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This is from Class Note from 6.042 ocw courses at MIT:

"Well Ordering Principle" section:

You can read the original here at page 1 and 2; Well Ordering Principle: http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics-for-computer-science-fall-2010/readings/MIT6_042JF10_chap03.pdf


In fact, looking back, we took the Well Ordering Principle for granted in proving that $\sqrt{2}$ is irrational. That proof assumed that for any positive integers $m$ and $n$, the fraction $\frac{m}{n}$ can be written in lowest terms, that is, in the form $\frac{m'}{n'}$ where $m'$ and $n'$ are positive integers with no common factors. How do we know this is always possible?

Suppose to the contrary that there were $m$, $n$ in $\mathbb{Z}^+$ such that the fraction $\frac{m}{n}$ cannot be written in lowest terms. Now let $C$ be the set of positive integers that are numerators of such fractions. Then $m$ in $C$, so $C$ is nonempty. Therefore, by Well Ordering, there must be a smallest integer, $m_0$ in $C$. So by definition of $C$, there is an integer $n_0 > 0$ such that the fraction $\frac{m_0}{n_0}$ cannot be written in lowest terms. This means that $m_0$ and $n_0$ must have a common factor, $p > 1$. But

$(\frac{m_0}{p}) / (\frac{n_0}{p}) = \frac{m_0}{n_0}$

so any way of expressing the left hand fraction in lowest terms would also work for $\frac{m_0}{n_0}$, which implies the fraction($\frac{m_0}{p}) / (\frac{n_0}{p})$ cannot be in written in lowest terms either.

So by definition of $C$, the numerator, $\frac{m_0}{p}$, is in $C$. But $\frac{m_0}{p} < m_0$, which contradicts the fact that $m_0$ is the smallest element of $C$. Since the assumption that $C$ is nonempty leads to a contradiction, it follows that $C$ must be empty. That is, that there are no numerators of fractions that can’t be written in lowest terms, and hence there are no such fractions at all.


I don't really understand the part where, the author says:

This means that $m_0$ and $n_0$ must have a common factor, $p > 1$ .

BECAUSE $\frac {m_0} {n_0}$ is a irreducible fraction, both $m_0$ and $n_0$ have no common factors other than 1. If they had common factors other than one, then $\frac {m_0} {n_0}$ would not be a irreducible fraction. I think that IT IS NOT THE CASE THAT $m_0$ and $n_0$ must have a common factor, $p > 1$.

Let that fraction be $2/3$, $2/3$ is irreducible. 2 and 3 have no common factors other than $1$.

I assume that if a fraction cannot be written in lowest terms, then the fraction is irreducible, i.e. something like one half.


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  • $\begingroup$ The author is showing that the assumption of the existence of a rational $q$ that cannot be written in lowest terms will result in a paradox: There exists a least $n_0\in \mathbb N$ such that $q=m_0/n_0$ for some $m_0\in \mathbb Z$. But the assumption that $q $ can't be written in lowest terms implies that $m_0/n_0$ is not in lowest terms, implying some integer $ p>1$ divides both $m_0$ and $n_0,$ which implies the desired paradox. $\endgroup$ – DanielWainfleet Jun 13 '17 at 1:05
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$m_0$ and $n_0$ are defined so that $m_0/n_0$ cannot be written in lowest terms. This means by definition that they have a common factor (since if they didn't, then $m_0/n_0$ would already be in lowest terms, so can be written in lowest terms). I think you're confusing 'irreducible' with 'cannot be written in lowest terms'. These are actually quite different and mutually exclusive. (The fact that the second property is a hypothetical one that the author is proving cannot occur, and that no such $m_0$ and $n_0$ actually exist probably makes it more difficult.)

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  • $\begingroup$ what does irreducible fraction mean? Isn't $4/3$ not possible of being written in lowest terms? $\endgroup$ – user420360 Jun 12 '17 at 3:05
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    $\begingroup$ Irreducible means the numerator and denominator have no common fact. In other words it means it is in lowest terms. 4/3 is reducible / in lowest terms, so it is possible to write it in lowest terms. I think you're confusing 'lowest' with 'lower'. If something can be written in lower terms (i.e. you can write 6/4 as 3/2) then it is reducible. $\endgroup$ – spaceisdarkgreen Jun 12 '17 at 3:32
  • $\begingroup$ For what it's worth I agree with Bill Dubuque below that this is an unnecessarily roundabout proof. Also it is prefaced with 'how do we know that every fraction can be written in lowest terms?' and says the proof answers this, which is preposterous. Understanding this proof won't give you any insight on fractions that you don't already have, but might give you good practice parsing abstract proofs. $\endgroup$ – spaceisdarkgreen Jun 12 '17 at 3:37
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The proof is needlessly complex. We claim that every fraction $f$ can be written irreducibly, i.e. $\,f = a/b\,$ where $a,b$ are coprime. Indeed, by well ordering, there exists a representation $ f = a/b $ with $\rm\color{#c00}{least}$ denominator $b$. If $a,b$ were not coprime then we could cancel a common factor $d>1$ from $a,b$ yielding an equal fraction with smaller denominator $b/d < b\,$ contra $\rm\color{#c00}{least}$ness of $b$.

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