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It seems to overlap with other content. Sorry for the confusion.

Young
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    If the sum is an integer, what does that tell you about the denominators? – Daniel Fischer Mar 13 '14 at 03:07
  • @BillDubuque Let's call them $x$ and $y$. $x = \lfloor x \rfloor + \frac{p}{q}$, with $\gcd (p,q) = 1$. $y = \lceil y\rceil - \frac{r}{s}$ with $\gcd(r,s) = 1$. $x+y \in \mathbb{Z}$ implies $d := \frac{p}{q} - \frac{r}{s} \in \mathbb{Z}$. Since $0 \leqslant \frac{p}{q} < 1$ and $0 \leqslant \frac{r}{s} < 1$, it follows that $-1 < d < 1$. Which means $\frac{p}{q} = \frac{r}{s}$. – Daniel Fischer Mar 13 '14 at 21:51
  • @BillDubuque $(d\in\mathbb{Z} \land -1 < d < 1) \implies (d = 0)$. $\frac{p}{q} - \frac{r}{s} = 0 \iff \frac{p}{q} = \frac{r}{s}$. Then we deduce $(p = r) \land (q = s)$ from $\gcd(p,q) = 1 = \gcd(r,s)$. – Daniel Fischer Mar 13 '14 at 22:04
  • @Daniel Thanks, now it is clear which proof you intend. It essentially invokes the uniqueness of least terms representations of fractions - what I call unique fractionization. This is equivalent to unique factorization, Euclid's Lemma, etc. Many of my prior posts discuss this further. Given that approach, one can skip your reduction to proper fractions and proceed directly as follows, where the second line of the proof has essentially inlined a proof of unique fractionizaton. – Bill Dubuque Mar 14 '14 at 01:50
  • $\dfrac{a}b=n-\dfrac{c}d=\dfrac{nd-c}d=:\dfrac{\bar c}d\ $ and $\ \color{#0a0}{(b,a)}=1=(d,c),\Rightarrow, \color{#c00}{(d,\bar c)}=(d,nd!-!c)=(d,c)=1.\ $ So $,ad = \bar cb^{\phantom{I^I}}!!!!,,$ by EL = Euclid's Lemma $\ \color{#0a0}{b\mid a},d,\Rightarrow, b\mid d,\ $ and $,\ \color{#c00}{d\mid \bar c},b,\Rightarrow,d\mid b,,$ hence $\ \color{#c0d}{b = d}.\ $

    Therefore $\ \dfrac{a}b\dfrac{c}d\in\Bbb Z,\Rightarrow,b\mid ac,\Rightarrow,b=1,$ by $,\color{}{(b,a)}=1=\color{}{\color{#b0d}{(b,c)=(d,c)}},,$ and EL. $\ $ QED $\ \ $

     – Bill Dubuque Mar 14 '14 at 02:09
  • @Daniel See my answer for a more direct way. – Bill Dubuque Mar 14 '14 at 18:33

5 Answers5

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This may not be the simplest proof, but I think it's pretty. Let your two rational numbers be $r,s$. Then the polynomial $f(x)=(x-r)(x-s)=x^2-(r+s)x+rs$ has integer coefficients by your hypotheses. By Gauss' Lemma, this polynomial must be reducible over the integers, and hence $r,s$ are both integers.

Mark S.
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vadim123
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Let $p,q$ be rational. Then $p+q=n\in\mathbb{Z}\implies p=n-q$. So let $\displaystyle q=\frac{a}{b}$ be in lowest terms. We then have $(n-\frac{a}{b})\frac{a}{b}=m\in\mathbb{Z}\implies na-\frac{a^2}{b}=mb\implies\frac{a^2}{b}\in\mathbb{Z}\implies \frac{a}{b}\in\mathbb{Z}$ since $\frac{a}{b}$ are coprime. So $q$, hence also $p$ are integers.

JLA
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  • This is essentially the common proof of the Rational Root Test, specialized to the case of a monic quadratic polynomial, here $, (x-p)(x-q), =, x^2-(p+q)x + pq \in \Bbb Z[x].\ \ $ – Bill Dubuque Mar 13 '14 at 16:17
  • @BillDubuque Maybe, but I don't see the motivation behind looking at that polynomial. – JLA Mar 13 '14 at 16:37
  • It is this: $,p+q,$ and $,pq,$ are both integers iff $,f(x) = (x-p)(x-q)$ has integer coefficients. By RRT, we know that rational roots of such a monic polynomial $\in\Bbb Z[x],$ must be integral. The conceptual foundations will be clearer when one learns about algebraic integers. – Bill Dubuque Mar 13 '14 at 16:59
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By plugging in $x=a$ and $x=b$, we see that $$ x^2-(a+b)x+ab=0 $$ As shown in this answer, a rational root of a monic polynomial with integer coefficients must be an integer.

Importing the Referenced Answer

It has been suggested that specializing the proof in the above referenced answer to quadratic polynomials might be useful.

Suppose $x=\frac pq$, where $ps+qr=1$, is a root of $x^2+mx+n=0$, where $m,n\in\mathbb{Z}$.
Subsitute $x=\frac{1-qr}{qs}$ $$ \frac{(1-qr)^2}{q^2s^2}+m\frac{1-qr}{qs}+n=0 $$ Multiply by $pqs^2$ $$ \left(\frac pq-2pr+pqr^2\right)+pms(1-qr)+npqs^2=0 $$ cancelling yields $$ \frac pq=2pr-pqr^2+pms(qr-1)-npqs^2 $$ In particular, $x=\frac pq\in\mathbb{Z}$.

Community
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robjohn
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  • Above is a perfect opportunity to specialize your linked Bezout-based proof to the quadratic case, which should prove more comprehensible than the general proof. Adding that above would be nice. With your new $\rm 100K$ powered Bezouka, can you Bezoutify the $1$-line proof in my answer? Congratulations! – Bill Dubuque Mar 14 '14 at 18:52
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Starting with the equation provided by Vadim above, $x^2-(r+s)x + rs=0$, take any rational root $a/b$, with $\gcd(a,b)=1$. We get, after clearing the denominator: $a^2 -b(r+s)x +rsb^2=0$,which can be rewritten to show $a^2$ is a multiple of $b$ which contradicts the fact that $a$ and $b$ are co-prime. (This is essentially the proof that a rational number that is an algebraic integer is an integer, for the quadratic case).

P Vanchinathan
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$\underbrace{\dfrac{a}b\!+\!\dfrac{c}d}_{\large (a,b)\,=\,1}\! =n\in\Bbb Z,\ \dfrac{a}b\not\in\Bbb Z\overset{\large \exists\, p\ {\rm prime}}\Rightarrow$ $\begin{array}{}p\nmid a\\ p\mid b\end{array}$ $\Rightarrow$ $\,\dfrac{a}b\dfrac{c}d = \dfrac{a}b\!\!\!\dfrac{\,(bn\!-\!a)}{b}\!\not\in\Bbb Z\,$ by $\,\begin{array}{} p\nmid a,\, bn\!-\!a\\ p\mid b\end{array}\ $ QED

Bill Dubuque
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