Here\'s the question. I don\'t know why but the hint really threw me for a loop.

Question

Here's the question. I don't know why but the hint really threw me for a loop.

Given a finite sequence of digits a1...an, with all the ai {0,1,2,3,4,5,6,7,8,9 }, prove that there exists a natural number N such that has precisely a1...an as the first n digits after the decimal point. [To be concrete (i.e. make the sequence of digits to be our very year. 2013): Show that there exists a natural number N such that has precisely 2013 after the decimal point.] Hint: You may want to use the fact that (which you will not need to prove):

Explanation / Answer

if we solve the inequality sqrt(n+1)-sqrt(n) < 1/2sqrt(n)

the solution obtained is n>0.8.

so this way it is proved that this inequality is valid for all integers.

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