Write down a few simple positive fractions.
Now create a new fraction, such that its numerator equals the sum of all the numerators you wrote down and the denominator equals the sum of all the denominators.
Example: you wrote down 1/3, 2/7, 4/15 so the new fraction is 7/25.
Prove: the new fraction is smaller than the largest on the initial list
and bigger than the smallest.
Let's suppose a1/b1 < a2/b2 <a3/b3 ... <an/bn
If n=2, (a1+a2)/(b1+b2)  a1/b1 = (a2b1  a1b2)/b1(b1+b2)
The above expression equals a2/b2  a1/b1 after multiplies by the positive number b1(b1+b2)/b1b2. a2/b2  a1/b1 > 0. Therefore (a1+a2)/(b1+b2)  a1/b1 > 0.
Now suppose (a2+a3+...+an)/(b2+b3+...+bn) > a2/b2 is already true.
(a1+a2+a3+...+an)/(b1+b2+b3+...+bn)  a1/b1 = [b1(a1+a2+a3+...+an)  a1(b1+b2+b3+...+bn)]/[b1(b1+b2+b3+...+bn)] = [b1(a2+a3+...+an)  a1(b2+b3+...+bn)]/[b1(b1+b2+b3+...+bn)]
The above expression equals (a2+a3+...+an)/(b2+b3+...+bn)  a1/b1 after multiplied by positive number [b1(b1+b2+b3+...+bn)]/[b1(b2+b3+...+bn)]
And now (a2+a3+...+an)/(b2+b3+...+bn)  a1/b1 > (a2+a3+...+an)/(b2+b3+...+bn)  a2/b2 > 0
End of proof. A little clumsy though.
Edited on January 3, 2018, 11:28 am
Edited on January 3, 2018, 11:30 am

Posted by chun
on 20180103 11:19:29 