Agent Smith
Full Member
- Joined
- Oct 18, 2023
- Messages
- 285
We have the set of naturals [imath]\mathbb{N}[/imath]
We have the set of rationals [imath]\mathbb{Q}[/imath]
We have the set of irrationals [imath]\mathbb{Q'}[/imath]
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We have the set of rational approximations of irrationals [imath]_\text{a} \mathbb{Q'}[/imath]
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The set [imath]\mathbb{Q'}[/imath] is in bijection with the set [imath]_\text{a} \mathbb{Q'}[/imath] (every irrational has its own rational approximation)
That is to say [imath]|\mathbb{Q'}| = |_\text{a} \mathbb{Q'}|[/imath] ... (A)
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[imath]_\text{a} \mathbb{Q'} \subseteq \mathbb{Q}[/imath]
[imath]|_\text{a} \mathbb{Q'}| \leq |\mathbb{Q}|[/imath] ... (B)
[imath]\therefore |\mathbb{Q'}| \leq |\mathbb{Q}|[/imath] ... from (A) and (B)
The cardinality of the irrationals is at most equal to the cardinality of the rationals.
![Thinking face :thinking: 🤔](https://cdn.jsdelivr.net/joypixels/assets/7.0/png/unicode/64/1f914.png)
We have the set of rationals [imath]\mathbb{Q}[/imath]
We have the set of irrationals [imath]\mathbb{Q'}[/imath]
---
We have the set of rational approximations of irrationals [imath]_\text{a} \mathbb{Q'}[/imath]
---
The set [imath]\mathbb{Q'}[/imath] is in bijection with the set [imath]_\text{a} \mathbb{Q'}[/imath] (every irrational has its own rational approximation)
That is to say [imath]|\mathbb{Q'}| = |_\text{a} \mathbb{Q'}|[/imath] ... (A)
---
[imath]_\text{a} \mathbb{Q'} \subseteq \mathbb{Q}[/imath]
[imath]|_\text{a} \mathbb{Q'}| \leq |\mathbb{Q}|[/imath] ... (B)
[imath]\therefore |\mathbb{Q'}| \leq |\mathbb{Q}|[/imath] ... from (A) and (B)
The cardinality of the irrationals is at most equal to the cardinality of the rationals.
![Thinking face :thinking: 🤔](https://cdn.jsdelivr.net/joypixels/assets/7.0/png/unicode/64/1f914.png)