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Title: Rational Numbers (Ordering)
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Series: Start Learning Numbers
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Parent Series: Start Learning Mathematics
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Chapter: Numbers
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YouTube-Title: Start Learning Numbers 11 | Rational Numbers (Ordering)
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Bright video: Watch on YouTube
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Dark video: Watch on YouTube
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Ad-free video: Watch Vimeo video
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Forum: Ask a question in Mattermost
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Quiz: Test your knowledge
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Dark-PDF: Download PDF version of the dark video
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Print-PDF: Download printable PDF version
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Thumbnail (bright): Download PNG
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Thumbnail (dark): Download PNG
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Subtitle on GitHub: sln11_sub_eng.srt missing
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Download bright video: Link on Vimeo
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Download dark video: Link on Vimeo
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Timestamps (n/a)
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Subtitle in English (n/a)
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Quiz Content
Q1: Consider the order relation $\leq$ on $\mathbb{Q}$. What does it make it to an ordering by definition?
A1: It’s a relation that is reflexive, antisymmetric, and transitive.
A2: It’s a relation that is reflexive, symmetric, and transitive.
A3: It’s a relation that is not reflexive but transitive.
Q2: Consider the order relation $\leq$ on $\mathbb{Q}$. What does it make it to a total order by definition?
A1: For every two elements $x \neq y$, we either have $x \leq y$ or $y \leq x$.
A2: $x \leq y$ and $y \leq x$ always imply $x = y$.
A3: If $x \leq y$, then $z \cdot x \leq z \cdot y$ for all $z \in \mathbb{Q}$
Q3: What is the Archimedean property for the rational numbers?
A1: For two positive numbers $x, \varepsilon \in \mathbb{Q}$, you can always find a natural number $n$ to make $n \cdot \varepsilon$ larger than $x$.
A2: For every $x \in \mathbb{Q}$, there is a number $\varepsilon > 0$ such that $x > \varepsilon$.
A3: For two positive numbers $x, \varepsilon \in \mathbb{Q}$, you can always find a rational number $q$ to make $q + \varepsilon$ larger than $x$.
A4: For every positive $x \in \mathbb{Q}$, there is a number $\varepsilon > 0$ such that $\varepsilon \cdot x > \varepsilon$.
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Last update: 2025-07
