Flaw in proof involving binomial theorem

Question

Suppose we have $-1 < x < 0$, also an irrational $r$. We have three claims:

1. $x^r$ is not defined on $\mathbb{R}$(pretty obvious right)
2. $(1+x)^r$ is defined on $\mathbb{R}$(once again, pretty obvious)
3. by binomial theorem

$$(x+1)^r = \sum_{k=1}^{\infty}{r \choose k}x^{r-k} $$ since $k$ is an integer, we have $r-k$ still being irrational, thus claim 1 implies that $x^{r-k}$ is undefined on $\mathbb{R}$ for all $k$, implying that every term in the summation will be undefined on $\mathbb{R}$. Thus the summation itself is undefined on $\mathbb{R}$. Then we have $(1+x)^r$ undefined on $\mathbb{R}$.

Clearly 3 contradicts 2 and 3 is wrong, but what's the flaw in my argument?

Answer 1

3. is wrong. The correct binomial expansion is $$ (1+x)^r=\sum_{k=0}^\infty\binom{r}{k}x^k,\quad |x|<1. $$