Real Numbers

Asked by Abhijeetoo7 | 2nd Dec, 2009, 10:58: PM

Expert Answer:

Let us assume an odd positive integer of the form 2n+1 whose square is (2n+1)^2 = 4n^2 +4n + 1 = 4(n^2+n) +1 and n^2+n will always be an even number as it is a product of two consecutive integer so  n^2+ n = 2k so the above expression becomes 4*2k+1 = 8k+1

Answered by  | 1st Jan, 2010, 12:54: PM

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