Question

What fraction of whole numbers is equal to 17100 and that has a denominator that is a power of 10?

Answer 1

17100 = 17100/1= 17100/1 × 10/10 = 171000/10= 17100/1 × 10²/10² = 1710000/100= 17100/1 × 10³/10³ = 17100000/1000...= 17100/1 × 10ⁿ/10ⁿThere is no upper limit to this sequence: n = 0, 1, 2, 3,....These numbers can be put in a one-to-one relationship with thecounting numbers 1, 2, 3, ...The counting numbers can also be put in a one-to-onerelationship with the whole numbersTherefore the sequence of fractions can be put in a one-to-onerelationship with the whole numbers.Therefore the fraction of whole numbers which are in thesequence is 1.-----------Let's try another approach:As 10 to some power is positive, there is no way we can change anegative number to make it positive by dividing by 10 to somepower, therefore the solution numbers are all positive.As negative whole numbers make up half of the whole numbers, thefraction of whole numbers which are equivalent to 17100 whendivided by a power of 10 is at most ½.The fraction of the ½ of the whole numbers which are positivecan now be considered:After the nth whole number has been found which matches thecriteria that is it equivalent to 17100 when it is divided by apower of 10, in total there are n numbers matching out of a totalof 17100 × 10ⁿ⁻³ numbers (the value of the nth number); thus:The first number which meets the criteria is 171/10⁻² → thefraction is 1/171The second number to meet the criteria is 1710/10⁻¹ → thefraction is 2/1710The third number to meet the criteria is 17100/10⁰ → thefraction is 3/17100The fourth number to match the criteria is 171000/10 → thefraction is 4/171000The fifth number to match the criteria is 1710000/10²→ thefraction is 5/1710000→ For the nth match, the fraction is: n/(17100 × 10ⁿ⁻³) =(1/17100) × n/10ⁿ⁻³This gives us a sequence of fractions:1/17100 × 1/10⁻², 1/17100 × 2/10⁻¹, 1/17100 × 3/10⁰, 1/17100 ×4/10¹, 1/17100 × 5/10², ...= 100000/17100000, 20000/17100000, 3000/17100000, 400/17100000,35/17100000, 6/17100000, ....Consider terms n and n+1:U{n} = (1/17100) × n/10ⁿ⁻³ = (1/17100) × 10 × n/10ⁿ⁻²U{n+1} = (1/17100) × (n+1)/10ⁿ⁻²U{n} - U{n+1} = (1/17100) × 10 × n/10ⁿ⁻² - (1/17100) ×(n+1)/10ⁿ⁻²= (10n - (n+1))/(17100 × 10ⁿ⁻²)= (9n - 1)/(17100 × 10ⁿ⁻²)As n ≥ 1,9n - 1 ≥ 9×1 - 1 = 8→ term n - term n+1 ≥ 8 > 0→ term n is larger than term n+1for all n ≥ 1Each of these terms is less than 1, and each term of thesequence is smaller than the previous one and so as n increases thevalue of the each term (the fraction of whole numbers which meetthe criteria) tends towards 0.→ The fraction of all whole numbers which equate to 17100 whendivided by a power of 10 is as near enough to zero as make noodds.ie the fraction is so small it is effectively none of the wholenumbers.-----------------------This is a problem of dealing with the infinite.