dripto biswas Posted January 17, 2013 Posted January 17, 2013 WHEN I ENTER ANY NUMBER GREATER THAN 1 IN A CALCULATOR AND KEEP SQUARE ROOTING THE SQUARE ROOT OF THE NUMBER, I WILL OBTAIN 1. ALSO IF THE SAME IS APPLIED TO ANY NUMBER GREATER THAN 0 BUT LESS THAN 1, THE RESULT IS 0.999999............. TO INFINITY. WT IS THE CAUSE AND IMPLICATION OF THESE RESULTS. PLEASE TELL ME AS SOON AS U CAN
x(x-y) Posted January 17, 2013 Posted January 17, 2013 (edited) If you square root a number, let's call it x where x > 0, and then square root it again - notice that you will obtain [latex]\sqrt{\sqrt{x}} = \left(x^{\frac{1}{2}}\right)^{\frac{1}{2}} = x^{\frac{1}{4}}[/latex] so, if you perform the square root on this variable x an "infinite amount of times" then you will obtain the term [latex]x^{\frac{1}{\infty}} = x^0 = 1[/latex] Thus, we can say that [latex]\displaystyle \lim_{n \to \infty} x^{\frac{1}{n}} = x^0 = 1[/latex] Edited January 17, 2013 by x(x-y) 1
mathematic Posted January 17, 2013 Posted January 17, 2013 WHEN I ENTER ANY NUMBER GREATER THAN 1 IN A CALCULATOR AND KEEP SQUARE ROOTING THE SQUARE ROOT OF THE NUMBER, I WILL OBTAIN 1. ALSO IF THE SAME IS APPLIED TO ANY NUMBER GREATER THAN 0 BUT LESS THAN 1, THE RESULT IS 0.999999............. TO INFINITY. WT IS THE CAUSE AND IMPLICATION OF THESE RESULTS. PLEASE TELL ME AS SOON AS U CAN The difference is an artifact of the calculator. .99999....... = 1. 1
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