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Can you express the fraction $\frac{1949}{1999}$ in the form $\frac 1a+\frac 1b$? Give reasons supporting your answer.


I think the only way to do this is keep trying numbers but then I will never get the answer. I cry every time.

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    $\begingroup$ Sure. How about $$\frac{1}{\left(\dfrac{1949+\sqrt{3790605}}{2}\right)}+\frac{1}{\left(\dfrac{1949-\sqrt{3790605}}{2}\right)}\,?$$ ;p $\endgroup$
    – user137731
    Oct 12, 2016 at 0:39
  • $\begingroup$ I asked the wrong question, but since I already got some answers, I'll just ask a new question. $\endgroup$
    – joonoon
    Oct 12, 2016 at 1:00
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    $\begingroup$ the ad hoc way: note that 1/3 + 1/3 is not enough, so you may assume a = 2, in which case it's clear that b will not be an integer. $\endgroup$
    – Woett
    Oct 13, 2016 at 13:51

2 Answers 2

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$$\frac 1a+\frac 1b = \frac{a+b}{ab},$$ so you would need $ab$ to divide $1999$. But…

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    $\begingroup$ What if $a+b/ab$ can be simplified to $1994/1999$? $\endgroup$
    – N. S.
    Oct 12, 2016 at 0:47
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    $\begingroup$ @N.S. Since 1999 is prime either a or b has to have 1999 as a factor $\endgroup$ Oct 12, 2016 at 0:49
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Given $$\frac 1a+\frac 1b=\frac {1949}{1999}$$ Combining the fraction gives $$\frac {a+b}{ab}=\frac {1949}{1999}$$ Setting terms gives the system $$\begin{cases}a+b=1949\\ab=1999\end{cases}$$ With $a,b$ can be solved by a quadratic. Namely $$b^2-1949b+1999$$ Where $a=1949-b$.

Since $1999$ cannot be factored, the roots are really ugly looking numbers. Namely, the two possible values of $b$ are $$b_1=\frac {1949+\sqrt{3790605}}2\\b_2=\frac {1949-\sqrt{3790605}}2$$and with the $a$ values as $$a_1=\frac {1949-\sqrt{3790605}}{2}\\a_2=\frac {1949+\sqrt{3790605}}2$$

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