I have the following task
$S_n: (1 + x)^n \geq 1 + nx$ for every $x \in \mathbb{R}$ with $x > −1$
I am wondering which number to take for the base case in the inductive proof, because when I choose $0$, I will not have shown it for the real numbers in $(0,-1)$. When showing it for $-0.999999$, I will have decreased the interval for numbers that I have not proven to $(-0.999999,-1)$ which still contains numbers that I have not shown....What should I do?
On
The proof is by induction on $n$; you cannot do induction on $x$.
The proposition you want to prove by induction is
for every $n\in\mathbb{N}$, for every $x>-1$, $(1+x)^n\ge 1+nx$
Thus the family of propositions to prove true is
$P(n)$: for every $x>-1$, $(1+x)^n\ge 1+nx$
The base case is $n=0$: $(1+x)^0=1$, $1+0x=1$. So $P(0)$ holds.
Induction step. Suppose $(1+x)^n\ge1+nx$. Then $$ (1+x)^{n+1}=(1+x)(1+x)^n\ge(1+x)(1+nx)=1+(n+1)x+nx^2\ge1+(n+1)x $$
You should not choose any $x$: the statements should hold for every $x>-1$.
Hint: Show something stronger: show that it holds for all $x\geq -1$ instead.