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 Simple fractions (Posted on 2018-01-03)
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.

Comments: ( Back to comment list | You must be logged in to post comments.)
 The First Half, Using Induction | Comment 1 of 3

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 2018-01-03 11:19:29

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