1. Archimedes in his book Measurement of a Circle presented, without a word of justification, the inequality \frac{265}{153}<\sqrt{3}<\frac{1351}{780} As an explanation of the probable steps leading to the right-hand bound,

show first that \sqrt{26^{2}-1}<\sqrt{26^{2}-1+\left(\frac{1}{52}\right)^{2}}=26-\frac{1}{52} and then \sqrt{3}=\frac{1}{15} \sqrt{26^{2}-1}<\frac{1}{15}\left(26-\frac{1}{52}\right)=\frac{1351}{780} Obtain the left-hand bound in a similar manner by replacing 1/52 by 1/51 . 2. To find the approximate value for 7, Archimedes consider the inscribed and circumscribed regular polygons of a circle with diameter 1. Let us denote by Pn and Pn the perimeters of the inscribed and circumscribed regular polygons of n sides. Notice the perimeter of the circle is T. Then we have p_{6}