# Tao Analysis. Definition of positive rational numbers.

Tao Analysis. Definition of positive rational numbers.

Definition $$4.2.6.$$ A rational number $$x$$ is said to be positive iff we have $$x = a/b$$ for some positive integers $$a$$ and $$b$$. It is said to be negative iff we have $$x = −y$$ for some positive rational $$y$$ ($$i.e., x$$ = $$(−a)/b$$ for some positive integers $$a$$ and $$b$$).

My question: Does not $$x = a/b$$ for some negative integers $$a$$ and $$b$$ satisfy? as in $$x = -2/-3$$. So why does this definition use iff? Thank you!

The original definition, made more verbose, is the following:

A rational number $$x$$ is said to be a positive rational number if and only if there exists some positive integer $$a$$ and some positive integer $$b$$ such that $$x=\frac ab$$. A rational number $$x$$ is said to be a negative rational number if and only if there exists a positive rational $$y$$ such that $$x=-y$$.

The key point to highlight here is "there exists". This means that the definition only "cares" about positive integers, and such a counter-example is not relevant.

If I have to explain this further, please tell me.

• So if it cares about negative integers, Can I conclude this "A rational number $x$ is said to be a positive rational number if and only if there exists some negative integer a and some negative integer b such that $x=\frac ab$"? Jul 30, 2022 at 16:21
• If you change it that way, it still works. In that case it no longer cares about positive integers, Jul 31, 2022 at 0:44

Definition $$4.2.6.$$ A rational number $$x$$ is said to be positive iff we have $$x = a/b$$ for some positive integers $$a$$ and $$b$$.

Why does this definition use iff?

Every definition in mathematics uses "if and only if"; try to come up with one that doesn't!

Does $$x = a/b$$ for some negative integers $$a$$ and $$b$$ not satisfy?

1. The above definition can be written $$\forall x{\in}\mathbb Q\;\left(x>0\iff\exists a{,}b{>}0\:\:x=\frac ab\right).$$

$$\implies$$ direction:

• Now, if $$x$$ equals $$\dfrac{-8}{-3}$$ or $$\dfrac{-24}{-9}$$ or $$\dfrac{16}{6}$$ (so, $$x$$ is positive), then $$x=\dfrac{\mathbf8}{\mathbf3},$$ that is, $$x=\frac ab\text{ for }\mathbf{some}\text{ pair of positive integers }a\text{ and }b;$$ here, we can choose $$(16,6)$$ instead of $$(8,3),$$ but $$(-8,-3)$$ won't work.

$$\impliedby$$ direction:

• If $$x$$ equals $$\dfrac{-8}{-3}$$ (so, $$x=\dfrac{\mathbf8}{\mathbf3},$$ that is, $$x=\dfrac ab\text{ for }\mathbf{some}\text{ pair of positive integers }a\text{ and }b$$), then $$x$$ is positive.
2. A less technical explanation.

Each positive rational number $$r$$ can be expressed as an non-simplified fraction $$F_1$$ of integers with numerator and denominator both negative; when these two negative signs are dropped, we obtain a non-simplified fraction $$F_2$$ of integers with numerator and denominator both positive; it is $$F_2,$$ not $$F_1,$$ that the above definition is using to define $$r$$ as positive.

This alternative definition is equivalent to the above one:

Definition $$4.2.6.$$Alt $$\;\;$$ A rational number $$x$$ is said to be positive iff we have $$x = a/b$$ for some negative integers $$a$$ and $$b.$$

If we adopt this definition, then we need to choose $$F_1,$$ not $$F_2.$$

• Thank you again @ryang. And what's this called if $\forall x{\in}\mathbb Q\;\left(x>0\iff\exists a{,}b{<}0\:\:x=\frac ab\right)$? Can we say "A rational number x is said to be positive iff we have x=a/b for some negative integers a and b"? Jul 30, 2022 at 14:28
• It's just writing out the theorem symbolically to clarify its structure. ⟺ means if and only if, A⟹B means B if A, while A⟸B means A if B. Please ignore this part if you have not encountered these symbols before; they are not essential to your Query or my Answer. Jul 30, 2022 at 14:36
• And another question: "here, we can choose (16,6) instead of (8,3), but (−8,−3) won't work.", you mean"$x=\dfrac{-8}{-3} \implies x=\dfrac{\mathbf8}{\mathbf3},$". Can we just say "$x=\dfrac{-8}{-3} \implies x=\dfrac{-8}{-3}$" to maintain the consistency with $x=\dfrac ab$ when $a$ and $b$ are negative integers. @ryang Jul 30, 2022 at 15:17
• Yes, in that case, just change the definition so that it requests for a negative $(a,b)$ pair instead of a positive $(a,b)$ pair! I've expanded my answer to elaborate on this. Jul 30, 2022 at 15:33