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What is the mathematically rigorous way to answer the question: "what is the largest fraction less than 1"? (or to explain why it cannot be answered in the manner worded).

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  • $\begingroup$ The answer should be "nothing". $\endgroup$ – mathlove Dec 23 '13 at 10:07
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Q: What is the largest fraction less than one?

A: There is no such fraction.


To see that this is the correct answer, note that for any fraction $\frac{p}{q}$ less than one, there is a slightly bigger fraction which is still less than one; in particular, $\frac{p+q}{2q}$ is a fraction such that $\frac{p}{q} < \frac{p+q}{2q} < 1$.

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For every fraction that we suppose to be the largest less than one, we can always find the arithmetic mean of this fraction and one, which is a rational number, too, and nearer to 1. So there is no such fraction.

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Here's another way to put it. Just consider the fractions $ \dfrac {x}{x+1} $. As $x$ gets larger, this fraction gets asymptotically closer to $1$, but never reaches it. Specifically, $$ \lim_{x \to \infty} \dfrac {x}{x+1} = 1 $$ but $ \dfrac {x}{x+1} \ne 1 $, for any $x$. Also note that, for integral values of the numerator and denominator (simplified), $\dfrac{x}{x+1}$ is the largest fraction less than $1$. Thus, it is impossible to find a definite largest rational number less than $1$.

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Q: "What is the largest element of the set $A = (0, 1) \cap \mathbb{Q}$ in a partially ordered set $(\le, \mathbb{R})$?"

A: "None, since if there's an element $p \in A$, then there always is an element $q = (1+p)/2 \in A$ which satisfies $q \not\le p$ (even thought $A$ has a least upper bound $1$)."

While the "usual" $\le$ does not satisfies the property "every non-empty set in $\mathbb{R}$ have a largest element", there exists an ordering (an ordering is a "rule" which orders elements in $\mathbb{R}$) which satisfies the property; which is the well-ordering theorem - "every set $A$ can be given an ordering which makes every non-empty subset of $A$ have a largest element".

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There isn't one. Suppose there were; let's call it $y$, where $y<1$.

Let $\epsilon = 1 -y$, the difference between $y$ and 1. $\epsilon$ is positive, and so $0 < \frac\epsilon2 < \epsilon$, and then $y < y+\frac\epsilon2 < y+\epsilon = 1$, which shows that $y+\frac\epsilon2$ is even closer to 1 than $y$ was.

So there is no number that is closest to 1. Whatever $y$ you pick, however close it is, there is another number that is even closer.

Consider the analogous question: “$x < \infty$; what is the greatest value of $x$?” There is no such $x$.

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