0

How would one go about setting out a formal proof that the product of two positive proper fractions will generate a positive number less than either of them?

(This is not homework, rather curiosity how to formally prove something that is generally taken as self-evident)

ose
  • 177

1 Answers1

2

The comments have proof simply. But I want to offer a more complicated proof (way more complicated than necessary), but that reveals the structure of "proper fractions" and how that structure leads to your property in question.

A positive proper fraction $A$ can be written in the form $$ A = \frac{a}{a + \epsilon} \mbox{ where } 0< a, \epsilon \in \mathbb{N}. $$ Note that $0 < A < 1$.

So consider two proper fractions, $A$ and $B$, and their product $P$: $$ P = AB = \frac{a}{a + \epsilon_1} \left( \frac{b}{b + \epsilon_2} \right) = \frac{ab}{ab + ( a\epsilon_2 + b\epsilon_1 + \epsilon_1 \epsilon_2)}. $$

Firstly we can see that the product is proper positive, $0 < P < 1$.

To check that $P<A$, note that $P/A = B < 1$. Similarly $P<B$ since $P/B = A < 1$.

NazimJ
  • 3,244
  • Thanks this is the type of proof I was looking for. I'm still not following the last line. I can see this proves that P < 1, but how does it prove that P < A and P < B? – ose Oct 06 '21 at 14:49
  • $P/A < 1$ implies $P<A$. Because a smaller number divided by a larger number is less than $1$ – NazimJ Oct 06 '21 at 14:50