- #1
issacnewton
- 1,035
- 37
Hello
I am trying to prove that $\mathbb{Q}$ is not a finite set. I proceed with path of proof by contradiction. Suppose that $\mathbb{Q}$ is a finite set. Then $\exists\; n \in \mathbb{N}$ such that $I_n \sim \mathbb{Q}$, where
\[ I_n = \{i \in \mathbb{Z^{+}} |\; i \leq n\} \]
This means that there is a bijection from $\mathbb{Q}$ to $I_n$. Now since $\mathbb{N} \subseteq \mathbb{Q}$, we can pair first n numbers in $\mathbb{N}$ with n members in $I_n$. Let $n_1$ be the highest of these n numbers. Then $n_1 + 1 \in \mathbb{Q}$, but we can't pair this number with anyone in $I_n$ since we have exhausted all of them. This means that we can't have a function from $\mathbb{Q}$ to $I_n$, hence there is no bijection from $\mathbb{Q}$ to $I_n$. This is a contradiction. So $\mathbb{Q}$ is not a finite set.
Is this an ok proof ?
I am trying to prove that $\mathbb{Q}$ is not a finite set. I proceed with path of proof by contradiction. Suppose that $\mathbb{Q}$ is a finite set. Then $\exists\; n \in \mathbb{N}$ such that $I_n \sim \mathbb{Q}$, where
\[ I_n = \{i \in \mathbb{Z^{+}} |\; i \leq n\} \]
This means that there is a bijection from $\mathbb{Q}$ to $I_n$. Now since $\mathbb{N} \subseteq \mathbb{Q}$, we can pair first n numbers in $\mathbb{N}$ with n members in $I_n$. Let $n_1$ be the highest of these n numbers. Then $n_1 + 1 \in \mathbb{Q}$, but we can't pair this number with anyone in $I_n$ since we have exhausted all of them. This means that we can't have a function from $\mathbb{Q}$ to $I_n$, hence there is no bijection from $\mathbb{Q}$ to $I_n$. This is a contradiction. So $\mathbb{Q}$ is not a finite set.
Is this an ok proof ?
