Question

show that every unit fraction i.e. a fraction of the form \(\frac{1}{n}\) can be written as the sum of two other unit fractions

Answer 1

are the two other unit fractions unique?

Answer 2

good question

Answer 3

let me see if i can find two different ways to do it for some unit fraction
seems kind of likely, not sure

Answer 4

a trivial way is \[\frac{1}{2n}+\frac{1}{2n}\]

Answer 5

so i assume they have to be different

Answer 6

ok cheater, let me rephrase
two different unit fractions

Answer 7

a unit fraction that looks like 1/[n(n+1)] can be wrriten as 1/n -1/(n+1)

Answer 8

then you win!

Answer 9

so n can be negative : eg: 3, -4

Answer 10

positive
@myininaya has it

Answer 11

does this have anything to do with egyptian fractions

Answer 12

hmm not that i know of

Answer 13

you need to add a slight restriction on the value of n (if the two other fractions are different)

Answer 14

reason I say that is because of this...
http://en.wikipedia.org/wiki/Unit_fraction

Answer 15

yes, i see that now
i guess \(n>1\)

Answer 16

\[n(n+1)=17 \\ n^2+n-17=0 \\ n=\frac{-1 \pm \sqrt{1-4(-17)}}{2(1)}=\frac{-1 \pm \sqrt{69}}{2} \\ \text{ choose } n=\frac{-1 + \sqrt{69}}{2} \\ \text{ so } n+1=\frac{-1 +\sqrt{69}}{2}+1 =\frac{1+\sqrt{69}}{2} \\ \frac{1}{17}=\frac{1}{\frac{-1 +\sqrt{69}}{2}}-\frac{1}{\frac{1+\sqrt{69}}{2}}\]
I guess that is still consider a sum (or difference) of unit fractions

Answer 17

we did't say n had to be an integer

Answer 18

you had it at step one

Answer 19

"A unit fraction is a rational number written as a fraction where the numerator is one and the denominator is a positive integer." ;)

Answer 20

\[\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\\
\frac{1}{n}=\frac{1}{n+1}+\frac{1}{n(n+1)}\]

Answer 21

oh wow I'm so dumb

Answer 22

I could have just solved my thingy for 1/n

Answer 23

now you can write 1/17 as a sum of actual unit fractions (now i know the definition of unit fractions which makes sense for that to be the definition)

Answer 24

i'm so dumb lol

Answer 25

no me more
let's argue with each other who is more dumb
it will be fun

Answer 26

another awesome question :)

Answer 27

no dumb jokes, too easy