer represented by the pair? – 2017-02-19
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    Rob I don't know how to that on the phone. I'm pretty don't know about MathJax. – 2017-02-19
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    (m,n)R(p,q) then m+q=p+n I know this then ? – 2017-02-19
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    You should learn MathJax: it is the standard way to format mathematics on MSE. I leave you to think a bit more about my second comment: what can you say about the possible ordering of $m$ and $n$ in $(m, n)$? – 2017-02-19
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    Where can i learn Mathjax? – 2017-02-20
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    Follow the help system when you ask or answer a question. – 2017-02-20

1 Answers 1

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When building the numbers the more difficult proofs are those on $\mathbb N$ since for the subsequent classes the proofs relies on a corresponding property on the earlier classes.

In this case the proof relies that for each pair $x$, $y$ of natural numbers exactly one of $x>y$, $x

The extension to $\mathbb Z$ first must define the relations on $\mathbb Z$ of course (or one does that at about the same time) and also negation. In doing so we need to know that for two numbers $x$ and $y$ we have that for each candidate of the equivalence class we have that $x_p+y_n$ and $x_n+y_p$ compares in the same way (that $x

We have that due to the property in $\mathbb N$ that exactly one of $x_p+y_n < x_n+y_p$, $x_p+y_n > x_n+y_p$ and $x_p+y_n = x_n+y_p$ is true. Especially we have exactly one of $x_p + 0_n > x_n+0_p$ (ie $x>0$), $x_p + 0_n < x_n + 0_p$ (ie $x<0$ or by using definition of negation immediately means $-x > 0$) or $x_p+0_n = x_n+0_p$ (ie $x=0$).