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I am asked to argue whether or not the following two functions are well-defined (textbook definition: a) define $y$ for all $x$ in domain, and b) any is mapped to exactly one y). Both of the below functions are functions from $\mathbb{Q}$ to $\mathbb{Q}$.

$$f\left(\frac{p}{q}\right) = \frac{p+1}{q}$$

$$g\left(\frac{p}{q}\right) = \frac{p+q}{p-q}$$

My argument is that since $0$ is a rational number, we can take, for $f$, $p=0$ and $q=x$ and the function will not be defined. Similarly, we can take $p=q=0$ for $g$, and the function, again, will not be defined.

But the argument seems to be too easy. Is there something I am missing that won't allow me to use these two counter examples?

Thanks!

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For the first one you probably means $p=0$ and $q=x$, otherwise you divide by $0$. For the second one, if $p=0$ and $q=0$, what does $\frac{p}{q}=\frac{0}{0}$ mean? –  N. S. Jun 3 '12 at 18:59
    
You are correct, typo. –  d185431 Jun 3 '12 at 19:06
    
As an addendum, the notion of a "homogeneous polynomial" is important to defining things this way. A homogeneous polynomial is one where the total degree of every monomial is the same, such as $x^2 + 2 xy + 3 y^2$. When you're defining $f(x/y)$ as a rational function in $x$ and $y$, the numerator and denominator must be homogeneous polynomials of the same degree. –  Hurkyl Jun 3 '12 at 19:33
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3 Answers

You might want to think of the question like this, with $r =\frac pq$

For the first case $f(r) = r + \frac 1q$

For the second case note that $q\neq 0 $ if $r$ is rational, and $f(r) = \frac {r+1}{r-1}$

A well-defined function is one for which you can determine a single defined value for each $r$.

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First note that in the definition of rationals, $q$ is not allowed to be zero i.e the numbers to be consider are of the form $\dfrac{p}{q}$, where $p \in \mathbb{Z}$ and $q \in \mathbb{Z} \backslash \{0\}$.

Consider the first function $$f \left( \dfrac{p}{q}\right) = \dfrac{p+1}{q}$$ Note that $$\dfrac{kp}{kq} = \dfrac{p}{q}$$ where $k \in \mathbb{Z} \backslash \{0\}$. Hence, for the function to be well defined we need $$f \left( \dfrac{p}{q}\right) = f \left( \dfrac{kp}{kq}\right) $$ i.e. $$\dfrac{p+1}{q} = \dfrac{kp+1}{kq}, \text{ for all }k \in \mathbb{Z}$$which is clearly not true. Hence, the first function is not well-defined.

The second function $$f \left( \dfrac{p}{q}\right) = \dfrac{p+q}{p-q}$$ is well-defined except for the rational number $1$ since $$f \left( \dfrac{p}{p}\right) = \dfrac{2p}{0} = \text{ not defined}$$ For all other rationals except $1$, note that $f \left( \dfrac{kp}{kq} \right)$ where $p \neq q$ and $k \in \mathbb{Z} \backslash \{0\}$, we get that $$f \left( \dfrac{kp}{kq} \right) = \dfrac{kp+kq}{kp-kq} = \dfrac{p+q}{p-q} = f \left( \dfrac{p}{q} \right)$$

Hence, the second function is well-defined for $x \in \mathbb{Q} \backslash \{1\}$.

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Your argument does not quite work because $\frac{0}{0}$ is not a rational number. What you want to look for is two different representations of the same fraction that give different answers. For instance

$$ f\left(\frac{1}{2}\right)=\frac{2}{2}=1, $$

but,

$$ f\left(\frac{2}{4}\right)=\frac{3}{4}. $$

The problem is that in $\mathbb{Q}$, $\frac{1}{2}=\frac{2}{4}$, but the function is not the same for both.

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