Proof by Induction with Two Basis I did an assignment where I used the proof by induction three times in the same manner, and now I have doubts creeping up whether this method was valid or not. I would be quite relieved if you could tell me if that's correct or not:

The sequence $(a_n)_{n\in\mathbb{N}}$ is given through $$a_1 = 1,\quad a_2 = \frac{1}{2},\quad a_{n+2}=a_na_{n+1} \text{ for } n\geq 1.$$

I want to show $A(n): a_n \geq a_{n+1} \forall\, n\geq 1$. So I did the base case as follows $$n=1:\quad a_1 = 1,\quad a_2=\frac{1}{2},\quad \therefore a_1=1 \geq \frac{1}{2} = a_2 \quad\checkmark$$
$$n=2:\quad a_2 = \frac{1}{2},\quad a_3=\frac{1}{2},\quad \therefore a_2=\frac{1}{2} \geq \frac{1}{2} = a_3 \quad\checkmark$$
Now comes the part where I'm insecure. For the induction step I assume that $A(n): a_{n} \geq a_{n+1}$ and $A(n+1): a_{n+1} \geq a_{n+2}$ are true and I want to deduce $A(n+2): a_{n+2} \geq a_{n+3}$. Because of the ordering axiom of $(\mathbb{R},+,\cdot\,)$ I think I can multiply $A(n)$ and $A(n+1)$ to get $$a_n \cdot a_{n+1} \geq a_{n+1} \cdot a_{n+2}$$
and with the definition of the recursive defined sequence $a_n$ I would get
$$a_{n+2} = a_n \cdot a_{n+1} \quad\geq\quad a_{n+1} \cdot a_{n+2} = a_{(n+1)+2} =  a_{n+3} \quad\checkmark$$
the induction step.
Do I get away with this?
 A: That works if you first prove (by induction) that all $a_n$ are positive. That will let you apply the ordering axiom as you stated.
For an example of why you need them positive,
$$5 \geq 4$$
$$-1 \geq -1$$
but it is not true that
$$-5 \geq -4$$

Strictly speaking, you are appealing to a slightly stronger induction principle than the one which is usually stated at school level. The weaker principle is "If $P(1)$, and if $P(n) \Rightarrow P(n+1)$, then $P(n)$ for all $n \in \mathbb{N}$." The stronger principle is "If $P(1)$ and $P(2)$, and $P(n), P(n+1) \Rightarrow P(n+2)$, then $P(n)$ for all $n \in \mathbb{N}$." Both principles are completely correct and true.
In fact, it gets more general: "If $P(1)$, and if $(\forall n < m)(P(m))$ implies $P(n)$, then $P(n)$ for all $n \in \mathbb{N}$." (Do you see why this includes both the one-step and the two-step induction principles as special cases?) This more general version is called "strong induction", and you can use the one-step induction principle to prove it. Likewise, it is possible to use the one-step induction principle to prove the two-step induction principle directly.
