 ... brought down, there is no remainder, the proposed number is a perfect square. But if there is a remainder, you have only found the root of the greatest perfect square contained in the given number, or the entire part of the root sought. For example,...  Elements of Algebra - Σελίδα 1161838 - 355 σελίδεςΠλήρης προβολή - Σχετικά με αυτό το βιβλίο
 | Charles Davies - 1835 - 378 σελίδες
...entire part of the root sought. For example, if it were required to extract the square root of 665, we should find 25 for the entire part of the root...of 25 the greatest perfect square contained in 665 1 that is, is 25 the entire part of the root ? To prove this, we will first show that, the difference... | |
 | Charles Davies - 1839 - 272 σελίδες
...if it were required to extract the square root of 665, we should find 25 for the entire part of llic root, and a remainder of 40, which shows that 665 is not a perfect square. Bufis the square of 25 the greatest perfect square contained in 665 ? that is, is 25 the entire part... | |
 | Charles Davies - 1841 - 264 σελίδες
...entire part of the root sought. For example, if it were required to extract the square root of 665, we should find 25 for the entire part of the root,...in 665 ? that is, is 25 the entire part of the root 1 To prove this, we will first show that, the difference between the squares of two consecutive numbers,... | |
 | William Scott - 1844 - 568 σελίδες
...numbers ; then (a+l)!=(a+l) (a+l)=as+2a+l (Art. 76), and a*=aXa=a! ; .-.(a+1)2— a"=2a+l. ' Whence the difference between the squares of two consecutive numbers is equal to twice the less number + 1 ; the greater, therefore, the number a the greater is the difference between (<z+l)2 and a2, and... | |
 | Charles Davies - 1848 - 300 σελίδες
...entire part of the root sought. For example, if it were required to extract the square root of 665, we should find 25 for the entire part of the root,...in 665 ? that is, is 25 the entire part of the root 1 To prove this, we will first show that, the difference between the squares of two consecutive numbers,... | |
 | Charles Guilford Burnham - 1850 - 350 σελίδες
...contained in 572 ; that _^ is, it is the entire part of the root. This . , • „„ may be shown, thus : The difference between ' the squares of two consecutive numbers, is equal to twice the less number, plus 1 . The 43 dift'erence between the squares of 8 and 9 is 17 = 8X2 + 1, and 23 X 2 + 1=47, which... | |
 | Charles Guilford Burnham - 1857 - 342 σελίδες
...greatest square contained in 572 ; that is, it is the entire part of the root. This may be shown, thus : The difference between „ the squares of two consecutive numbers, is — equal to twice the less number, plus 1. The " difference between the squares of 8 and 9 is 17=8x2 + 1, and 23 x 2 + 1=47, which is... | |
 | Charles Davies - 1860 - 412 σελίδες
...perfect square. But is the square of 12 the greatest perfect square contained in 168? That is, is 12 the entire part of the root? To prove this, we will...of two consecutive numbers, is equal to twice the lest number augmented by 1. Let a represent the less number, and a + 1, the greater. Then, (a + I)2... | |
 | Charles Davies - 1860 - 328 σελίδες
...entire part of the root sought. For example, if it were required to extract the square root of 605, we should find 25 for the entire part of the root, and a remainder of 40, which shows that 605 is not a perfect square. But is the square of 25 the greatest perfect square contained in 665 ?... | |
 | Charles Davies - 1861 - 322 σελίδες
...entire part of the root sought. For example, if it were required to extract the square root of 665, we should find 25 for the entire part of the root,...the root? To prove this, we will first show that, thi difference between the squares of two consecutive numbers, it equal to twice the less number augmentel... | |
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