<p>There\'s a solved problem in cramster.com for the book mathematical structure

Question

<p>There's a solved problem in cramster.com for the book mathematical structures for computer science&#160; by Judith Gersting chapter 2.1 number 55 and I just want a detailed explaination about why it used (n-1)^2 to prove "For every positive integer n, n + 1/n &gt;= 2." Here's a link <a href="/solution/solution/1040768">http://www.cramster.com/solution/solution/1040768</a>.</p>

Explanation / Answer

first let me prove n+1/n >= 2, n+1/n >= 2 subtracting both side by 2. => n+1/n-2 >= 0 => (n^2+1-2n)/n >= 0 ((n-1)^2)/n >=0 since square of every number is always positive (except complex numbers). so for n>0,we have (n-1)^2 >= 0 which is true for all n>0. hence we can say that n+1/n >=2. in the solution ,at the starting step (n-1)^2 is chosen ,because after squaring and then dividing by n we can get the form n+1/n. also, for form (n+k/n) for k>=1,n>0,the whole square term can be get from (n-1)^2. let's take the example of n+k/n >= 2 for k >= 1,n>0 so we have, n^2 - 2n + k >= 0 which is, (n-1)^2 +(k-1) >=0 since k>=1,and (n-1)^2 is already positive so we can say that n+k/n >= 2. one other reason in choosing (n-1)^2 in these types of proofs we can see that n^2 - 2n (which occurs in all n+k/n types of proof) can be written as [(n-1)^2 -1].

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