# How do you prove that a sequence is well defined?

I'm not sure what it even means for a sequence to be well defined. I'm supposed to prove that the sequence $x_{n+2} = \frac{x_{n+1}}{x_{n}}$ where $x_{1} = a$ and $x_{2} = b$ for $a>0 , b>0$ is well defined.

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More details, please! In general "well-defined" means that some object was given a description up to some arbitrary choices, but that the arbitrary choices have no material effect. – Kevin Carlson Oct 21 '12 at 6:31

You have a recurrence $x_{n+2}=f(x_{n+1},x_n)$ with initial values $x_1=a$ and $x_2=b$. To show that the resulting sequence $\langle x_n:n\in\Bbb Z^+\rangle$ is well-defined, you must show that the recurrence and initial values actually do determine a unique real number $x_n$ for each $n\ge 1$. If the function $f(x,y)$ is defined for all real numbers $x$ and $y$, there is no problem. In this case, however, the recurrence is

$$x_{n+2}=\frac{x_{n+1}}{x_n}\;,\tag{1}$$

and the function $f$ is $f(x,y)=\dfrac{x}y$, which is not defined when $y=0$.

Specifically, if the recursion $(1)$ every produced a $0$ output, so that some $x_n=0$, you’d have

$$x_{n+1}=\frac{x_n}{x_{n-1}}=\frac0{x_{n-1}}=0$$

and then $$x_{n+2}=\frac{x_{n+1}}{x_n}=\frac00\;,$$ which is not well-defined.

You’re being asked to prove that if the initial values $a$ and $b$ are positive, this never happens. In other words, you want to show that if $a,b>0$, then the recurrence $(1)$ never produces a term $x_n=0$.

Think about it for a moment: if you divide one positive number by another, what kind of number do you get as quotient? Now convert that insight into a proof by induction that if $a,b>0$, then every $x_n$ is ... what?

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In the definition of the sequence you divide by $x_n$. So for well-definedness you better make sure this is not zero.

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You simply have to prove that the sequence returns a real number for any given value of $n$ i.e. $$\forall n \in \mathbb N, a_n \in \mathbb R$$ Or you can also prove by contradiction to show that the sequence isn't well defined as @Brian M. Scott did.

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