# Can you express this in form $\frac1a+\frac1b$

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.

• Sure. How about $$\frac{1}{\left(\dfrac{1949+\sqrt{3790605}}{2}\right)}+\frac{1}{\left(\dfrac{1949-\sqrt{3790605}}{2}\right)}\,?$$ ;p
– user137731
Oct 12, 2016 at 0:39
• I asked the wrong question, but since I already got some answers, I'll just ask a new question. Oct 12, 2016 at 1:00
• 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. Oct 13, 2016 at 13:51

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

• What if $a+b/ab$ can be simplified to $1994/1999$? Oct 12, 2016 at 0:47
• @N.S. Since 1999 is prime either a or b has to have 1999 as a factor Oct 12, 2016 at 0:49

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 $$a_1=\frac {1949-\sqrt{3790605}}{2}\\a_2=\frac {1949+\sqrt{3790605}}2$$