well defined functions

[logistic_guy]

Determine whether the following functions \(\displaystyle f\) are well defined:

\(\displaystyle \bold{(a)} \ f \ : \ \mathbb{Q} \rightarrow \mathbb{Z}\) defined by \(\displaystyle f\left(\frac{a}{b}\right) = a\).

\(\displaystyle \bold{(b)}\ f \ : \ \mathbb{Q} \rightarrow \mathbb{Q}\) defined by \(\displaystyle f\left(\frac{a}{b}\right) = \frac{a^2}{b^2}\).

 

[khansaheb]

Determine whether the following functions \(\displaystyle f\) are well defined:

\(\displaystyle \bold{(a)} \ f \ : \ \mathbb{Q} \rightarrow \mathbb{Z}\) defined by \(\displaystyle f\left(\frac{a}{b}\right) = a\).

\(\displaystyle \bold{(b)}\ f \ : \ \mathbb{Q} \rightarrow \mathbb{Q}\) defined by \(\displaystyle f\left(\frac{a}{b}\right) = \frac{a^2}{b^2}\).

show us your effort/s to solve this problem.

 

[logistic_guy]

\(\displaystyle \bold{(a)}\)
Let us pick any rational number, say \(\displaystyle \frac{5}{6}\), and apply the definition.

\(\displaystyle f\left(\frac{5}{6}\right) = 5\)

We also know that \(\displaystyle \frac{5}{6} = \frac{5}{6} \times \frac{2}{2} = \frac{10}{12}\), so:

\(\displaystyle f\left(\frac{10}{12}\right) = 10\)

Since \(\displaystyle f\left(\frac{5}{6}\right) \neq f\left(\frac{10}{12}\right)\), then \(\displaystyle f\) is not well defined.

 

[logistic_guy]

\(\displaystyle \bold{(b)}\)

If I choose the same numbers as in \(\displaystyle \bold{(a)}\), I get:

\(\displaystyle f\left(\frac{5}{6}\right) = \frac{5^2}{6^2} = \frac{25}{36}\)

And

\(\displaystyle f\left(\frac{10}{12}\right) = \frac{10^2}{12^2} = \frac{100}{144} = \frac{50}{72} = \frac{25}{36}\)

They are equalbut this is not enough. I need to prove this for any arbitrary rational number.

I will let \(\displaystyle \frac{x}{y} = \frac{m}{n}\)

\(\displaystyle f\left(\frac{x}{y}\right) = \frac{x^2}{y^2}\)

And

\(\displaystyle f\left(\frac{m}{n}\right) = \frac{m^2}{n^2}\)

Since \(\displaystyle \frac{x^2}{y^2} = \frac{m^2}{n^2}\)

It means

\(\displaystyle f\left(\frac{x}{y}\right) = f\left(\frac{m}{n}\right)\)

Then,

\(\displaystyle f\) in this case is well defined.