Notation for the product of $n$ rational numbers (proof by induction)

Question

I have a proof I'm trying to solve: For any natural number $n$, the product of $n$ rational numbers is rational.

The base case is fairly easy. When $n = 1, 1 =\dfrac{a}{b}.$ $1$ is rational when $a = b$ and $a,b$ are elements of the set $\mathbb{N}$.

I have something like this. $(1 \cdot \dfrac{1}{2} \cdot \dfrac{1}{3} \cdot \dfrac{1}{4} \cdot \dfrac{1}{5} \cdot 2 \cdot \dfrac{1}{50}) \cdot n + 1 = \dfrac{r}{s}.$ The terms in parentheses are $n = \dfrac{a}{b}$ by our inductive step. Since $n + 1$ and $a$ are natural numbers, $(n+1) \cdot a$ is also a natural number so therefore the product of $\dfrac{a}{b} \cdot (n+1)$ is also a rational number.

I haven't written the proof yet because I'd like it to be a little cleaner and write the product down as a notation. I'm not really sure what to put for $i$ though. So far, I have (product notation) of $i = 1$ to $n$ is equal to $\dfrac{a}{b}$.

Sorry if it may not be clear, I'm not sure how to input notation yet and this is my first post. Thanks!

Answer 1

Begin with the trivial case: $n = 0$. The product of no numbers is $1$, a rational.

Let $n = 1$. Simply choose a rational number.

Let $n = 2$. Let the first rational be $a_1/b_1$. The second $a_2/b_2$. Then, $a_1/b_1 \times a_2/b_2 = (a_1b_1)/(a_2b_2)$, a rational.

Answer 2

What you'd like to prove (by induction) is that if $r_1, r_2, r_3,\ldots$ is a sequence of rational numbers, then so is $R_1, R_2, R_3,\ldots$, where

$$R_n=\prod_{k=1}^nr_k$$

The thing to note is that the product notation is itself defined inductively, that is,

$$\prod_{k=1}^1r_k=r_1$$

and

$$\prod_{k=1}^{n+1}r_k=\left(\prod_{k=1}^nr_k\right)r_{n+1}$$

The induction is now straightforward: the base case $R_1=r_1$ is rational by assumption, and, if $R_n$ is rational (i.e., the inductive hypothesis) then so is $R_{n+1}=R_nr_{n+1}$, since $r_{n+1}$ is rational by assumption, and the product of two rational numbers is rational.

Answer 3

How to write this out: For $n\in \Bbb N$ let $S(n)$ be the sentence $\{x_j: n\ge j\in \Bbb N\}\subset \Bbb Q\implies\prod_{j=1}^n x_j\in \Bbb Q.$

(i). $S(1) \iff (\{x_1\}\subset \Bbb Q \implies \prod_{j=1}^1 x_j\in \Bbb Q) \iff ( x_1\in \Bbb Q \implies x_1=$ $\prod_{j=1}^1x_1\in \Bbb Q).$

So $S(1)$ is true.

(ii). If $n\in \Bbb N$ and $\{x_1,...,x_{n+1}\}\subset \Bbb Q$ then $S(n)\implies q=\prod_{j=1}^nx_j\in \Bbb Q\implies \prod_{j=1}^{n+1}x_j=qx_{n+1}\in \Bbb Q.$

Because $(q\in \Bbb Q \land x_{n+1}\in \Bbb Q)\implies (xq_{n+1}\in \Bbb Q). $ And I assume that we are not required to prove this.

Therefore $S(n)\implies S(n+1)$ for each $n\in \Bbb N.$

(iii). By (i) and (ii), induction on $n\in \Bbb N$ implies that $S(n)$ is true for all $n\in \Bbb N.$

Remark. The definition of $S(n)$ is not really a "sentence about sentences" but is just the definition of a typographical abbreviation "$S(n)$" for a sequence of key-strokes so that we don't have write out the whole sequence all the time.