How does one write $0$ as a fraction in lowest terms?

Question

In Spivak's Calculus, two established properties of real numbers are:

(1) If $a$ is any number, then $a+0=0+a=a$

(2) For every number $a$, there is a number $-a$ such that $a+(-a)=(-a)+a=0$.

Looking at (2) and letting $a:=0$, we have:

$0 +(-0)=0$

Applying (1), where $a:=(-0)$, we have $-0=0$

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Given the above claim, my question is about the minimum value that the following function obtains on the closed interval $[0,1]$:

$ f(x)= \begin{cases} 1&\text{if}\, x \text{ is irrational}\\ 1/q&\text{if}\, x=p/q \text{ in lowest terms}\\ \end{cases} $

My solution manual says that, "There is no minimum".

However, given that $0 = -0$, I have a moment of doubt with this statement.

Specifically, $0 = -0 \implies \frac{0}{q}=-\frac{0}{q} \implies \frac{0}{q} = \frac {0}{-q}$

Letting $q=1$, I feel like an argument could be made that the minimum is $-1$ for this function along this interval.

Do we simply assert that the convention for $0$ expressed as a fraction in lowest terms is $\frac{0}{1}$?

Answer 1

On page $97$, Spivak writes

$p/q$ is in lowest terms if $p$ and $q$ are integers with no common factor [other than $\pm1$] and $q>0$.

We can write $0$ as $\dfrac{0}{n}$ for any $n\in\mathbb{Z^+}$. Since $n$ is a common factor of $0$ and $n$, the fraction is not in lowest terms unless $n=1$.

Answer 2

It is not a separate convention. By definition lowest terms means the denominator has minimal absolute value and is positive. The only denominator that meets this criterion with a numerator of $0$ is $1$. Ergo $0/1$.

Answer 3

I understand your confusion but you have written:

$\frac{p}{q} = -\frac{p}{q}$ which only holds true when $p = 0$ and $q \ne 0$

or in your words:

$\frac{0}{q} = -\frac{0}{q}$ which is completely true when $q \ne 0$

Now, on substituting $q = 1$ as you suggested we get:

$-\frac{0}{1} = 0 \ne -1$

[Side note: The reason I have added the condition $q \ne 0$ in the first statement is because in any $\frac{p}{q}$ for if $q = 0$ then the output is undefined. Check dividing by zero for more information.]

Answer 4

Yes, it's a convention to write fractions using positive denominators. The author arguably should have defined "in lowest terms" to clarify this, but most mathematical writing is not entirely precise and sometimes you have to make reasonable assumptions when interpreting it. Note that they're defining a function here, so we have to choose one denominator $q$ for each rational $x$. Even without the convention, it's hard to imagine they meant for the function to pick the negative denominator.