The Product of Rational Numbers

[abbeyofthelema]

I've been grinding my brain at this problem because I am trying to figure out if the product of two rational numbers is always, never, or sometimes rational. a rational number would either have to terminate, or be infinitely periodic, so i would say that the product of two rational numbers is always rational, but i can't say this for sure :(

 

[whozum]

I can't prove it, but its true.

 

[abbeyofthelema]

coolio

well thank you, is there any particular reason that you believe it to be true?

 

[Galileo]

I've been grinding my brain at this problem because I am trying to figure out if the product of two rational numbers is always, never, or sometimes rational. a rational number would either have to terminate, or be infinitely periodic, so i would say that the product of two rational numbers is always rational, but i can't say this for sure :(

Simply use the definition of a rational number.

A rational number can be written in the form:

\frac{a}{b},
with a and b integers and b not equal to zero.

Suppose you have two rational numbers. Compute their product (and their sum while you're at it). Is the result again of the above form?

 

[Data]

Yes. If you want to try to prove something, then 100% of the time your best bet for a first step is to write down the definitions of the things you're working with

The fact that rationals happen to be the set of all reals with periodic limiting behaviour in their decimal representations is a derived property. The definition of a rational number is just that it can be represented as a quotient of integers.

 

[abbeyofthelema]

definitely

that's great. so then the product of two rational numbers must always be rational :)

 

[danne89]

Def:
\frac{a}{b} ~ , ~ ~ a, b \in \mathbb{N}
It's easy to see that a product of two natrual number must be natrual, thus
\frac{a}{b} ~ * \frac {c}{d} ~ = ~ \frac{ac}{bd}, ~ ~ a, b, c, d \in \mathbb{N}
Setting ac to a and bd to b in the def. finish the proof.